Find the perimeter and the
area of this triangle
18.5cm
8.5cm
4cm

Answer:

Step-by-step explanation:

g(f(x)) = x2  -1 + 2 = x2 + 1

The measure of the angle that defines the orange arc is 144°.

We know that if we have a circle of radius R, and there is an arc defined by an angle θ, then the area of that arc is given by:

A = (θ/360°)*pi*R^2

Where pi = 3.14

Here we know that the diameter is 10 miles, then the radius is:

R = 10mi/2 = 5mi

We can see that the shaded area is a = 10*pi  mi²

Then we can write the equation:

(θ/360°)*pi*(5mi)^2 = 10*pi  mi²

We can divide both sides by pi to get:

(θ/360°)*(5mi)^2 = 10  mi²

We can solve this for θ.

(θ/360°)*(5mi)^2 = 10  mi²

(θ/360°)25 mi² = 10  mi²

(θ/360°) = (10/25)

θ = (10/25)*360° = 144°

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The probability that when n=6,  [tex]Y_{1}[/tex] =3, [tex]Y_{2}[/tex]=1, [tex]Y_{2}[/tex]=2 is 0.0822.

We have to find the probability that n=6,  [tex]Y_{1}[/tex] =3, [tex]Y_{2}[/tex]=1, [tex]Y_{2}[/tex]=2

A rat has three possible exits to come out from the maze

The pmf of multinomial distribution is

P[tex](y_{1}, y_{2}, y_{3}......, y_{k} ) = \frac{n!}{{y_{1!} }y_{2}!....y_{k}! } p^y_1 1p^y_2 2......p^y_k k[/tex]

The probability of choosing one of the ways is 1/3

Then [tex]p_{1}[/tex]= 1 / 3,[tex]q_{1}[/tex] = 1-(1/3) = 2/3

        [tex]p_{2} = 1/3, q_{2} = 2/3\\ p_{3} = 1/3, q_{3} = 2/3[/tex]

[tex]P(y_{1}, y_{2}, y_{3} ) = P(1,2,3)\\ = \frac{6!}{3! 2! 1!} (\frac{1}{3} )3(\frac{1}{3} )2(\frac{1}{3} )1[/tex]

                      =60(0.00137)

                      =0.0822

Hence, the probability that when n=6,  [tex]Y_{1}[/tex] =3, [tex]Y_{2}[/tex]=1, [tex]Y_{2}[/tex]=2 is 0.0822.

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The correct question is:

A learning experiment requires a rat to run a maze (a network of pathways) until it locates one of three possible exits. Exit 1 presents a reward of food, but exits 2 and 3 do not. (If the rat eventually selects exit 1 almost every time, learning may have taken place.) Let [tex]Y_{i}[/tex] denote the number of times exit i is chosen in successive runnings. For the following, assume that the rat chooses an exit at random on each run.

a Find the probability that n = 6 runs result in [tex]Y_{1} = 3, Y_{2} = 1, and Y_{3} = 2.[/tex]

The transformation from f(x) to g(x) is a horizontal shift to the right by 2 units to obtain the graph of g(x).

In mathematics, transformation refers to a change in position, shape, size, or orientation of a figure or a function. There are various types of transformations, including translation, rotation, reflection, dilation, and more.

A translation is a transformation in which a figure is moved to a new position on the coordinate plane, while keeping its original size and orientation intact. A rotation is a transformation in which a figure is turned about a fixed point, called the center of rotation. A reflection is a transformation in which a figure is flipped over a line, called the line of reflection. A dilation is a transformation in which a figure is enlarged or reduced, while keeping its shape intact.

Transformations are used in various areas of mathematics, including geometry, engineering, computer graphics, and more. They provide a way to model real-world objects and processes and to solve problems related to size, position, and orientation. Understanding transformations is a fundamental aspect of mathematical skills and is crucial in many areas of study and research.

