The number of adult tickets sold is 55 and the number of student tickets sold is 2.
Let the number of student tickets be x and the number of adult tickets be y.
Step 1: Constructing the equations
Given,Student tickets cost $5 and adult tickets cost $7.The dance team sold 57 tickets and made $395. In order to find the number of student and adult tickets sold, we need to construct two equations.Using the given information, we can write the following equations:
x + y = 57 (Equation 1)
5x + 7y = 395 (Equation 2)
Step 2: Solving the equations We need to solve the equations we have constructed to find the values of x and y. We can do this using the elimination method by multiplying the first equation by 5 and subtracting the second equation from it.
5x + 5y = 285 (Multiplying Equation 1 by 5)
5x + 7y = 395 (Equation 2)2y = 110
(Subtracting Equation 2 from Equation 1)
y = 55 (Dividing by 2)
Now we can substitute y = 55 in Equation 1 to find x:
x + 55 = 57 (Substituting y = 55) x = 2
Therefore, the number of adult tickets sold is y = 55 and the number of student tickets sold is x = 2.
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State the Differentiation Part of the Fundamental Theorem of Calculus. Then find a d/dx integral^x_2 cos(t^4) dt. b Find d/dx integral^6_x cos (squareroot s^4 + 1)ds. C Find d/dx integral^2x + 1_2 In(t + 1)dt. d Find d/dx integral^x_-x z + 1/z + 2 dz. e Find d/dx integral^2_-3x 2^t2 dt.
Thus, Differentiation Part of the Fundamental Theorem of Calculus:
a) sin(t^4)/4
b) sin(sqrt(s^4 + 1))/sqrt(s^4 + 1)
c) (t + 1)ln(t + 1) - (t + 1)
d) (1/2)ln|z + 2| + z
e) (1/ln2)(sqrt(pi)/2)erfi(sqrt(ln2)t)
The Differentiation Part of the Fundamental Theorem of Calculus states that if f(x) is a continuous function on the interval [a,b] and F(x) is an antiderivative of f(x), then:
d/dx integral^b_a f(t) dt = f(x)
Using this theorem, we can find the derivatives of the given integrals as follows:
a) d/dx integral^x_2 cos(t^4) dt
= cos(x^4) [by applying the Differentiation Part of the FTC and noting that the antiderivative of cos(t^4) is sin(t^4)/4]
b) d/dx integral^6_x cos (squareroot s^4 + 1)ds
= -cos(sqrt(x^4 + 1)) [by applying the Differentiation Part of the FTC and noting that the antiderivative of cos(sqrt(s^4 + 1)) is sin(sqrt(s^4 + 1))/sqrt(s^4 + 1)]
c) d/dx integral^2x + 1_2 In(t + 1)dt
= In(x + 1) [by applying the Differentiation Part of the FTC and noting that the antiderivative of ln(t + 1) is (t + 1)ln(t + 1) - (t + 1)]
d) d/dx integral^x_-x z + 1/z + 2 dz
= 0 [by applying the Differentiation Part of the FTC and noting that the antiderivative of z + 1/(z + 2) is (1/2)ln|z + 2| + z]
e) d/dx integral^2_-3x 2^t2 dt
= -6x2^(9x^2) [by applying the Differentiation Part of the FTC and noting that the antiderivative of 2^(t^2) is (1/ln2)(sqrt(pi)/2)erfi(sqrt(ln2)t)]
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Si un empleado gana unos 33. 500pesos diarios ¿Cuanto ganara en 30 dias ?,¿Cuanto ganara en 1 año?
An employee will earn 12,060,000 pesos in a year if he/she earns 33,500 pesos per day.
If an employee earns 33,500 pesos per day, he/she will earn 1,005,000 pesos in 30 days and 12,060,000 pesos in one year.
The calculation of earnings of an employee can be calculated using the following formula:
Salary = daily wage x number of working days in a month/year
Let us calculate the salary of the employee in 30 days:
Salary for 30 days = 33,500 pesos/day x 30 days
= 1,005,000 pesos
An employee will earn 1,005,000 pesos in 30 days if he/she earns 33,500 pesos per day.
Let's calculate the salary of the employee in a year:
Salary for 1 year = 33,500 pesos/day x 365 days
= 12,227,500 pesos
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You may freely use techniques from one-variable calculus, such as L'Hôpital's rule. Consider f(x, y). f(x, y) = (xy^3) / (x^2 + y^6) if (x, y) ≠ (0, 0) 0 if (x, y) = (0, 0)
(a) Compute the limit as (x, y) → (0, 0) of f along the path x = 0. (If an answer does not exist, enter DNE.)
(b) Compute the limit as (x, y) → (0, 0) of f along the path x = y3. (If an answer does not exist, enter DNE.)
(c) Show that f is not continuous at (0, 0). Since the limits as (x, y) → (0, 0) of f along the paths x = 0 and x = y3 ,are equal? or are not equal? or DNE?
f is not continuous at (0, 0).
Using L'Hopital's rule (a) Limit along x=0 is o (b) Limit along [tex]x = y^3[/tex] is 1/2 (c) Limits along paths x = 0 and[tex]x = y^3[/tex] are not equal, f is not continuous at (0,0)
A mathematical method called L'Hopital's rule is used to determine the limit of an indeterminate form of a fraction of two functions at a specific location. It claims that, in some circumstances, the limit of the ratio of two functions can be discovered by taking the derivative of the numerator and denominator individually, evaluating the resulting quotient at the point of interest, and repeating this process for the other function. This rule can be used in calculus to evaluate limits that are challenging or impossible to solve via direct substitution.
Using L'Hopital's rule :
(a) To compute the limit as (x, y) → (0, 0) of f along the path x = 0, we can substitute x = 0 into the function f(x, y):
[tex]f(x, y) = (0 * y^3) / (0^2 + y^6) = 0 / y^6 = 0[/tex]
The limit as (x, y) → (0, 0) along the path x = 0 is 0.
