Drag the tiles to the correct boxes to complete the pairs. Not all tiles will be used.
Match each area to its corresponding radius or diameter of the circle.(All areas are approximate.)
area: 221.5584 square units
area: 78.5 square units
area: 452.16 square units
area: 36.2984 square units
area: 314 square units
area: 886.2336 square units
radius: 12 units
arrowBoth
diameter: 16.8 units
arrowBoth
radius: 3.4 units
arrowBoth
diameter: 10 units
arrowBoth

The number of possible outcomes when you roll the die would be 6

A die is known to have only 6 faces. The 6 faces are numbered from number 1 to number 6

Such that the sample space that we would have would be given as

SS = {1, 2, 3, 4, 5, 6}

Hence the number of outcomes that we would be able to have from one die would be given as 6

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The model would be 2 inches wide.

The ratio of the scale of an original thing to a new object that is a representation of it but of a different size is known as a scale factor (bigger or smaller).

Given:

A model is being built of a car.

The car is 12 feet long and 6 feet wide.

The length of the model is 4 inches.

The scale factor x is,

= 4/12

= 1/3

The width of the model is,

= 6 x 1/3

= 2 inches.

Therefore, the width is 2 inches.

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Probability of survival is 1/2, since only one of the two chambers is loaded. probability of survival is 4/5 is obtained by not re-spinning the chamber.

If I were faced with the choice, I would not choose either option as both choices involve significant risk. Instead, I would opt to not play the game at all as survival cannot be guaranteed.

In scenario mathematically, we can calculate the probabilities of survival for both options.

Let's assume that the chambers are numbered 1, 2, 3, 4, 5, and 6, with chamber 1 being the one that is lined up with the barrel after Bayes' first trigger pull.

If you choose to not re-spin the chamber and directly pull the trigger, the probability of survival is 4/5, since there are 4 remaining chambers that are not loaded.

If you choose to re-spin the chamber, the probability of survival:

The probability that the chamber aligned with the barrel is chamber 3 or 4 is 1/2. Given that chamber 3 or 4 is aligned with the barrel, the probability of survival is 1/2, since only one of the two chambers is loaded.

Therefore, the overall probability of survival if you choose to re-spin the chamber is (1/2) * (1/2) = 1/4.

Thus, the higher probability of survival is obtained by not re-spinning the chamber, with a probability of 4/5.

In the case with only one bullet, the probability of survival would be 5/6, making it a less dangerous scenario than this scenario with two bullets.

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_____The given question is incomplete, the complete question is given below:

bayes is playing russian roulette. the revolver has six chambers. he puts two bullets in two adjacent chambers, spin the cylinder, hold the gun to his head, and pull the trigger. it clicks and you survived.

so it is now the second try: may put the gun straight to your head and pull the trigger, or re-spin the gun before you do the same.

What is your choice and why? How does this differ from the case with only one bullet?

(a) A distance r from the cylinder axis is pr/2∈₀ and (b) expression for E when r > R is πR²lρ/2πϵ₀

(a) Consider a Gaussian surface in the form of a cylinder with radius r and length A, coaxial with the charged cylinder. An “end view” of the Gaussian surface is shown as a dashed circle. The charge enclosed by it is

q = ρV = πr²lp

V = volume of cylinder

If ρ is positive, the electric field lines are radially outward, normal to the Gaussian surface, and distributed uniformly along with it. Thus, the total flux through the Gaussian cylinder is  Φ = E(2πrl). Now, Gauss’ law leads to :

2π∈₀rlE = πr²lp

E = pr/2∈₀

(b) ) Next, we consider a cylindrical Gaussian surface of radius r > R. If the external field [tex]E_{ext}[/tex] then the flux is  Φ=2πϵ₀[tex]E_{ext}[/tex]

​ The charge enclosed is the total charge in a section of the charged cylinder with length A. That is,  q=πR²lρ. In this case, Gauss’ law yields :

2πϵ₀[tex]E_{ext}[/tex] = πR²lρ

[tex]E_{ext}[/tex] = πR²lρ/2πϵ₀

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The work required to pump out half of the water from the aquarium is approximately 2000 J.

The volume of the aquarium is V = (6 m) x (1 m) x (1 m) = 6 cubic meters. Since the density of water is 1000 kg/m3, the mass of the water in the aquarium is m = ρV = 1000 kg/m3 x 6 m3 = 6000 kg.

To pump out half of the water, we need to remove 1/2 x 6000 kg = 3000 kg of water. Since work is force times distance, we need to calculate the force required to lift this mass of water a distance of 1 m (the height of the aquarium).

