The average rainfall in Phoenix is 8.29 inches per year. The table shows recent data on the difference in annual rainfall from the average. Phoenix Annual Total Rainfall Year Rainfall compared to average yearly rainfall 2008 +6.57 inches 2009 –2.68 inches 2010 +12.26 inches 2011 –4.38 inches 2012 –4.46 inches Which list represents the years from driest to wettest?

Answer:

20% robability that the tourist will be able to talk with a randomly encountered resident of the region, given that the resident speaks German

Step-by-step explanation:

We use the conditional probability formula to solve this question. It is

[tex]P(B|A) = \frac{P(A \cap B)}{P(A)}[/tex]

In which

P(B|A) is the probability of event B happening, given that A happened.

[tex]P(A \cap B)[/tex] is the probability of both A and B happening.

P(A) is the probability of A happening.

In this problem, we have that:

Event A: Speaking German

Event B: Speaking English

3% speak both English and German

This means that [tex]P(A \cap B) = 0.03[/tex]

15% speak German

This means that [tex]P(A) = 0.15[/tex]

So

[tex]P(B|A) = \frac{0.03}{0.15} = 0.2[/tex]

20% robability that the tourist will be able to talk with a randomly encountered resident of the region, given that the resident speaks German

Answer:

Step-by-step explanation:

Hello!

You have two variables of interest

X₁: failure stress of a NiCrAlZr coating after nine 1-hr cycles.

X₂: failure stress of a NiCrAlZr coating after six 1-hr cycles.

a)

To be able to estimate the difference between the means using a confidence interval, you need that both variables have a normal distribution and to determine whether or not the population variances are equal.

If the population variances are equal, σ₁²=σ₂², you can use a pooled variance t-test

If the population variances are different, σ₁²≠σ₂², you have to use Welch's t-test

Using α: 0.05

The normality test for X₁ shows a p-value of 0.7449 ⇒ You can assume it has a normal distribution.

The normality test for X₂ shows a p-value of 0.9980 ⇒ You can assume it has a normal distribution.

The F-test for variance homogeneity shows a p-value of 0.6968 (H₀:σ₁²=σ₂²) ⇒You can assume both population variances are equal.

b) and c)

You need to test if both population means are the same, the hypotheses are:

H₀: μ₁=μ₂

H₁: μ₁≠μ₂

α: 0.05

[tex]t= \frac{(X[bar]_1-X[bar]_2)-(Mu_1-Mu_2)}{Sa*\sqrt{\frac{1}{n_1} +\frac{1}{n_2} } } ~~t_{n_1+n_2-2}[/tex]

[tex]Sa= \sqrt{\frac{(n_1-1)S^2_1+(n_2-1)S^2_2}{n_1+n_2-2} } = \sqrt{\frac{8*4.28+5*5.62}{9+6-2} }= \sqrt{4.7953}= 2.189= 2.19[/tex]

[tex]t_{H_0}= \frac{(X[bar]_1-X[bar]_2)-(Mu_1-Mu_2)}{Sa*\sqrt{\frac{1}{n_1} +\frac{1}{n_2} } } = \frac{(16.36-11.48)-0}{2.19*\sqrt{\frac{1}{9} +\frac{1}{6} } } = 4.23[/tex]

The distribution of this test is a t with 13 degrees of freedom and the test is two-tailed, so to calculate the p-value you have to do the following:

P(t₁₃≤-4.23)+P(t₁₃≥4.23)= P(t₁₃≤-4.23)+[1-P(t₁₃<4.23)]=  0.000492 + (1-0.999508)= 2*0.000492= 0.000984≅ 0.001

The p-value: 0.001 is less than α: 0.05, the decision is to reject the null hypothesis.

I hope it helps!

Answer:

Part 1

(a) 0.28434

(b) 0.43441

(c) 29.9 mm

Part 2

(a) 0.97722

Step-by-step explanation:

There are two questions here. We'll break them into two.

Part 1.