(b) To compute the limit as (x, y) → (0, 0) of f along the path[tex]x = y^3[/tex], we can substitute x = y^3 into the function f(x, y):
[tex]f(x, y) = (y^3 * y^3) / (y^6 + y^6) = y^6 / (2y^6) = 1/2[/tex]
The limit as (x, y) → (0, 0) along the path[tex]x = y^3[/tex] is 1/2.
(c) Since the limits as (x, y) → (0, 0) of f along the paths x = 0 and[tex]x = y^3[/tex] are not equal (0 ≠ 1/2), f is not continuous at (0, 0).
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A state fisheries commission wants to estimate the number of bass caught in a given lake during a season in order to restock the lake with the appropriate number of young fish. The commission could get a fairly accurate assessment of the seasonal catch by extensive "netting sweeps" of the lake before and after a season, but this technique is much too expensive to be done routinely. Therefore, the commission samples a number of lakes and record the seasonal catch (thousands of bass per square mile of lake area) and size of lake (square miles). A simple linear regression was performed and the following R output obtained.Estimate Std. Error t value Pr(>|t|)(Intercept) 2.5463 0.4427 5.7513 0.0000size 0.0667 0.3672 0.1818 0.8578The response variable is ____.a. size of lakeb. seasonal catch
The response variable in the given linear regression output is seasonal catch, as indicated by the coefficient estimate and standard error of the variable "size."
The response variable in this simple linear regression is the seasonal catch (thousands of bass per square mile of lake area). In a linear regression, the response variable is the variable we are trying to predict or estimate based on the values of other variables. In this case, we are trying to estimate the seasonal catch of bass in the lake based on the size of the lake. So, the correct answer is b. seasonal catch.
The response variable in the given linear regression output is seasonal catch, as indicated by the coefficient estimate and standard error of the variable "size."
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Evaluate 1 sit dc +.24 as a power series centered at 0. Write out the first four nonzero terms (not counting the integration constant), as well as the full series with summation notation. For which z is the representation guaranteed to be valid?
The representation is guaranteed to be valid for values of dc + 0.24 such that |dc + 0.24| < 1, or -1.24 < dc < 0.76.
We know that the power series representation of the function f(z) = 1/(1-z) is:
f(z) = ∑(n=0 to infinity) z^n
If we substitute z = dc + 0.24 into this power series, we get:
f(dc + 0.24) = ∑(n=0 to infinity) (dc + 0.24)^n
To get this in a form we can work with, we can expand the binomial term using the binomial theorem:
f(dc + 0.24) = ∑(n=0 to infinity) [(d^0 * 0.24^n)/0! + (d^1 * 0.24^(n-1))/1! + (d^2 * 0.24^(n-2))/2! + ...] * dc^n
We can simplify this expression by writing out the first few terms explicitly:
f(dc + 0.24) = 1 + (dc + 0.24) + (dc + 0.24)^2 + (dc + 0.24)^3 + ...
The first four nonzero terms are:
1 + (dc + 0.24) + (dc^2 + 0.48dc + 0.0576) + (dc^3 + 0.72dc^2 + 0.2688dc + 0.031104)
The full series with summation notation is:
∑(n=0 to infinity) [(d^0 * 0.24^n)/0! + (d^1 * 0.24^(n-1))/1! + (d^2 * 0.24^(n-2))/2! + ...] * dc^n
The representation is guaranteed to be valid for values of z such that |z| < 1, since this is the radius of convergence of the power series for 1/(1-z).
Therefore, the representation is guaranteed to be valid for values of dc + 0.24 such that |dc + 0.24| < 1, or -1.24 < dc < 0.76.
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(a) Give pseudocode for an algorithm that finds the first repeated integer in given a sequence of integers. (b) Analyze the worst-case time complexity of the algorithm you devised in part (a).
(a) Pseudocode for the algorithm that finds the first repeated integer in a given sequence of integers is as follows:
1. Initialize an empty set called "visited".
2. Traverse the given sequence of integers.
3. For each integer in the sequence, check if it is already in the "visited" set.
4. If the integer is in the "visited" set, return it as the first repeated integer.
5. Otherwise, add the integer to the "visited" set.
6. If there is no repeated integer, return "None".
(b) The worst-case time complexity of the algorithm is O(n), where n is the length of the sequence of integers.
Therefore, the time complexity of the algorithm increases linearly with the size of the input sequence.
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Write 2/3 and 3/4 as a pair of fractions with a common denominater
To express 2/3 and 3/4 as a pair of fractions with a common denominator, we can find the least common multiple (LCM) of the denominators and then adjust the numerators accordingly.
To begin, we need to find the least common multiple (LCM) of the denominators, which in this case is 12. Next, we convert 2/3 and 3/4 to fractions with a common denominator of 12.
For 2/3, we multiply both the numerator and denominator by 4 to get 8/12. Since 2 multiplied by 4 is 8, and 3 multiplied by 4 is 12.
For 3/4, we multiply both the numerator and denominator by 3 to get 9/12. Since 3 multiplied by 3 is 9, and 4 multiplied by 3 is 12.
Now, we have 8/12 and 9/12 as a pair of fractions with a common denominator of 12. These fractions can be compared or used in further calculations since they have the same denominator.
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pls help lol my grade’s a 62 rn & grades are almost due !
The triangle in the image is a right triangle. We are given a side and an angle, and asked to find another side. Therefore, we should use a trigonometric function.
Trigonometric Functions: SOH-CAH-TOA
---sin = opposite/hypotenuse, cosine = adjacent/hypotenuse, tangent = opposite/adjacent
In this problem, looking from the angle, we are given the adjacent side and want to find the opposite side. This means we should use the tangent function.
tan(40) = x / 202
x = tan(40) * 202
x = 169.498
x (rounded) = 169 meters
Answer: the tower is 169 meters tall
Hope this helps!