The force required is F = mg = (3000 kg) x (9.8 m/s2) = 29,400 N. The work done is W = Fd = (29,400 N) x (1 m) = 29,400 J.

To approximate the required work by a Riemann sum, we can divide the height of the aquarium into n subintervals of width Δx, and choose a sample point xi* in each subinterval.

The force required to lift the water in each subinterval is approximately constant, so the work required to lift the water in each subinterval is approximately F(xi*)Δx. The total work required is therefore approximately given by the Riemann sum:

nlim ∑ F(xi*)Δxi= nlim ∑ (1000 x 6 x Δxi x xi*)i=1

Taking the limit as n goes to infinity, this Riemann sum becomes the integral:

∫0^1 1000 x 6 x x dx

Evaluating this integral gives:

∫0^1 6000 x^2 dx = [2000 x^3]0^1 = 2000 J

Therefore, the work required to pump out half of the water from the aquarium is approximately 2000 J. This approximation becomes more accurate as the number of subintervals n becomes larger.

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The maximum number of students the school can add if it wants to maintain a ratio of student to teachers 20 : 1 is 210.

Ratio defines the relationship between two quantities where it tells how much one quantity is contained in the other.

The ratio of a and b is denoted as a : b.

Given that,

Ratio of student to teachers now = 150 : 18

Ratio of student to teachers to maintain = 20 : 1

For 1 teacher , number of students needed = 20

For 18 teachers, number of students needed = 20 × 18 = 360

Present number of students for 18 teachers = 150

Maximum number of students who can be added = 360 - 150 = 210

Hence school can add a maximum of 210 students in order to maintain a ratio of student to teachers 20:1.

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Answer:

57327392393629324

Step-by-step explanation:

The range of the graph is,

⇒ Range = { y | - 1 < y < 3 }

The domain of the graph is,

⇒ Domain = { x | - 4 < x < 2 }

Mathematical expression is defined as the collection of the numbers variables and functions by using operations like addition, subtraction, multiplication, and division.

Given that;

The graph is shown in figure.

Now, We know that;

The range of the graph is the value of y where the function of graph is defined.

Hence, The range of the graph is,

⇒ Range = { y | - 1 < y < 3 }

And, The domain of the function is defined the value of y where the function of graph is defined.

Hence, The domain of graph is,

⇒ Domain = { x | - 4 < x < 2 }

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Answer:

55 L of water

Step-by-step explanation:

11-L of 60% acid contains 0.6 × 11 L = 6.6 L acid

Let x = amount of pure water.

Let y = total amount produced of 90% water solution.

A 90% water solution is a 10% acid solution.

Amounts of solutions:

x + 11 = y

Amounts of acid:

6.6 = 0.1y

y = 66

x + 11 = 66

x = 55

Answer: 55 L of water

55 L water has 0 acid.

11 L 60% acid solution has 6.6 L acid.

66 L of 10% acid solution has 6.6 L acid

The Equation of line perpendicular to line y = x + 1 and passes through (8,-3) is y= -x+ 5.

A line's  b is determined by how its y coordinate changes in relation to how its x coordinate changes. y and x are the net changes in the y and x coordinates, respectively. Therefore, it is possible to write the  change in y coordinate with respect to the change in x coordinate as,

m = Δy/Δx where, m is the slope

Given:

Equation: y= x+ 11

Now, the slope of perpendicular have the slope equal to negative reciprocal so that the product of slope of perpendicular line is -1.

Then, the slope of perpendicular line is, m= -1.

As, line passes through (8, -3) then the slope intercept form

y= mx+ b

-3 = (-1)(8)+ b

-3 = -8 +b

b= 5

Thus, the Equation of line is y= -x+ 5.

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▶️Watch help video

It's on you ss "▶️Watch help video"

The solution is, the nth term of the arithmetic sequence 4,7,10,13 is 1 + 3n.

An arithmetic progression or arithmetic sequence is a sequence of numbers such that the difference between the consecutive terms is constant.

For instance, the sequence 5, 7, 9, 11, 13, 15.. . is an arithmetic progression with a common difference of 2.

The nth term of AP : a_n = a + (n – 1) × d

here, we have,

the arithmetic sequence 4,7,10,13

so. we get,

1st term = 4 = a

common difference = 3 = d

i.e.  the nth term of the arithmetic sequence is, a_n = 4+(n-1)3

                                                                                      =1+3n

Hence, The solution is, the nth term of the arithmetic sequence 4,7,10,13 is 1 + 3n.

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From the following statements: statement 1 and statement 6th is false, rest are true about matrix and vector.