The transformation from f(x) = ln(x) to g(x) = ln(x + 2) is a horizontal shift to the right by 2 units. This is because when you replace x with x + 2 in the logarithmic function, you are shifting the graph to the right by 2 units. The x-intercepts of the graph of f(x) will be at x = 1 (since ln(1) = 0), whereas the x-intercepts of the graph of g(x) will be at x = -2 (since ln(2) = 0). So, in essence, the transformation is simply a horizontal shift of the graph of f(x) to the right by 2 units to obtain the graph of g(x).

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Answer: 10

Step-by-step explanation: beacause you add both of the sides

Answer:

Solve for B:

[tex]B = -b+\frac{5A}{6h}[/tex]

Step-by-step explanation:

[tex]A=1.2h(b+B)[/tex]

Use the distributive property to multiply [tex]1.2h by b+B[/tex]

[tex]A=1.2hb+1.2hB[/tex]

Swap sides so that all variable terms are on the left hand side.

[tex]1.2hb+1.2hB=A[/tex]

Subtract [tex]1.2hb[/tex] from both sides.

[tex]1.2hB= A - 1.2hb[/tex]

The equation is in standard form.

[tex]\frac{6h}{5} B=-\frac{6bh}{5} +A[/tex]

Divide both sides by [tex]1.2h[/tex].

[tex]\frac{5*(\frac{6h}{5})B }{6h} =\frac{5(-\frac{6bh}{5}+A) }{6h}[/tex]

Dividing by [tex]1.2h[/tex] undoes the multiplication by [tex]1.2h[/tex].

[tex]B = \frac{5(-\frac{6bh}{5}+A) }{6h}[/tex]

Divide [tex]A - \frac{6hb}{5}[/tex] by [tex]1.2h[/tex].

[tex]B = -b+\frac{5A}{6h}[/tex]

Solve For B:

[tex]B = -b+\frac{5A}{6h}[/tex]

We can use the Law of Sines to find the measure of angle W:

sin(W)/w = sin(V)/v

sin(W) = w*sin(V)/v

sin(W) = 7.2*sin(8°)/6.2

sin(W) ≈ 0.1001

Taking the inverse sine of both sides, we get:

W ≈ 5.78° or W ≈ 174.22°

Since W is an angle in a triangle, it must be between 0° and 180°. Therefore, the only possible value for ∠W is:

W ≈ 5.78° (to the nearest tenth of a degree).

Answer:

there are 1456 different ways that the six medals can possibly be given out.

Step-by-step explanation:

There are different ways to approach this problem, but one common method is to use combinations.

First, we need to determine how many ways we can choose three girls out of 14, and three boys out of 4. This can be calculated using combinations:

Number of ways to choose 3 girls out of 14: C(14,3) = 364

Number of ways to choose 3 boys out of 4: C(4,3) = 4

Now, we can use the multiplication principle to determine the total number of ways to give out the six medals:

Total number of ways = number of ways to choose 3 girls * number of ways to choose 3 boys = 364 * 4 = 1456

Therefore, there are 1456 different ways that the six medals can possibly be given out.

The statement that is true about the data shown in the stemplot is that the most common speed limit is 55.

What is stemplot ?
A stemplot, also known as a stem-and-leaf plot, is a type of chart used to display data. It shows the distribution of a set of values by separating each value into a stem and a leaf.

Given by the question:
Based on the values given, the stemplot titled "speed limit" could be constructed as follows:

1 | 0

4 | 5 5 5 5

5 | 0 5 5 5 5 5

6 | 0 0 0 5 5 5 5

7 | 0 0

The stem represents the tens digit of each value, and the leaves represent the ones digit. For example, the first value of 10 has a stem of 1 and a leaf of 0.

Based on the stemplot, we can make the following observations:

Therefore, the statement that is true about the data shown in the stemplot is that the most common speed limit is 55.

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Answer: Function A is greater.

Step-by-step explanation:

just find the slope...

Slope for Function C = [tex]\frac{1}{3}[/tex] = [tex]0.33333[/tex] repeat...

and for Function A = [tex]\frac{5}{2}[/tex] = [tex]2.5[/tex]

to find the slope we use; [tex]s=[/tex] [tex]\frac{y}{x}[/tex].