Answer:
170 meters
Step-by-step explanation:
The three sides of a right triangle are named hypotenuse, adjacent side and opposite side and the angle the adjacent side makes with they hypotenuse is θ (see Figure 1)
In this description the terms
Opposite --> side opposite to the angle θ
Adjacent --> side adjacent to the angle θ
Hypotenuse --> longest side of the right triangle
The relationship between the ratio of the shorter sides and and the angle θ in the figure is given by the formula
[tex]\mathrm {\tan(\theta) = \dfrac{Opposite \; side}{Adjacent \;side}}[/tex]
We can view the Eiffel Tower as the opposite side, the distance from the base to the surveyor location as the adjacent side (see the second figure)
If we let h = height of the Eiffel Tower in meters , opposite side length = h m
The adjacent side length = 202 meters
The angle θ = 40°
Applying the tan formula we get
[tex]\tan(40^\circ) = \dfrac{h}{202}\\\\\textrm{Multiplying both sides by 202, }\\202 \tan(40^\circ) = h\\\\\\h = 202 \tan(40^\circ) \\\textrm{Using a calculator we get}\\\\h = 169.5\; meters[/tex]
Rounded to the nearest meter, the height = 170 meters
3. Triangle ABC is an equilateral triangle (all angles are equal and all side lengths are equal) and triangle CDE is a right triangle. If the length of side AE is 20 units, what is the length of side BD.
A - 12 units
B - 14 units
C - 16 units
D - 18 units
E - 20 units
The correct option is B - 14 units.
Triangle ABC and triangle CDE are shown below. Side AE is a hypotenuse to the right triangle CDE whose right angle is at vertex D. We are to find the length of side BD. We will proceed to solve this problem by making use of Pythagorean Theorem.Let a side of the equilateral triangle ABC be x units, then the other two sides of the triangle will also be x units. Since it is an equilateral triangle, all its sides are equal, and all its angles are equal.
Therefore, angle BAC = angle ABC = angle ACB = 60°.Also, we know that side AE is 20 units. Let the length of side BD be y units.Now, we will use Pythagorean Theorem to find the length of side BD:BD2 + DE2 = BE2DE = AE - AD = 20 - xNow, since triangle ABC is an equilateral triangle, we have:x2 + x2 - 2xcos 60° = 20²2x2 - x² = 400∴ 3x² = 400∴ x² = 400/3∴ x = (400/3)1/2Putting this value of x into the expression for DE, we get:DE = 20 - x = 20 - (400/3)1/2Now, we can substitute the value of DE and x into the expression for BD2:BD2 + (20 - (400/3)1/2)2 = (400/3)This expression simplifies to:BD2 = (400/3) - 400 + 400/3∴ BD2 = 400/3 - 400/3∴ BD2 = 400/9∴ BD = (400/9)1/2.
Therefore, the length of side BD is:BD = (400/9)1/2 ≈ 6.66 units (rounded off to 2 decimal places)In option A, the length of side BD is given as 12 units. But, as we have shown, the length of side BD is approximately 6.66 units. Hence, option A is incorrect.The correct option is B - 14 units.Answer: B - 14 units.
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A circle with a center of (0, 0) and passes through (0, -3). find the area and circumferences of this circle
The circle with a center at (0, 0) and passing through (0, -3) has an area and circumference that can be calculated. The area can be found using the formula A = πr^2, and the circumference can be found using the formula C = 2πr, where r is the radius of the circle.
Given that the center of the circle is at (0, 0) and it passes through (0, -3), we can determine that the radius of the circle is 3 units. The distance between the center (0, 0) and the point on the circle (0, -3) gives us the radius.
To find the area of the circle, we use the formula A = πr^2. Substituting the radius, we have A = π(3^2) = 9π square units.
To find the circumference of the circle, we use the formula C = 2πr. Substituting the radius, we have C = 2π(3) = 6π units.
Therefore, the area of the circle is 9π square units, and the circumference of the circle is 6π units.
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What is 4x+3 answer for math homework please answer or else
The the answer to the expression 4x + 3 is simply 4x + 3 itself.
4x + 3 is an algebraic expression that represents a polynomial. It can be simplified or evaluated depending on the given problem. If there are no instructions given, then we assume that the expression is to be simplified. Hence, we must combine like terms. 4x and 3 cannot be combined as they are not like terms. Therefore, the expression is already in its simplest form.
All algebraic expressions are not polynomials, though. But algebraic expressions are what all polynomials are. The distinction is that algebraic expressions also include irrational numbers in the powers, whereas polynomials only include variables and coefficients with the mathematical operations (+, -, and ).Additionally, algebraic expressions may not always be continuous (for example, 1/x2 - 1), whereas polynomials are continuous functions (for example, x2 + 2x + 1).
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Darren bought a toy. He sold the toy to peter for 5/4 the price he paid for it. Peter then sold the toy to Allen for 2/5 less than what he paid for it. Allen paid 12. 45 for the tou. How much did darren pay for the toy
Darren paid $16.6 for the toy.
To find out how much Darren paid for the toy, we'll follow these steps:
Let's assume Darren paid "x" amount for the toy.
Peter bought the toy from Darren for 5/4 of the price Darren paid, which means Peter paid (5/4) * x.
Allen bought the toy from Peter for 2/5 less than what Peter paid. So, Allen paid
(1 - 2/5) * (5/4) * x.
We know that Allen paid $12.45 for the toy, so we can set up the equation:
(1 - 2/5) * (5/4) * x = 12.45.
Simplifying the equation, we get
(3/5) * (5/4) * x = 12.45.
Multiplying the fractions and solving for x, we find
x = (12.45) * (4/3) = 16.6.
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Sequences by recurrence relations The following sequences, defined by a recurrence relation, are monotonic and bounded, and therefore converge by Theorem 10.5. a. Examine the first three terms of the sequence to determine whether the sequence is nondecreasing or nonincreasing. b. Use analytical methods to find the limit of the sequence
For the given sequence, aₙ₊₁=1/2(aₙ+(2/aₙ)); a₀=2, the sequence is non- increasing and the limit of the sequence is 2/√3.
a.