1. if the augmented matrix has a pivot position in every row, then the system is inconsistent, this statement is false.

2.  a vector is a linear combination of the columns of a matrix if and only if the equation has at least one solution, this statement is true.

3. if the columns of an matrix span , then the equation is consistent for each in b in R^m, this statement is true.

4. any linear combination of vectors can always be written in the form Ax for suitable matrix A and vector x, this statement is true.

5. if the system Ax = b is inconsistent, then b is not in the column space of A, this statement is true.

6. The equation Ax = b is referred to as a vector equation, this statement is false.

7. if A is an matrix and if the equation Ax = b is inconsistent for some in , then the RRef of A cannot have a pivot position in every row, this statement is true.

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In the case of a one-dimensional normal distribution with K classes, the Bayes classifier is not linear but is in fact quadratic.

We begin by noting that the Bayes classifier for the one-dimensional normal distribution is given by:

[tex]h(x) = argmaxk P(Ck|x) = argmaxk P(x|Ck)P(Ck)[/tex]

We can rewrite this as:

[tex]h(x) = argmaxk (1/√2πσk) exp(-1/2σ2k(x-uk)2) P(Ck)[/tex]

We can see that this is a quadratic equation in the form: ax2 + bx + c = 0. We can illustrate this by substituting the values for a, b, and c:

[tex]a = -1/2σ2k[/tex]

b = 0

[tex]c = ln(P(Ck)) - 1/2σ2k u2k[/tex]

We can see that this equation is not linear and is instead quadratic. Therefore, we can conclude that in the case of a one-dimensional normal distribution with K classes, the Bayes classifier is not linear but is in fact quadratic.

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Diego paid $25.25 to travel 13 miles.

An equation is an expression that shows the relationship between two or more numbers and variables using mathematical operations. An equation can be linear, quadratic, cubic and so on, depending on the degree of the variable.

The slope intercept form of the linear equation is:

y = mx + b

where m is the slope and b is the initial value.

Let y represent the total cost for travelling x miles.

City Cabs charges a $2.50 pickup fee and $1.75 per mile traveled. Hence:

y = 1.75x + 2.5

If the cab ride is $25.25:

25.25 = 1.75x + 2.5

x = 13 miles

He travelled 13 miles.

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The kind of sample described in the given information is a stratified random sample.

What is stratified random sample ?

A stratified random sample is a type of probability sampling method used to obtain a representative sample of a population by dividing the population into smaller, more homogeneous groups called strata, and then selecting a random sample from each stratum.

This method is used when the population has certain characteristics or subgroups that are of interest to the researcher, and the goal is to ensure that the sample accurately represents each subgroup in proportion to its size in the population.

The process of selecting a stratified random sample involves dividing the population into strata based on some relevant characteristic, such as age, gender, education level, income, or geographic location. Then, a random sample is selected from each stratum, using a simple random sampling method. The sample size from each stratum is proportional to the size of that stratum in the population.

The advantage of a stratified random sample is that it can provide a more accurate representation of the population than other types of sampling methods, as it ensures that each subgroup is adequately represented in the sample.

According to given information :

The population of interest is people age 60 or older from the three surrounding counties. To obtain a representative sample of the population, the director of the hospital pharmacy has divided the population into three strata (i.e., the three surrounding counties) and selected a random sample of 100 people age 60 or older from each stratum. This ensures that the sample includes a proportionate number of participants from each county, which should help to reduce the potential for sampling bias that might result from selecting participants from only one county.

Therefore, this is an example of a stratified random sample.

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Answer:

4. Yes it is proportional

5.  -2          -1            0        1          2

     1             3           5        7         9  

6.    -8           -4             8          (I don't know pick ur own numbers I guess)

      -8         -6.5            -2  

Step-by-step explanation:

the line is a bit off look at my old answer.

The value of x is x= π/8

The area of mathematics that deals with particular angles' functions and how to use those functions in calculations. There are six popular trigonometric functions for an angle. Sine (sin), cosine (cos), tangent (tan), cotangent (cot), secant (sec), and cosecant are their respective names and acronyms (csc).

Given:

2tan x/1-tan² x= 1

We know that tan π/4= 1

So, 2tan x/1-tan² x= tan π/4

Also, we know that tan 2x=  2tan x/1-tan² x

Again, 2tan x/1-tan² x= tan π/4

tan 2x= tan π/4

Now comparing we get

2x= π/4

x= π/8

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Answer:

The answer to the expression, (x + 6)^2 is [tex]x^2+12x+36[/tex].