This is a normal distribution problem healthy children having the size of their left atrial diameters normally distributed with

Mean = μ = 26.4 mm

Standard deviation = σ = 4.2 mm

a) proportion of healthy children have left atrial diameters less than 24 mm

P(x < 24)

We first normalize/standardize 24 mm

The standardized score for any value is the value minus the mean then divided by the standard deviation.

z = (x - μ)/σ = (24 - 26.4)/4.2 = -0.57

The required probability

P(x < 24) = P(z < -0.57)

We'll use data from the normal probability table for these probabilities

P(x < 24) = P(z < -0.57) = 0.28434

b) proportion of healthy children have left atrial diameters between 25 and 30 mm

P(25 < x < 30)

We first normalize/standardize 25 mm and 30 mm

For 25 mm

z = (x - μ)/σ = (25 - 26.4)/4.2 = -0.33

For 30 mm

z = (x - μ)/σ = (30 - 26.4)/4.2 = 0.86

The required probability

P(25 < x < 30) = P(-0.33 < z < 0.86)

We'll use data from the normal probability table for these probabilities

P(25 < x < 30) = P(-0.33 < z < 0.86)

= P(z < 0.86) - P(z < -0.33)

= 0.80511 - 0.37070 = 0.43441

c) For healthy children, what is the value for which only about 20% have a larger left atrial diameter.

Let the value be x' and its z-score be z'

P(x > x') = P(z > z') = 20% = 0.20

P(z > z') = 1 - P(z ≤ z') = 0.20

P(z ≤ z') = 0.80

Using normal distribution tables

z' = 0.842

z' = (x' - μ)/σ

0.842 = (x' - 26.4)/4.2

x' = 29.9364 = 29.9 mm

Part 2

Population mean = μ = 65 mm

Population Standard deviation = σ = 5 mm

The central limit theory explains that the sampling distribution extracted from this distribution will approximate a normal distribution with

Sample mean = Population mean

¯x = μₓ = μ = 65 mm

Standard deviation of the distribution of sample means = σₓ = (σ/√n)

where n = Sample size = 100

σₓ = (5/√100) = 0.5 mm

So, probability that the sample mean distance ¯x for these 100 will be between 64 and 67 mm = P(64 < x < 67)

We first normalize/standardize 64 mm and 67 mm

For 64 mm

z = (x - μ)/σ = (64 - 65)/0.5 = -2.00

For 67 mm

z = (x - μ)/σ = (67 - 65)/0.5 = 4.00

The required probability

P(64 < x < 67) = P(-2.00 < z < 4.00)

We'll use data from the normal probability table for these probabilities

P(64 < x < 67) = P(-2.00 < z < 4.00)

= P(z < 4.00) - P(z < -2.00)

= 0.99997 - 0.02275 = 0.97722

Hope this Helps!!!

Answer:

the answer is B

Step-by-step explanation:

The expression that is equivalent to [tex](256 x^{16})^{\frac{1}{4} }[/tex] is [tex]4x^4[/tex].

The given parameters:

The given expression can be simplified by applying laws of indicial expressions as follows;

[tex](256 x^{16})^{\frac{1}{4} } = (2^8 x^{16})^{\frac{1}{4} } \\\\[/tex]

The expression can be simplified further as follows;

[tex](2^8 x^{16})^{\frac{1}{4} } = (2^8)^{\frac{1}{4} } \times (x^{16}) ^{\frac{1}{4} } = (2^2) \times (x ^4) = 4x^4\\\\[/tex]

Thus, the expression that is equivalent to [tex](256 x^{16})^{\frac{1}{4} }[/tex] is [tex]4x^4[/tex].

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Tony can get 12 combined outcomes when he throws a dice and a coin, which are combinations of six numbers (1-6) of the dice and two possible outcomes of the currency (heads or tails).

In the given scenario, Tony is throwing a dice and a coin. The dice can land on any number from 1 to 6, and the coin can land on either heads (H) or tails (T). The combined outcomes can be listed as follows:

So, there are a total of 12 possible outcomes.

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The possible outcomes of throwing a dice and flipping a coin are listed. The format is [Dice number][Coin result].

The outcomes of Tony throwing the dice once and throwing a coin to get heads or tails are:

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[tex]7x^{2} +33x-10=0[/tex]

[tex]\implies 7x^{2} +35x-2x-10=0\\\implies7x(x+5)-2(x+5)=0\\\implies (7x-2)(x+5)=0[/tex]

[tex]7x - 2 = 0\\\implies \boxed{\bold{x = \frac{2}{7} }}[/tex]

[tex]x+5=0\\\implies \boxed{\bold{x=-5}}[/tex]

Answer:

Upper bond is 0.398

Step-by-step explanation:

See attached file

Answer:

781250 Square Meters

Step-by-step explanation:

Let the dimensions of the rectangular plot be x and y

Farmer Ed wants to enclose three sides of a rectangular plot with a fencing of 2500 meters.