To determine whether the sequence is non-decreasing or non-increasing, we need to examine the signs of aₙ₊₁ − aₙ for all n. So, let's find the first few terms of the sequence:
a₁ = 1/2(a₀ + 2/a₀) = 1/2(2 + 1) = 3/2
a₂ = 1/2(a₁ + 2/a₁) ≈ 1.5288
a₃ = 1/2(a₂ + 2/a₂) ≈ 1.4991
Since a₃ < a₂, the sequence is non-increasing.
b.
To find the limit of the sequence, we can use the fact that it is bounded and monotonic, and apply Theorem 10.5. Let L be the limit of the sequence, then taking the limit of both sides of the recurrence relation, we get:
L = 1/2(L + 2/L)
Multiplying both sides by 2L, we get:
2L² = L² + 4
Simplifying, we get:
L² = 4/3
Taking the positive square root, since L is nonnegative, we get:
L = 2/√3
Therefore, the limit of the sequence is 2/√3.
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19 . find the particular solution to the differential equation y′=3x3 that passes through (1,4.75), given that y=c 3x44 is a general solution.
To find the particular solution to the differential equation y′=3x3 that passes through (1,4.75), we need to use the given general solution y=c 3x44.
First, we differentiate the general solution to get y′=12cx33.
Next, we substitute the point (1,4.75) into the equation:
4.75 = c 3(1)^4 + C
where C is the constant of integration.
Simplifying this equation, we get:
4.75 = 3c + C
To find the value of C, we need to solve for it. We can do this by using the fact that the particular solution passes through the point (1,4.75). Substituting these values into the equation above, we get:
4.75 = 3c + C
4.75 = 3c + C
4.75 - 3c = C
So C = 4.75 - 3c.
Now we substitute this value of C back into the general solution to get the particular solution:
y = c 3x44
y = (4.75 - 3c) 3x44
Therefore, the particular solution to the differential equation y′=3x3 that passes through (1,4.75), given that y=c 3x44 is a general solution, is y = (4.75 - 3c) 3x44.
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Use intercepts to help sketch the plane. 2x+5y+z=10
To sketch the plane, we start at the x-intercept (5, 0, 0), then draw a line to the y-intercept (0, 2, 0), and finally connect to the z-intercept (0, 0, 10). This forms a triangle in three-dimensional space that represents the plane 2x+5y+z=10.
To use intercepts to help sketch the plane 2x+5y+z=10, we first need to find the x, y, and z intercepts.
To find the x-intercept, we set y and z equal to zero:
2x + 5(0) + 0 = 10
2x = 10
x = 5
So the x-intercept is (5, 0, 0).
To find the y-intercept, we set x and z equal to zero:
0 + 5y + 0 = 10
5y = 10
y = 2
So the y-intercept is (0, 2, 0).
To find the z-intercept, we set x and y equal to zero:
0 + 0 + z = 10
z = 10
So the z-intercept is (0, 0, 10).
Now we can plot these three points on a three-dimensional coordinate system and connect them to form a triangle, which represents the plane.
To sketch the plane, we start at the x-intercept (5, 0, 0), then draw a line to the y-intercept (0, 2, 0), and finally connect to the z-intercept (0, 0, 10). This forms a triangle in three-dimensional space that represents the plane 2x+5y+z=10.
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For cones with radius 6 units, the equation V=12h relates the height h of the cone, in units, and the volume of the cone, in cubic units
The volume of the cone is 48 cubic units when the height of the cone is 4 units.
The given equation V = 12h represents the volume of cones with a radius of 6 units.
The volume of a cone is given by the formula V = (1/3)πr²h, where r is the radius of the cone, h is the height of the cone and π is the value of pi which is approximately equal to 3.14.
Given that radius, r = 6 units. Therefore, the formula for the volume of the cone can be written as
V = (1/3)π(6)²h= 12h cubic units
As per the problem, this relation is used to find the volume of cones with a radius of 6 units. For instance, if the height of the cone is 4 units, then using the formula above, the volume of the cone can be calculated by substituting h = 4 units.V = 12 × 4= 48 cubic units
Therefore, the volume of the cone is 48 cubic units when the height of the cone is 4 units.
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Use the method of iteration to find a formula expressing S nas a function of n for the given recurrence relation and initial conditions. b. S n=−S n−1+10;S 0=−4
The formula expressing [tex]S_n[/tex] as a function of n for the recurrence relation [tex]S_n=-S_{n-1}+10[/tex] and initial condition [tex]S_0=-4[/tex] is [tex]S_n = 5n-4[/tex] if n is even and [tex]S_n = -5n+14[/tex] if n is odd.
if n is even, and[tex]S_n = 5n - 4[/tex] if n is odd.
The given recurrence relation is:
[tex]S_n = -S_{n-1} + 10[/tex]
And the initial condition is:
[tex]S_0 = -4[/tex]
To use the method of iteration, we start by substituting n-1 for n in the recurrence relation:
[tex]S_{n-1} = -S_{n-2} + 10[/tex]
Next, we can substitute this expression into the original recurrence relation:
[tex]S_n = -(-S_{n-2} + 10) + 10[/tex]
Simplifying this, we get:
[tex]S_n = S_{n-2}[/tex]
We can continue this process of substitution, getting:
[tex]S_{n-2} = -S_{n-3} + 10[/tex]
Simplifying, we get:
[tex]S_n = S_{n-3} - 10[/tex]
Substituting again:
[tex]S_{n-3} = -S_{n-4} + 10[/tex]
Simplifying:
[tex]S_n = S_{n-4} - 20[/tex]
We can see a pattern emerging: each time we substitute, we go back two steps and subtract 10 or 20.