Step-by-step explanation:

We can solve this problem by breaking (x + 6)^2 into (x + 6) (x + 6).  Now, we can use FOIL to figure out the expression that is equivalent to the problem.

(x + 6) (x + 6)

First, we multiply the Xs together.

[tex]x*x=x^2[/tex]

Next, is the Outer values.

[tex]x*6=6x[/tex]

Third, we multiply the Inner values.

[tex]6*x=6x[/tex]

Finally, we can multiply the Last values.

[tex]6*6=36[/tex]

Now, we can put it all together.

[tex]x^2+6x+6x+36[/tex]

We still have one more step left.  Add like terms.

[tex]x^2+12x+36[/tex]

The answer is [tex]x^2+12x+36[/tex].

Answer:

the value of b is 12.

Step-by-step explanation:

To find the value of b, we can use the vertex form of a quadratic function, which is f(x) = a(x - h)^2 + k, where (h, k) is the vertex of the parabola.

Since we know the vertex of this parabola from the table, we can use that information to write the equation in vertex form: f(x) = a(x + 6)^2 + 8

Expanding this, we get f(x) = a(x^2 + 12x + 36) + 8

Comparing this with the standard form, we see that a = 1, b = 12, and c = 8.

Therefore, the value of b is 12.

Answer:

so

2×4.57 = (3V/2pi)^1/3

Volume is approximately = 1600

The order of the fraction from least of greatest are

The fractions in order from least to greatest.

7/15 3/11 13/5 4/2 4/3

converting to decimals

7/15 = 0.4667

3/11 = 0.2727

13/5 = 2.6

4/2 = 2

4/3 = 1.3333

The order is written as follows

3/11 = 0.2727

7/15 = 0.4667

4/3 = 1.3333

4/2 = 2

13/5 = 2.6

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[tex]~~~~~~~~~~~~\textit{distance between 2 points} \\\\ \stackrel{ origin }{(\stackrel{x_1}{0}~,~\stackrel{y_1}{0})}\qquad (\stackrel{x_2}{-15}~,~\stackrel{y_2}{36})\qquad \qquad d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2} \\\\\\ d=\sqrt{(~~-15 - 0~~)^2 + (~~36 - 0~~)^2} \implies \implies d=\sqrt{( -15 )^2 + ( 36 )^2} \\\\\\ d=\sqrt{ 225 + 1296 } \implies d=\sqrt{ 1521 }\implies d=39[/tex]

The True- False of given statements of Solutions of Equations are justified.

1.13 False. The solution set to a system of three equations in three unknowns can be a point, a line, or a plane. It depends on the system of equations and how they intersect in three-dimensional space.

1.14 False. A system of linear equations can have infinitely many solutions, one solution, or no solutions. It depends on the coefficients of the equations and the rank of the coefficient matrix.

1.15 False. The solution set to a consistent rank 2 linear system in four unknowns would be a plane in four-dimensional space, not a line.

1.16 False. A system of four equations in four unknowns may not have a solution, or it may have infinitely many solutions or one solution. It depends on the coefficients of the equations and the rank of the coefficient matrix.

1.17 False. A system of four equations in four unknowns can have infinitely many solutions or one solution, but it cannot have at most one solution.

1.18 True. The rank of a system is always less than or equal to the number of equations in the system.

1.19 (a) False. If (i) has exactly one solution, it does not necessarily mean that (ii) will have exactly one solution. It depends on the coefficients of the equations in (ii).

(b) False. If the solution set of (i) is a line, it does not necessarily mean that the same is true for (ii). It depends on the coefficients of the equations in (ii).

(c) True. If (i) has no solutions, then the same is true for (ii), since (ii) is equivalent to (i).

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The value of y is 45.

Those triangles look the same but are different in size.

And in similar triangles,

the corresponding sides are in proportion to each other and the corresponding angles are equal to each other.

Given:

The shape of the triangle is given in the image.

Assuming the shape is ABC.

And the segment that is parallel to the base side of the triangle is DE.

BC II DE.

So, the triangle ΔADE is similar to ΔABC.

So, the corresponding sides have a constant proportion.

So, y/15 = (20 + 10)/10

y = 15 x 3

y = 45.

Therefore, y = 45.

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the counts that make this sample fail the large counts condition for the chi-square goodness-of-fit test are

B) The observed count of orange candies and

C) The observed count of yellow candies.

Given,

Terrell's company sells candy in packs that are supposed to contain 50% red candies, 25% orange, and 25% yellow.

The large counts condition for a chi-square goodness-of-fit test requires that the expected count for each category is at least 5.