Using standard deviation, the probability that the service technician will have to work overtime to fix the machine today is 0.5.

The degree of variance or dispersion in a set of data values is measured by standard deviation. It gauges how far the data values depart from the data set's mean (average).

The mean of the data set is first determined, then the standard deviation. The difference between each data point and the mean is then squared for each data point. After dividing the total number of data points by the sum of these squared differences, minus one, the standard deviation is calculated by taking the square root of this result.

a. The expected number of emails that are tracked can be found by multiplying the total number of emails by the probability that an email is tracked:

Expected number of tracked emails = 50 x 0.4 = 20

Therefore, we can expect that 20 out of 50 received emails will be tracked.

b. To find the variance and standard deviation for the number of tracked emails, we can use the formula:

Variance = np(1-p)

Standard deviation = sqrt(np(1-p))

where n is the number of trials (in this case, 50) and p is the probability of success (0.4).

Variance = 50 x 0.4 x (1 - 0.4) = 12

Standard deviation = sqrt(50 x 0.4 x (1 - 0.4)) ≈ 3.46

Therefore, the variance for the number of tracked emails is 12 and the standard deviation is approximately 3.46.

c. The probability distribution for the duration of a service call is:

Duration (hours) Probability

1                               0.25

2                               0.25

3                               0.25

4                               0.25

d. The probability that a service call will take three hours is 0.25, as shown in the probability distribution table.

e. To find the probability that the service technician will have to work overtime, we need to calculate the probability that a service call will take longer than two hours, since the technicians usually get off at 5:00 P.M. and it is currently 3:00 P.M.

The probability that a service call will take longer than two hours is:

P(call takes 3 hours or 4 hours) = P(call takes 3 hours) + P(call takes 4 hours) = 0.25 + 0.25 = 0.5

Therefore, the probability that the service technician will have to work overtime to fix the machine today is 0.5.

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The equation which models the problem is m - $23.94 = $12.75 and the value of m is $36.69.

A mathematical statement containing two algebraic expressions on two sides of an equal to sign is defined as an equation.

Given,

Cost of a reusable water bottle that Amanda bought = $23.94

Money remaining after buying the bottle = $12.75

Let m represent how much money Amanda had to start.

Then the equation can be written as,

m - $23.94 = $12.75

We have to solve the equation.

Adding $23.94 on both sides,

m = $12.75 + $23.94

m = $36.69

Hence the money at the start is $36.69.

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Answer:

Least common denominator is 3.

Step-by-step explanation:

2/3 = 2/3 x 1/1 = 2/3

1/3 = 1/3 x 1/1 = 1/3

The function representing the number of bacteria present each day is f(t) = 2032[tex](0.85)^{t}[/tex].

Both the set X and the set Y are referred to as the function's domain and codomain, respectively.

Beginning in 2032, there will be a 15% daily decline in the number of microorganisms.

a function that displays the daily average amount of microorganisms.

f(x) = 2032[tex](1 - .15)^{t}[/tex]

f(x) = 2032[tex](0.85)^{t}[/tex]

Therefore f(t) = 2032[tex]0.85^{t}[/tex] is the correct answer.

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The value of the game is equal to -20 and is the least loss for ABC (which is the maximum gain for XYZ).

The Minimax theorem, which states that in a two-person zero-sum game, each player minimizes the maximum loss conceivable, can be used to determine the best course of action for both stores.

ABC retailer:

If XYZ decides to use newspaper advertising, ABC could see a maximum loss of -80.

If XYZ chooses radio advertising, ABC might lose as much as -20.

If XYZ chooses television advertising, ABC will ultimately lose 50.

As a result, selecting the advertising medium that results in the lowest possible loss, which is radio advertising with a loss of -20.

XYZ retailer:

If ABC chooses newspaper advertising, XYZ might earn up to 40.

If ABC chooses radio advertising, XYZ might earn a maximum of 15.

If ABC chooses television advertising, XYZ might earn up to $20.