So we can write the general formula for [tex]S_n[/tex] in terms of [tex]S_0[/tex] as follows:
If n is even:
[tex]S_n = S_0 + 10\times (n/2)[/tex]
If n is odd:
[tex]S_n = -S_0 - 10\times ((n-1)/2)[/tex]
Using the initial condition [tex]S_0 = -4,[/tex] we can simplify these formulas:
If n is even:
[tex]S_n = -4 + 10\times (n/2) = 5n - 4[/tex]
If n is odd:
[tex]S_n = 4 - 10\times ((n-1)/2) = -5n + 14.[/tex]
The formula expressing [tex]S_n[/tex] as a function of n for the given recurrence relation and initial conditions is: [tex]S_n = 5n - 4[/tex]
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To use the method of iteration, we need to repeatedly apply the recurrence relation to the initial condition and previous terms until we reach the nth term.
Starting with S0 = -4, we can find S1 by plugging in n=1 into the recurrence relation:
S1 = -S0 + 10 = -(-4) + 10 = 14
Using S1, we can find S2:
S2 = -S1 + 10 = -(14) + 10 = -4
We can continue this process to find the first few terms:
S3 = -S2 + 10 = -(-4) + 10 = 14
S4 = -S3 + 10 = -(14) + 10 = -4
Notice that S2 and S4 are the same value, and S1 and S3 are the same value. This suggests that the sequence alternates between two values: -4 and 14.
We can write this as a formula:
S(n) = -4 if n is even
S(n) = 14 if n is odd
Alternatively, we could write it as:
S(n) = (-1)^n * 9 + 5
This formula also produces alternating values of -4 and 14, and can be derived using the method of recurrence relations.
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Your math teacher is planning a test for you. The test will have 30 questions. Some of the questions will be worth 3 points, and the others will be worth 4 points. There will be a total of 100 points on the test. How many 3-point questions and how many 4-point questions will be on the test?
a. Identify the problem: ______
b. Let the number of 3-point questions = x and the number of 4-point questions = y. Write the two equations for the system. I
c. Use subsititution to solve for y in the first equation.
d. Substitute the value for y into the second equation to solve for x.
e. There will be 3-point questions and 4-point questions.
f. Check your solution by substituting the values into both equations.
There will be 20 3-point questions and 10 4-point questions on the test.
a. Identify the problem: Determine the number of 3-point and 4-point questions on the test.
b. Let the number of 3-point questions = x and the number of 4-point questions = y. Write the two equations for the system:
x + y = 30 (equation 1, representing the total number of questions)
3x + 4y = 100 (equation 2, representing the total points on the test)
c. Use substitution to solve for y in the first equation:
y = 30 - x
d. Substitute the value for y into the second equation to solve for x:
3x + 4(30 - x) = 100
3x + 120 - 4x = 100
-x = -20
x = 20
e. There will be 20 3-point questions and 30 - 20 = 10 4-point questions.
f. Check the solution by substituting the values into both equations:
20 + 10 = 30 (equation 1 is satisfied)
3(20) + 4(10) = 100 (equation 2 is satisfied)
Therefore, there will be 20 3-point questions and 10 4-point questions on the test.
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A truck's 42-in.-diameter wheels are turning at 505 rpm. Find the linear speed of the truck in mph: miles/hour Write answer as an exact expression using pi for a. No need to simplify
The linear speed of the truck is 199.5π/88 mph.
The circumference of each wheel is:
C = πd = π(42 in.) = 42π in.
The distance the truck travels in one revolution of the wheels is equal to the circumference of the wheels. Therefore, the distance the truck travels in one minute is:
d = 42π in./rev × 505 rev/min = 21159π in./min
To convert this to miles per hour, we need to divide by the number of inches in a mile and the number of minutes in an hour:
d = 21159π in./min × (1 mile/63360 in.) × (60 min./1 hour) = 199.5π/88 miles/hour
So, the linear speed of the truck is 199.5π/88 mph.
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What is the median of -18,-18,-14,-13,12,13,14,16?
The median of the given set of numbers is -13.5.
To find the median of a set of numbers, you need to arrange them in ascending order and then determine the middle value. If there is an even number of values, the median is the average of the two middle values.
Let's arrange the numbers in ascending order:
-18, -18, -14, -13, 12, 13, 14, 16
The set has 8 elements, so it has an even number of values. The middle two values are -14 and -13. To find the median, we take the average of these two values:
Median = (-14 + -13) / 2 = -27 / 2 = -13.5
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A set of plastic spheres are to be made with diameter of 16 cm_ If the manufacturing process is accurate to mm, what is the propagated error in volume of the spheres? Error cm3
The propagated error in volume of the spheres is[tex]181.16 cm^3[/tex].
To find the propagated error in volume of the spheres, we need to first calculate the volume of one sphere using the given diameter of 16 cm.
The formula for the volume of a sphere is: [tex]V = (4/3)\pi r^3[/tex], where r is the radius of the sphere.
The diameter is given as 16 cm, so the radius (r) would be half of that, which is 8 cm.
Substituting this value in the formula, we get: [tex]V = (4/3)\pi (8)^3 = 2144.66 cm^3[/tex] (rounded to 2 decimal places).
Now, we need to find the propagated error in volume due to the manufacturing process being accurate to mm.
Since the diameter is given accurate to mm, the maximum error in the diameter could be half of a mm (0.5 mm). This means the diameter could be anywhere between 15.5 cm and 16.5 cm.
To find the maximum possible error in volume, we need to calculate the volume using the maximum diameter of 16.5 cm:
V = [tex](4/3)\pi (8.25)^3 = 2325.82 cm^3[/tex](rounded to 2 decimal places). [tex]181.16 cm^3[/tex]
The difference between the maximum volume and the actual volume is:
[tex]2325.82 cm^3 - 2144.66 cm^3 = 181.16 cm^3[/tex](rounded to 2 decimal places).
Therefore, the propagated error in volume of the spheres is[tex]181.16 cm^3[/tex].
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Consider the two court cases discussed in this module. Why is the analysis used in the first court case an example of statistical inference, but the analysis in the second court case is not? The first case draws a conclusion based on probability. The first case involves 1025 students instead of only 88 students. The first case uses proportion of matches on wrong answers instead of all answers. D Question 2 2 pts 3 I9060 E- E R SATIRERER
The analysis used in the first court case is an example of statistical inference and the second court case is not an example of statistical inference.