Therefore: Perimeter, P=x+2y=2500

We want to find the largest area that can be enclosed.

Area of the plot, A(x,y)=xy

Substitute x=2500-2y

A(y)=(2500-2y)y

[tex]A(y)=2500y-2y^2[/tex]

To maximize A, we first find its derivative

[tex]A'(y)=2500-4y\\$Setting A'=0\\2500-4y=0\\2500=4y\\y=625 meters\\Recall: x=2500-2y\\x=2500-2(625)=1250meters[/tex]

The largest area that can be enclosed(at x=1250m,y=625m) is:

1250 X 625

=781250 Square Meters

Answer:

  x = -2 or +7

Step-by-step explanation:

We presume you want to solve ...

  18^(x^2 +4x +4) = 18^(9x +18)

Equate exponents of the same base and put the quadratic in standard form.

  x^2 +4x +4 = 9x +18

  x^2 -5x -14 = 0

  (x -7)(x +2) = 0

Values of x that make these factors zero are 7 and -2.

The solutions are x=-2 or x = 7.

Answer:

c

Step-by-step explanation:

got it right on edge

Answer:

E(Y) = $0.5

Var(Y)  = 14.25

you should pay the same amount  $0.5

Step-by-step explanation:

E(Y) =  = Σ(YP)

P = probability of each outcomes.

Var(Y) = Σ[tex]Y^{2}[/tex]p − (μ x μ)

E(Y) = (2 x 0.25) +(6 x 0.25) + (0.5 x (-3)) = $0.5

Var(Y) = ([tex]2^{2}\\[/tex]x 0.25) + ([tex]6^{2}[/tex] x 0.25) +([tex]-3^{2}[/tex] x 0.5) - ([tex]0.5^{2}[/tex])

       = 14.5 - 0.25

Var(Y)  = 14.25

for the difference between the payoff and cost of playing to have mean 0, you should pay the same amount  $0.5

The mean of your winnings in this coin tossing game is $0.50 and the variance is $3.75. To break even on average, the cost of playing the game should be $0.50.

To calculate the mean and variance of your winnings in this coin tossing game, we first need to compute the expected value and probabilities of each outcome.

Both coins coming up heads (HH) or tails (TT) each have a probability of 0.25 (1/2 * 1/2), while the coins not matching (HT or TH) has a probability 0.5.

The expected winnings, E(Y), can be calculated using the formula E(Y) = (sum of (value x probability)).

So, E(Y) = $2(0.25) + $6(0.25) + (-$3)(0.50) = $0.50 + $1.50 - $1.50 = $0.50.

The variance can be calculated using the formula Var(Y) = E(Y^2) - [E(Y)]^2.

First, find E(Y^2) = ($2^2x0.25) + ($6^2x0.25) + ((-$3)^2x0.50) = $4. Then, calculate [E(Y)]^2 = ($0.50)^2 = $0.25. Substituting in the formula gives Var(Y) = $4 - $0.25 = $3.75.

If your net winnings are to have mean 0, the cost of playing the game should be equivalent to the expected winnings, which is $0.50.

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Answer:

Step-by-step explanation:

Answer:

19.5 inches or 19.685 inches for length

Step-by-step explanation:

50•0.39. 50÷3

19.5. 19.685

how to convert centimeters to inches:

multiply the centimeter by 0.39. example 10•0.39= 3.9 inches.

If you want centimeters to inches as length, then divide the centimeters by 3. example: 10÷3= 3.3333333333 inches

Answer:

X is a solution of the system.

Step-by-step explanation:

To verify that the vector X is a solution of the given system:

[tex]\frac{dx}{dt}=3x-5y \\\frac{dy}{dt}=5x-7y\\X=\left(\begin{array}{c}1&1\end{array}\right)e^{-2t}[/tex]

Writing the system in the form X'=AX for some coefficient matrix A, one obtains the following.