Therefore, by selecting the advertising medium that offers the highest possible gain, which is newspaper advertising with a gain of 40, XYZ optimizes the greatest gain.

Therefore, the value of the game is equal to -20 and is the least loss for ABC (which is the maximum gain for XYZ).

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Correct question:
In a small town, there are two discount stores ABC and XYZ. They are the only

stores that handle the festival goods. The total number of customers is equally divided

between the two because the price and quality of goods sold are equal. Both stores

have good reputations in the community, and they render equally good customer

services. Assume that a gain of customer by ABC is a loss to XYZ and vice versa.

Both stores plan to run annual pre-Christmas sale during the first week of December.

Sales are advertised through the local newspaper, radio and television media. With the

aid of advertising the payoff for ABC store is constructed and given below.

XYZ store

News paper radio Television

News paper 30 40 -80

ABC radio 0 15 -20

Television 90 20 50

Find optimal strategies for both stores and the value of the game.

On average, ABC can expect to earn 15 more customers than XYZ during the pre-Christmas sale.

In mathematics and computer science, a graph is a collection of points (called vertices or nodes) that are connected by lines or curves (called edges). Graphs can be used to represent many different types of relationships or structures, including social networks, transportation networks, electrical circuits, and mathematical functions.

A graph is usually represented visually as a set of points on a plane or in space, with lines or curves connecting them. The points can represent any type of object or entity, such as cities, people, or data points, while the edges represent the connections or relationships between them. Edges can be directed (with arrows indicating a one-way relationship) or undirected (with no preferred direction).

To find the optimal strategies for both stores and the value of the game, we can use the graphical method of solving 2-player zero-sum games.

First, we can create a payoff matrix using the given information:

markdown

Copy code

     |  Newspaper |  Radio  | Television |

--------------------------------------------

ABC   |     30     |    15   |     20     |

XYZ   |     40     |   -15   |     50     |

Note that the payoff for XYZ in the radio column is negative, indicating a loss.

Next, we can plot the payoffs on a graph, with ABC's strategies on the x-axis and XYZ's strategies on the y-axis. We can also draw a line at the minimum value of each column (the "minimax" line) and a line at the maximum value of each row (the "maximin" line).

The graph should look like this:

lua

Copy code

           -15      15       20

             |-------|--------|

         30  |   30  |   15   |

             |       |        |

             |-------|--------|

         50  |   40  |  -15   |

             |       |        |

             |-------|--------|

               40       50

To find the optimal strategies, we need to find the intersection of the minimax and maximin lines. In this case, the intersection is at the point (Radio, Television) = (15, 50), so the optimal strategies are for ABC to advertise on the radio and XYZ to advertise on television.

The value of the game is the payoff at the intersection of the minimax and maximin lines, which is 15. This means that on average, ABC can expect to earn 15 more customers than XYZ during the pre-Christmas sale.

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an = 3 + (15 - 1) * 4 = 3 + 14 * 4

= 3 + 56

= 59

So, the 15th term in the sequence is 59.

2. To find the sum of an arithmetic series with first term a1, last term an, and number of terms n, we use the formula: Sn = n/2 * (a1 + an). Plugging in the values, we get:

Sn = 20/2 * (5 + 119)

= 20/2 * 124

= 620

So, the sum of the series is 620.

3. To find the sum of an arithmetic series with first term a1, common difference d, and number of terms n, we use the formula: Sn = n/2 * (2a1 + (n - 1)d). Plugging in the values, we get:

Sn = 15/2 * (2 * 12 + (15 - 1) * 6) = 15/2 * (24 + 84)

= 15/2 * 108

= 810

So, the sum of the series is 810.

4.To find the nth term in a geometric sequence with first term a1, common ratio r, and nth term an, we use the formula: an = a1 * r⁽ⁿ⁻¹⁾. Plugging in the values, we get:

64 = 2 * r⁽⁶⁻¹⁾

64 = 2 * r⁵

32 = r⁵

r = [tex]2^{(5^{(1/5)} )}[/tex]

So, the 6th term in the sequence is 64.