The analysis used in the first court case is an example of statistical inference because it involves drawing a conclusion based on probability. It utilizes statistical techniques to make inferences about the entire population based on a sample.
In this case, the conclusion about cheating on the test was made by comparing the proportion of matches on wrong answers between the two groups.
On the other hand, the analysis in the second court case is not an example of statistical inference.
This is because it does not involve drawing conclusions based on probability or using statistical techniques to make inferences about a larger population. The fact that the second case involves 1025 students instead of only 88 students does not necessarily make it an example of statistical inference.
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Given a Binomial Asset Pricing model and M[nt] the Symmetric Random Walk up to time [nt] for t ≥ 0, we want to prove that the distribution of Sn(t) = S(0)e − [nt] 1+[nt]2 1 + σ √ n [nt]+M[nt] 2 1 − σ √ n [nt]−M[nt] 2 converges to the distribution of S(t) = S(0)e σW(t)− 1 2 σ 2 t 1. Compute Zn(t) = ln(Sn(t)) and Z(t) = ln(S(t)). 2. Using the Taylor series of expansion of f(x) = ln(1 + x) at the order 2, find an approximation of Zn(t) as a function of the Scaled Symmetric Random Walk W(n) (t) = 1 √ n M[nt] . 3. Use the fact W(n) (t) converges to the Brownian motion W(t) to compute Z(t) = lim n→+[infinity] Zn(t) and conclude
The distribution of S(t) in the Black-Scholes model, where the underlying asset follows a geometric Brownian motion.
We have:
Zn(t) = ln(Sn(t))
= ln(S(0)) − [n t] + ln(1 + σ √[n t] [M[n t] 2 − [n t]]) − ln(1 − σ √[n t] [M[n t] 2 − [n t]])
= ln(S(0)) − [n t] + ln(1 + σ √[n t] W(n)(t)) − ln(1 − σ √[n t] W(n)(t))
where we have used M[n t] = W(n)(t)√[n t] and the fact that ln(1 + x) ≈ x − x^2/2 for small x.
Using the Taylor series expansion of ln(1 + x) at the order 2, we have:
ln(1 + σ √[nt] W(n)(t)) ≈ σ √[nt] W(n)(t) − σ^2/2 [nt] W(n)(t)^2
ln(1 − σ √[nt] W(n)(t)) ≈ −σ √[nt] W(n)(t) − σ^2/2 [nt] W(n)(t)^2
Substituting these into the expression for Zn(t) yields:
Zn(t) ≈ ln(S(0)) − [nt] + σ √[nt] W(n)(t) − σ^2/2 [nt] W(n)(t)^2 − (−σ √[nt] W(n)(t) − σ^2/2 [nt] W(n)(t)^2)
= ln(S(0)) − [nt] + σ^2 [nt] W(n)(t)^2
Taking the limit as n → ∞, we have:
Z(t) = lim n→∞ Zn(t)
= ln(S(0)) − tσ^2/2
This means that the distribution of Zn(t) converges to a normal distribution with mean ln(S(0)) − tσ^2/2 and variance σ^2t. Since Zn(t) approximates ln(Sn(t)), the distribution of Sn(t) converges to a lognormal distribution with mean S(0) e^(−tσ^2/2) and variance S(0)^2 (e^(σ^2t) − 1).
This is the distribution of S(t) in the Black-Scholes model, where the underlying asset follows a geometric Brownian motion.
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Like bias and confounding, effect modification is a natural phenomenon of scientific interest that the investigator needs to eliminate.a. Trueb. False
The given statement is False.
Effect modification, also known as interaction, is not a phenomenon that needs to be eliminated. Instead, it is a phenomenon that the investigator needs to identify and account for in data analysis.
Effect modification occurs when the relationship between an exposure and an outcome differs depending on the level of another variable, known as the effect modifier. Failing to account for effect modification can lead to biased estimates and incorrect conclusions.
Therefore, it is essential for investigators to assess for effect modification and report findings accordingly. This can involve stratifying the data by the effect modifier and analyzing each stratum separately or including an interaction term in the statistical model.
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calculate the taylor polynomials 2 and 3 centered at =0 for the function ()=7tan().
The taylor polynomials for 2 is [tex]7 + 7x^2[/tex] and for 3 is [tex]7x + (7/3)x^3.[/tex]
What is the taylor polynomials for 2 and 3?To find the Taylor polynomials for a function, we need to calculate the function's derivatives at the point where we want to center the polynomials. In this case, we want to center the polynomials at x=0.
First, let's find the first few derivatives of[tex]f(x) = 7tan(x):[/tex]
[tex]f(x) = 7tan(x)[/tex]
[tex]f'(x) = 7sec^2(x)[/tex]
[tex]f''(x) = 14sec^2(x)tan(x)[/tex]
[tex]f'''(x) = 14sec^2(x)(2tan^2(x) + 2)[/tex]
[tex]f''''(x) = 56sec^2(x)tan(x)(tan^2(x) + 1) + 56sec^4(x)[/tex]
To find the Taylor polynomials, we plug these derivatives into the Taylor series formula:
[tex]P_n(x) = f(0) + f'(0)x + (f''(0)x^2)/2! + ... + (f^n(0)x^n)/n![/tex]
For n=2:
[tex]P_2(x) = f(0) + f'(0)x + (f''(0)x^2)/2![/tex]
[tex]= 7tan(0) + 7sec^2(0)x + (14sec^2(0)tan(0)x^2)/2[/tex]
[tex]= 7 + 7x^2[/tex]
So the second-degree Taylor polynomial centered at x=0 for f(x) is [tex]P_2(x) = 7 + 7x^2.[/tex]
For n=3:
[tex]P_3(x) = f(0) + f'(0)x + (f''(0)x^2)/2! + (f'''(0)x^3)/3![/tex]
[tex]= 7tan(0) + 7sec^2(0)x + (14sec^2(0)tan(0)x^2)/2 + (14sec^2(0)(2tan^2(0) + 2)x^3)/6[/tex]
[tex]= 7x + (7/3)x^3[/tex]
So the third-degree Taylor polynomial centered at x=0 for f(x) is [tex]P_3(x) = 7x + (7/3)x^3.[/tex]
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A certain sports car comes equipped with either an automatic or a manual transmission, and the car is available in one of four colors. Relevant probabilities for various combinations of transmission type and color are given in the table below.COLORTRANSM?SS?ON TYPE white blue black redA 13 10 11 11M 15 07 15 18Let A = {automatic transmission}, B = { black } , and C = { white }. a) Calculate P(A), P(B), and P(A ? B). b) Calculate both P(A | B) and P(B | A), and explain in context what each of these probabilities represent. c) Calculate and interpret P(A | C) and P(A | C').