[tex]X'=\left(\begin{array}{cc}3&-5\\5&-7\end{array}\right)X[/tex]

[tex]For\:X=\left(\begin{array}{c}1&1\end{array}\right)e^{-2t} , X'=\left(\begin{array}{c}-2&-2\end{array}\right)e^{-2t}[/tex]

Similarly:

[tex]AX=\left(\begin{array}{cc}3&-5\\5&-7\end{array}\right)\left(\begin{array}{c}1&1\end{array}\right)e^{-2t}=\left(\begin{array}{c}-2&-2\end{array}\right)e^{-2t}[/tex]

Since the above expressions are equal, [tex]X=\left(\begin{array}{c}1&1\end{array}\right)e^{-2t}[/tex] is a solution of the given system.

Final answer:

To verify that the vector X is a solution of the given system, we can substitute X into the differential equations and check if they hold true.

Explanation:

To verify that the vector X is a solution of the given system, we can substitute X into the differential equations and check if they hold true. The given system of equations is dx/dt = 3x - 5y and dy/dt = 5x - 7y. Let's substitute X = (1, 1, e^(-2t)) into these equations:

dx/dt = 3(1) - 5(1) = -2

dy/dt = 5(1) - 7(1) = -2

Since the values match, we can conclude that X = (1, 1, e^(-2t)) is a solution of the given system.

Answer: 23/111

Step-by-step explanation:

Answer:

23/111

Step-by-step explanation:

Answer:

30

Step-by-step explanation:

Answer:

not satisfactional do not vary

Step-by-step explanation:

The subtraction is one of the four basic arithmetic operations in mathematics. The operation subtraction is used here to determine the number of students enrolled in the Saturday morning class. The number of students is 26.

The operation subtraction represents the process of removing objects from a collection. The minus sign implies the subtraction. We can also describe subtraction as the decreasing or removing physical and abstract quantities.

The subtraction method consists of three parts of numbers, they are minuend, subtrahend and difference. The number in a subtraction sentence from which we subtract another number is the minuend, and after that it is subtrahend and last number is the difference.

Here the total present students = 30

Number of absents = 4

So number of presents = 30 - 4 = 26

Thus the students who are enrolled is 26.

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Your question is incomplete most probably your full question was:

In a class the number of students is actually 30, 4 of them were absent on saturday. A dance teacher asks, "How many students are enrolled in the Saturday?

Answer:

x₂ = 0,59 sec

x₁  = 0,8475 sec

Step-by-step explanation:

h(t) = -16*t² + 23*t + 5

h(t) is the trajectory of the ball, the curve is a parable opens downwards

if we force h(t) = 13 feet, we get;

h(t) =  13

13 =  -16*t² + 23*t + 5   ⇒  -16*t² + 23*t - 8 = 0

or    16*t² - 23*t + 8 = 0

The above expression is a second degree equation, we proceed to solve it for t

x =  [ 23  ±  √529 - 512 ] /32

x = [ 23 ± √17 ] /32

x₁ =  [ 23 + 4,12 ]/32    ⇒  x₁  = 27,12/32    ⇒   x₁  = 0,8475 sec

x₂ =   [ 23 - 4,12 ]/32   ⇒  x₂ = 18,88 /32   ⇒   x₂ = 0,59 sec

Answer:

The confidence interval for the mean is given by the following formula:

[tex]\bar X \pm z_{\alpha/2}\frac{\sigma}{\sqrt{n}}[/tex]   (1)

Or equivalently:

[tex]\bar X \pm ME[/tex]

For this case we have the interval given (3.9, 7.7) and we want to find the margin of error. Using the property of symmetry for a confidence interval we can estimate the margin of error with this formula:

[tex]ME= \frac{Upper -Lower}{2}= \frac{7.7-3.9}{2}= 1.9[/tex]

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

[tex]\bar X[/tex] represent the sample mean for the sample  

[tex]\mu[/tex] population mean (variable of interest)

Solution to the problem

The confidence interval for the mean is given by the following formula:

[tex]\bar X \pm z_{\alpha/2}\frac{\sigma}{\sqrt{n}}[/tex]   (1)

Or equivalently:

[tex]\bar X \pm ME[/tex]

For this case we have the interval given (3.9, 7.7) and we want to find the margin of error. Using the property of symmetry for a confidence interval we can estimate the margin of error with this formula:

[tex]ME= \frac{Upper -Lower}{2}= \frac{7.7-3.9}{2}= 1.9[/tex]

Answer:

5 pints of blue and 2 pints of yellow make 7 pints of green. (5 + 2 = 7)

You must do this four times to get 28 pints of green. (4  · 7 = 28)

So you need 20 (4 · 5) pints of blue and 8 (4 · 2) pints of yellow.