5. To find the sum of a finite geometric series with first term a1, common ratio r, and number of terms n, we use the formula: Sn = a1 * (1 - rⁿ) / (1 - r). Plugging in the values, we get:

Sn = 2 * (1 - 4⁶) / (1 - 4)

= 2 * (1 - 4096) / -3

= 2 * (-4095) / -3

= 2730

So, the sum of the series is 2730.

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The proof is completed as follows

       Statement                             Reason

NL/AB =  MN/BC = LM/CA          ratios of side of similar triangles are equal

Δ ABC is similar to Δ LNM         Definition of similar triangles

Similar triangles are triangles that have the same shape but may differ in size. That is to say, they have the same angles but their sides may be scaled differently.

When two triangles are similar, their corresponding angles are congruent and the corresponding sides are in proportion to each other.

This means that if you know the lengths of any two sides of a similar triangle, you can use those ratios to find the lengths of the other sides.

In the proof, the equation NL/AB =  MN/BC = LM/CA means that proportions are equal

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The proof of 8 sin26. sin34. sin60. sin86 = root3 sin78 is given below.

Equations using trigonometric functions that hold true for all possible values of the variables are known as trigonometric identities.

Given:

An equation,

8 sin26. sin34. sin60. sin86 = root3 sin78

LHS = 8 sin26. sin34. sin60. sin86

      = 0.9085192402

      ≈ 0.9 to the nearest tenth.

And RHS = 3 sin78

               = 0.89023679976

               ≈ 0.9 to the nearest tenth.

Therefore, 8 sin26. sin34. sin60. sin86 = root3 sin78.

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Yes , the arithmetic sequence ( written as sequences of the form (a , a+k , a+2k , a+3k for some constants a and k ) is a subspace of V .

To determine whether subsets of V in the arithmetic sequences are subspaces of V, we check if they satisfy the three conditions for a subset to be a subspace:

(i) The subset must contain the zero vector.

(ii) The subset must be closed under vector addition.

(iii) The subset must be closed under scalar multiplication.

Let S be a subset of V consisting of arithmetic sequences.

Now we check the three conditions :

(i) We find an arithmetic sequence in S that has all its terms equal to zero. The only arithmetic sequence that satisfies this is the sequence (0, 0, 0, 0, ...), which is in S.

So , the subset "S" contains zero vector.

(ii) Next we need to show that if u and v are two arithmetic sequences in S, then their sum "u + v" is also an arithmetic sequence in S.

Let U = (a, a+k, a+2k, a+3k, ...) and V = (b, b+l, b+2l, b+3l, ...) be two arithmetic sequences in S.

Then, their sum is ⇒ u+v = (a+b, a+k+b+l, a+2k+b+2l, a+3k+b+3l, ...).

This is also an arithmetic sequence,

So , subset "S" is closed under vector addition.

(iii) Next , we show that if U is an arithmetic sequence in S and C is a scalar, then CU is also an arithmetic sequence in S.

Let U = (a, a+k, a+2k, a+3k, ...) be an arithmetic sequence in S, and let c be a scalar.

we get , CU = (ca, ca+ck, ca+2ck, ca+3ck, ...) is also an arithmetic sequence,

So , subset "S" is closed under scalar multiplication.

we see that , S satisfies all three conditions, it is a subspace of V.

Therefore, any subset of V consisting of arithmetic sequences is a subspace of V.

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The given question is incomplete , the complete question is

Let V be the space of all infinite sequences of real numbers , Which of the subsets of V in The arithmetic sequences [i.e. sequences of the form (a , a+k , a+2k , a+3k for some constants a and k] are subspaces of V ?

The youngest age for which the average income of a lawyer is $275,000 is 27.91 year.

Quadratic Equation helps solve quadratic equations. First, put the equation into the form ax²+bx+c=0. where a, b, and c are the coefficients. Then plug these coefficients into the equation.