To determine which counts in Terrell's sample fail this condition, we first need to calculate the expected counts for each color:

Expected count of red candies = 0.5 x 16 = 8

Expected count of orange candies = 0.25 x 16 = 4

Expected count of yellow candies = 0.25 x 16 = 4

The observed count of red candies meets the large counts condition because the expected count and the observed count are both 8. However, the observed counts of orange and yellow candies do not meet the large counts condition because the expected count for each is 4, but the observed count is less than 5.

Therefore, the counts that make this sample fail the large counts condition for the chi-square goodness-of-fit test are

B) The observed count of orange candies and

C) The observed count of yellow candies.

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the expected orange

and

the expected yellow

The mean (μ) and standard deviation (σ) represent the parameters of a normal distribution, so in this case, the parameters of the distribution are: μ = 80 transactions and σ = 10 transactions

A normal distribution, also known as a Gaussian distribution, is a continuous probability distribution that has a bell-shaped curve. This type of distribution is often used to model real-world phenomena that are expected to be distributed in a symmetrical fashion around a central value.

The central value of a normal distribution is the mean (μ), which represents the average of all the observations in the distribution. The spread or dispersion of the distribution is measured by the standard deviation (σ), which indicates how much the observations deviate from the mean.

In the given problem, the distribution of the number of transactions per day at an ATM is assumed to be normal with a mean of 80 transactions and a standard deviation of 10 transactions.

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A. (iv) No the case that P(A ∩ B) = 0.5 is not possible. Since B is contained in the event A ∩ B, it must be the case that P(A ∩ B) ≤ P(B). However, 0.5 > 0.4 violates this requirement. (iv)

B. To find the probability that the selected student has at least one of these two types of cards, we can use the formula:

P(A ∪ B) = P(A) + P(B) - P(A ∩ B)

Substituting the values, we get:

P(A ∪ B) = 0.7 + 0.4 - 0.3 = 0.8

Therefore, the probability that the selected student has at least one of these two types of cards is 0.8.

C. The probability that the selected student has neither type of card can be calculated as the complement of the event that the student has at least one of these two types of cards. Therefore,

P(neither A nor B) = 1 - P(A ∪ B) = 1 - 0.8 = 0.2

D. The event that the selected student has a Visa card but not a MasterCard can be written as A ∩ B'. We can calculate its probability as:

P(A ∩ B') = P(A) - P(A ∩ B) = 0.7 - 0.3 = 0.4

E. To calculate the probability that the selected student has exactly one of the two types of cards, we can use the formula:

P(exactly one of A or B) = P(A ∪ B) - P(A ∩ B)

Substituting the values, we get:

P(exactly one of A or B) = 0.8 - 0.3 = 0.5

Therefore, the probability that the selected student has exactly one of the two types of cards is 0.5.

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There were 44 children and 365 adults.

Let x = the number of children and

      y = the number of adults.

First equation: Children cost $4 and Adults cost $6.00.

In order to find the amount of money a group of children will cost, we multiply the number of children, x, by 4.

This is represented by 4x.

For adults, who cost 6 dollars to enter, we will use 6y.

The total amount of money made on the given day was $1574. To get this amount, we must add 4x and 6y.

14x+6y=1574 --------(1)

Second equation:

Total amount of people on the given day is 321.

To get this number, we must add together x and y, or the number of children and adults.

x + y = 321-----(2)

System of equations:

14x+6y=1574

x + y = 321

Let's use the substitution of the x variable to solve the system.

x + y = 321

y = 321 - x

Substitute y= 321 - x into the first equation.

14x+6(321 - x)=1574

14x + 1926 -6x = 1574

8x = - 352

x = 44

Substitute x = 44 back into y = 321 - x

y = 321 - (- 44)

y=365

There were 44 children and 365 adults.

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A and B are mutually exclusive, meaning that a student can only be a senior or a freshman, but not both.

If events A and B are independent, the occurrence of one event does not affect the probability of the other event occurring. Mathematically, we can write this as:

P(A and B) = P(A) x P(B)

In our case, event A represents the probability that a student is a senior and event B represents the probability that a student is a freshman. Since a student can only be one of these options, the events are mutually exclusive. Mathematically, we can write this as:

P(A and B) = 0

Therefore, we can conclude that events A and B are mutually exclusive.

It's important to note that events can also be neither independent nor mutually exclusive. If events A and B are not independent or mutually exclusive, then the occurrence of one event affects the probability of the other event occurring. In this case, the formula for the probability of both events occurring is:

P(A and B) = P(A) + P(B) - P(A or B)

However, since events A and B are mutually exclusive, we do not need to use this formula.

Therefore, the correct option is (a).

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