P(B) = P(black and A) + P(black and M) = (11+15+15)/80 = 41/80
P(A ? B) = P(black and A) = 41/80
we have P(A) = 1, P(B) = 41/80, and P(A ? B) = 41/80.
P(B | A) = P(A and B) / P(A) = (11+15+15) / (13+10+11+11+15+7+15+18) = 41/80. This represents the probability of a randomly selected black car having an automatic transmission.
P(A | C') = P(A and C') / P(C') = (10+11+15+18) / (10+11+15+18+7+11+11+15) = 54/73. This represents the probability of a randomly selected non-white car having an automatic transmission.
a) From the table, we can calculate the following probabilities:
P(A) = P(A and white) + P(A and blue) + P(A and black) + P(A and red) = (13+10+11+11+15+7+15+18)/80 = 80/80 = 1
P(B) = P(black and A) + P(black and M) = (11+15+15)/80 = 41/80
P(A ? B) = P(black and A) = 41/80
So, we have P(A) = 1, P(B) = 41/80, and P(A ? B) = 41/80.
b) We can calculate the following conditional probabilities:
P(A | B) = P(A and B) / P(B) = (11+15+15) / (11+10+11+15+7+15+18) = 41/77. This represents the probability of a randomly selected car having an automatic transmission, given that it is black.
P(B | A) = P(A and B) / P(A) = (11+15+15) / (13+10+11+11+15+7+15+18) = 41/80. This represents the probability of a randomly selected black car having an automatic transmission.
c) We can calculate the following conditional probabilities:
P(A | C) = P(A and C) / P(C) = (13+15) / (13+10+11+15) = 28/49. This represents the probability of a randomly selected white car having an automatic transmission.
P(A | C') = P(A and C') / P(C') = (10+11+15+18) / (10+11+15+18+7+11+11+15) = 54/73. This represents the probability of a randomly selected non-white car having an automatic transmission.
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The probability values are
(a) P(A) = 9/20, P(B) = 13/50, P(A and B) = 11/100(b) P(A | B) = 11/26, P(B | A) = 11/45(c) P(A | C) = 13/28, P(A | C') = 4/9How to calculate the probabilitiesGiven that
COLOR
TRANSMISSION TYPE white blue black red
A 13 10 11 11
M 15 07 15 18
Also, we have
A = Automatic transmissionB = BlackC = WhiteFor the probabilities, we have
(a) P(A) = (13 + 10 + 11 + 11)/(13 + 10 + 11 + 11 + 15 + 07 + 15 + 18)
P(A) = 9/20
P(B) = (11 + 15)/100
P(B) = 13/50
P(A and B) = 11/100
(b) P(A | B) = P(A and B)/P(B)
P(A | B) = (11/100)/(13/50)
P(A | B) = 11/26
This means that the probability that a car is automatic given that it is black is 11/26
P(B | A) = P(A and B)/P(A)
P(B | A) = (11/100)/(9/20)
P(B | A) = 11/45
This means that the probability that a car is black given that it is automatic is 11/45
(c) P(A | C) = P(A and C)/P(C)
Where P(A and C) = 13/100 and P(C) = 28/100
So, we have
P(A | C) = (13/100)/(28/100)
P(A | C) = 13/28
This means that the probability that a car is automatic given that it is white is 13/28
P(A | C') = P(A and C')/P(C')
Where P(A and C') = 32/100 and P(C') = 72/100
So, we have
P(A | C') = (32/100)/(72/100)
P(A | C') = 4/9
This means that the probability that a car is automatic given that it is not white is 4/9
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Microwaves of wavelength 3. 00 cm are incident normally on a
row of parallel metal rods. The separation of the rods is 8. 00 cm.
The first order diffraction maximum is observed at an angle of 22. 0°
to the direction of the incident waves.
What is the angle between the first and second order diffraction
maxima?
The angle between the first and second order diffraction maxima for microwaves of wavelength 3.00 cm that are incident normally on a row of parallel metal rods with a separation of 8.00 cm and the first order diffraction maximum observed at an angle of 22.0° to the direction of the incident waves is 37.0°.
Explanation:The grating equation is given as;dsinθ = nλWhere;d = separation between the slitsθ = angle between the incident direction and the direction of the diffracted light
λ = wavelength of lightn = order of diffraction
When the first order diffraction maximum is observed at an angle of 22.0°, we have;d sin 22.0° = λ …(1)
Also, when the second order diffraction maximum is observed at an angle θ2, we have;d sin θ2 = 2λ …(2)
Dividing equation (2) by equation (1);
d sin θ2/d sin 22.0°
= 2λ/λsin θ2/sin 22.0°
= 2sin θ2/sin 22.0°
= 2 × 3.00 c/sin 22.0°θ2 = sin⁻¹(2sin 22.0°)θ2 = 37.0°
Therefore, the angle between the first and second order diffraction maxima is 37.0°.
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In the cinema below
a) what is the angle of elevation from Row A to the bottom of the screen?
b) what is the angle of depression from Row P to the bottom of the screen?
Give your answers to 1 d.p.