(-b±√(b²-4ac))/(2a) . See examples of solving various equations using formulas

The average annual income, I, in dollars, of a lawyer with an age of x years is modeled with with the following function:

I = - 425x² + 45500x - 650000 .......... (1)

If the average annual income of the lawyer at the age of x years is $275000, then from the equation (1) we can write

- 425x² + 45500x - 650000 = 250000

I=-425x^(2) + 45,500x-650,000

275000 = -425x^(2) + 45,500x-650,000

-425x^(2) + 45,500x -650000-275000=0

425x^(2) -45,500x +925000=0

x = 900/17 ± 10√(1810)/17

= 77.96 and 27.91

here we are not accepting the ans 77.96 because we have already less value which is 27.91 so final ans will be 27.91

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Answer:

Since the second box has inside dimensions four times those of the first box, its volume is $(4a)^3 = 64a^3$ times as much as the volume of the first box, whose volume is $a^3$. Therefore, the answer is (D) 64.

The relationship between the two variables is y=4/3 x+3.

The general equation of a straight line is y=mx+c, where m is the gradient, and y = c is the value where the line cuts the y-axis. This number c is called the intercept on the y-axis.

The coordinate points are (0, 3) and (3, 7)

Here, slope (m) = (7-3)/(3-0)

= 4/3

Substitute m=4/3 and (x, y)=(0, 3) in y=mx+c, we get

3=4/3(0)+c

c=3

So, the equation is y=4/3 x+3

Therefore, the relationship between the two variables is y=4/3 x+3.

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The magnitude of the acceleration due to gravity would decrease with increasing height.

If the two shells were fired simultaneously with the same speed, their initial velocity would be the same. However, the shell that hits ship A, which is closer, would take less time to reach its target compared to the shell that hits ship B.

Since the magnitude of the acceleration due to gravity is constant, the vertical motion of the shells would be described by the following equation:

h = vi * t + (1/2) * g * [tex]t^2[/tex]

where h is the height of the shell, vi is the initial velocity, t is the time, and g is the acceleration due to gravity.

The horizontal motion of the shells would be described by the following equation:

d = vi * t

where d is the horizontal distance travelled by the shell.

By solving the above equations, we can determine the time taken by each shell to reach its target and therefore, the time difference between the two shells.

Note that this analysis assumes that air resistance is negligible and that the magnitude of the acceleration due to gravity is constant. In reality, air resistance would play a role in the motion of the shells, and the magnitude of the acceleration due to gravity would decrease with increasing height.

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Answer:

Rachels age = 3 years

Greg's age = 9 years

Step-by-step explanation:

Present age:

                         Let Rachels age = x years

              5 time of Rachel's age = 5*x = 5x

                          6 less than 5x  = 5x - 6

                                Greg's age = (5x - 6) years

Age after four years:

                    Rachel's age = (x + 4) years

                      Greg's age =  5x - 6 + 4

                                          = ( 5x - 2) years

          Sum of their ages = 20

           x + 4 + 5x - 2        = 20

                  x + 5x + 4 - 2 = 20

   Combine like terms,

                            6x  + 2 = 20

Subtract 2 from both sides,

                                    6x = 20 - 2

                                    6x = 18

Divide both sides by 6,

                                      x = 18 ÷ 6

                                       x = 3

Rachel's age = 3 years

Greg's age = 5*3 - 6

                   = 15 - 6

                   = 9 years

The scale factor of figure A to figure B is 5/3.

A triangle is a three-sided polygon that consists of three edges and three vertices.

Two triangles are said to be similar if their corresponding angles are congruent and the corresponding sides are in proportion .

A and B are similar triangles.

We need to find the scale factor.

The ratio between the scale of a given original object and a new object

Scale factor is 15/9=35/21=40/24

5/3=5/3=5/3

Hence, 5/3 is the scale factor of figure A to figure B.

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The median did not change because the 5th and 6th term of data is same.

The mean is the average value which can be calculated by dividing the sum of observations by the number of observations

Mean = Sum of observations/the number of observations

Median represents the middle value of the given data when arranged in a particular order.