Screen
2.5 m
5.6 m
12°
Row A
19.6 m
Row P
Not drawn accurately
Step-by-step explanation:
remember, the sum of all angles in a triangle is always 180°.law of sine :a/sin(A) = b/sin(B) = c/sin(C)with a, b, c being the sides, and A, B, C being the corresponding opposite angles.law of cosine :c² = a² + b² - 2ab×cos(C)with a, b, c being the sides, and C is the opposite angle of side c (whatever side we choose to be c).sin(90) = 1a)
it all starts with the right-angled triangle at the bottom, under the seat row plane. it gives us the length of the tilted line from the front wall to row A, which is the baseline (Hypotenuse) for that triangle.
we know the bottom line (5.6 m). we know the angle at the left vertex (12°), and because the angle on the ground right underneath row A is 90°, the angle at row A is
180 - 90 - 12 = 78°
Hypotenuse/sin(90) = bottom line/sin(78)
Hypotenuse = 5.6/sin(78) = 5.725107331... m
the outside angle at the bottom left vertex is the inside angle of the same vertex for the triangle above the tilted floor. and that is the complementary angle to 12° (= 90-12 = 78°).
so the length of the line of sight from row A to the bottom of the screen (= side c) is then for the triangle above the tilted floor :
c² = 2.5² + 5.725107331...² - 2×2.5×5.72...×cos(78) =
= 33.07527023...
c = 5.751110347... m
so, we see, the length of the line of sight is slightly different to the length of the tilted floor. it is not an isoceles triangle.
the angle at the vertex at the bottom of the screen we get with the same method (this time we have all sides and need the angle) :
5.725107331...² = 2.5² + 5.751110347...² - 2×2.5×5.75...×cos(C)
cos(C) = -(5.725107331...² - 2.5² - 5.751110347...²)/(2×2.5×5.75...) = 0.227727026...
C = 76.8367109...°
the angle of elevation is then based on a horizontal line from row A
180 - 90 - 76.8367109... = 13.1632891...° ≈ 13.2°
b)
now we need to do the same things for row P.
the bottom line is now 19.6 m.
the angles still the same as before for the bottom triangle :
12° at the left bottom vertex, 90° in the ground under row P, 78° at the vertex directly at row P.
the length of the tilted floor (Hypotenuse) is then
Hypotenuse/sin(90) = 19.6/sin(78) = 20.03787566... m
the outside angle at the bottom left vertex is also the same as before. the complementary angle to 12° (= 90-12 = 78°).
so the length of the line of sight from row P to the bottom of the screen (= side c) is then for the triangle above the tilted floor :
c² = 2.5² + 20.03787566...² - 2×2.5×20.03...×cos(78) =
= 386.9359179...
c = 19.67068677... m
the angle at the vertex at the bottom of the screen we get with the same method (this time we have all sides and need the angle) :
20.03787566...² = 2.5² + 19.67068677...² - 2×2.5×19.75...×cos(C)
cos(C) = -(20.03787566...² - 2.5² - 19.67068677...²)/(2×2.5×19.67...) = -0.084700073...
C = 94.85877813...°
the angle of depression is then based on a horizontal line from row P
94.85877813... - 90 = 4.858778132...° ≈ 4.9°
why does this look different to the case in a) ?
because we are looking down instead of up, we have to compare it now to the outside supplementary angle at the bottom vertex of the screen (we are building another triangle on top of the line of sight) :
180 - 94.85877813... = 85.14122187...°
and our angle of depression is
180 - 90 - 85.14122187... = 4.858778132...° (see above).
The angle of elevation from Row A to the bottom of the screen is 78⁰.
The angle of depression from Row P to the bottom of the screen is 7.5⁰.
What is the angle of elevation?The angle of elevation from Row A to the bottom of the screen is calculated as follows;
from row A to the bottom of the screen, is a straight line;
angle elevation of row A to bottom of screen = 90 - 12⁰ = 78⁰
The length of row A to row P is calculated as;
cos 12 = L/19.6 m
L = 19.6 m x cos (12)
L = 19.2 m
The angle of depression from Row P to the bottom of the screen is calculated as follows;
sinθ = 2.5 m / 19.2 m
sinθ = 0.1302
θ = sin⁻¹ (0.1302)
θ = 7.5⁰
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[5 pts] suppose that you toss a fair coin repeatedly. show that, with probability one, you will toss a head eventually. hint: introduce the events an = {"no head in the first n tosses"}, n = 1,2,....
If you toss a fair coin repeatedly. show that, with probability one, you will toss a head eventually.
To show that with probability one, you will eventually toss ahead, we need to show that the probability of never tossing a head is zero. Let's define the event An as "no head in the first n tosses."
Then, we have P(A1) = 1/2, since there is a 1/2 probability of getting tails on the first toss. Similarly, we have P(A2) = 1/4, since the probability of getting two tails in a row is (1/2) * (1/2) = 1/4.
More generally, we have P(An) = (1/2)^n, since the probability of getting n tails in a row is (1/2) * (1/2) * ... * (1/2) = (1/2)^n.
Now, we can use the fact that the sum of a geometric series with a common ratio r < 1 is equal to 1/(1-r) to find the probability of never tossing a head:
P("never toss a head") = P(A1 ∩ A2 ∩ A3 ∩ ...) = P(A1) * P(A2) * P(A3) * ... = (1/2) * (1/4) * (1/8) * ... = ∏(1/2)^n
This is a geometric series ith a common ratio r = 1/2, so its sum is:
∑(1/2)^n = 1/(1-1/2) = 2
Since the sum of the probabilities of all possible outcomes must be 1, and we have just shown that the sum of the probabilities of never tossing a head is 2, it follows that the probability of eventually tossing a head is 1 - 2 = 0.
Therefore, with probability one, you will eventually toss a head.
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A polygon is shown below . Write down the sum of its exterior angles. Work out the size of angle x
Answer:
use 360°/ n
Step-by-step explanation:
where n is the number of sides
did you understand like that