Given that;

The median of data set with nine numbers= 36

After the addition of tenth value median= 36

Now,

Median is the middle term of data

Here, the middle value and 6th term must be same.

So, 5th term= 6th term= 36

Therefore, the median of the given data set is same.

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The effective rate for a nominal rate of 5% compounded continuously over a time period of 1 year is 5.13%.

To find the effective rate corresponding to a nominal rate compounded continuously, we use the formula:

r_effective = [tex]e^{(r\_nominal * t)} - 1[/tex]

where r_nominal is the nominal rate, and t is the time period.

For a nominal rate of 5% compounded continuously, the effective rate can be calculated as follows:

r_effective = [tex]e^{(0.05*t)}-1[/tex]

Note that the time period t is usually expressed in years. For example, if we want to find the effective rate for a time period of 1 year, we have:

r_effective = [tex]e^{(0.05*t)}-1[/tex]

= 1.051296 - 1

= 0.051296 or 5.13%

So, the effective rate for a nominal rate of 5% compounded continuously over a time period of 1 year is 5.13%.

It's important to understand the difference between nominal and effective rates. The nominal rate is the rate that is advertised or quoted, while the effective rate takes into account the frequency of compounding. The effective rate is a more accurate representation of the true interest rate because it shows the actual amount of interest earned over a given time period.

Continuous compounding means that interest is calculated and added to the principal continuously, rather than at regular intervals. This results in a higher effective rate compared to the same nominal rate compounded at regular intervals.

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The area of a sector of a circle with radius 100 meters and central angle 1/8 radian is approximately 625 * π square meters.

The area of a sector of a circle with radius 100 meters and central angle 1/8 radian can be found using the formula:

A = [tex](\theta/2\pi )*\pi r^{2}[/tex]

where A is the area of the sector, Θ is the central angle in radians, π is Pi (approximately 3.14), and r is the radius of the circle.

Plugging in the given values, we have:

A = [tex](\frac{\frac{1}{8}}{{2\pi }} )*\pi *100^{2}[/tex]

A =[tex](1/16)*\pi *100^{2}[/tex]

A = [tex](\pi /16)*100^{2}[/tex]

A = (π/16) * 10000

A = 625 * π square meters

So, the area of the sector of the circle with radius 100 meters and central angle 1/8 radian is approximately 625 * π square meters.

The area of a sector of a circle is a portion of the circle's area enclosed by two radii and an arc. The central angle of the sector determines the fraction of the circle's circumference that the arc represents, and therefore the fraction of the circle's area that is enclosed by the sector.

To find the area of a sector, we first need to find the central angle in radians. The central angle is the angle formed by two radii at the center of the circle that intercept the circumference of the circle at the endpoints of the arc.

The formula for the area of a sector is given by:

A = [tex](\theta/2\pi )*\pi r^{2}[/tex]

where A is the area of the sector, Θ is the central angle in radians, π is Pi (approximately 3.14), and r is the radius of the circle.

In the given problem, the radius of the circle is 100 meters and the central angle is 1/8 radian. We plug these values into the formula and simplify to get the final answer.

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Estimate of the total distance traveled in the first 6 hours of travel for a particle v(t)=−t^2 +4t+6 is 36m.

We have v(t)=−t² +4t+6

For finding the distance we have to find the integration of the given equation:

s = ∫ -t² +4t+6

= -t³/3 + 4t²/2 + 6t

= -t³/3 + 2t²/ + 6t

For the value of t = 6 hours

= -6³/3 + 2(6)² + 6(6)

= -72 + 72 + 36

= 36

Estimate of the total distance traveled in the first 6 hours is 36 m.

When a moving object has a positive velocity, its position is constantly rising. (We focus on the case when velocity is always positive; we will soon explore cases where velocity is negative.) We have shown that the area under the velocity curve represents the precise distance travelled whenever is constant on an interval. Finding the areas of rectangles that closely resemble the area under the velocity curve allows us to calculate the total distance travelled when is not constant.

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