Step-by-step explanation:

Answer:

140 pints

Step-by-step explanation:

Answer:

P(X>4)= 0.624

Step-by-step explanation:

n = 10

p= 0.5 ,q= 1 - p = 0.5

Two fifth of 10 = 2/5 x 10 =4

It means that we have to find probability P(X>4).

P(X>4)= 1 -P(X=0)-P(X=1)-P(X=2)-P(X=3)-P(X=4)

We know that

P(X>4)= 1 -P(X=0)-P(X=1)-P(X=2)-P(X=3)-P(X=4)

P(X>4)= 1 -0.0009 - 0.0097 - 0.043 - 0.117-0.205

The probability that more than 12 of the entry forms will include an order is 0.624.

It is defined as the ratio of the number of favorable outcomes to the total number of outcomes, in other words, the probability is the number that shows the happening of the event.

We have:

The Magazine Mass Marketing Company has received 16 entries in its latest sweepstakes.

The probability that more than 12 of the entry forms will include an order can be found as:

P(X > 4) = 1 - P(X = 0) - P(X = 1) - P(X = 2) - P( X = 3) - P(X = 4)

P(X > 4) = 1 - P(X = 0) - P(X = 1) - P(X = 2) - P(X = 3) - P(X = 4)

P(X > 4) = 1 - 0.0009 - 0.0097 - 0.043 - 0.117 - 0.205

P(X > 4) =  0.624

Thus, the probability that more than 12 of the entry forms will include order is 0.624.

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Answer:

We conclude that the Redwood trees have an average height greater than 240 feet.

Step-by-step explanation:

We are given that a random sample of 47 California Redwood trees was taken and their heights measured. The sample mean average height was 248 feet with a standard deviation of 26 feet.

We have to test the claim that Redwood trees have an average height greater than 240 feet.

Let [tex]\mu[/tex] = mean weight bag filling capacity of machine.

SO, Null Hypothesis, [tex]H_0[/tex] : [tex]\mu[/tex] [tex]\leq[/tex] 240 feet   {means that the Redwood trees have an average height smaller than or equal to 240 feet}

Alternate Hypothesis, [tex]H_A[/tex] : [tex]\mu[/tex] > 240 feet   {means that the Redwood trees have an average height greater than 240 feet}

The test statistics that will be used here is One-sample t test statistics as we don't know about the population standard deviation;

                        T.S.  = [tex]\frac{\bar X -\mu}{\frac{s}{\sqrt{n} } }[/tex]  ~ [tex]t_n_-_1[/tex]

where, [tex]\bar X[/tex] = sample mean average height = 248 feet

             s = sample standard deviation = 26 feet

             n = sample of trees = 47

So, test statistics  =  [tex]\frac{248-240}{\frac{26}{\sqrt{47} } }[/tex]  ~ [tex]t_4_6[/tex]   

                               =  2.109

Now at 5% significance level, the t table gives critical value of 1.6792 at 46 degree of freedom for right-tailed test. Since our test statistics is higher than the critical value of t as 2.109 > 1.6792, so we have sufficient evidence to reject our null hypothesis as it will fall in the rejection region due to which we reject our null hypothesis.

Therefore, we conclude that the Redwood trees have an average height greater than 240 feet.

Answer:

108

Step-by-step explanation:

Answer:

The answer is ten bouquets

Step-by-step explanation:

1 dozen is 12 if you were multiply 12 by 10 you would get 120. Now multiply 10 by twelve instead you get the same answer

Final answer:

When the price of each bouquet increased from $10 to $12, the flower stand sold 10 bouquets to make the same total sales of $120 as the previous day.

Explanation:

The question at hand involves a basic mathematical calculation of how changes in price affect the number of units sold to achieve the same total revenue. If a flower stand sold a dozen bouquets for $10 each yesterday, they made $120 in total sales (12 bouquets  imes $10 per bouquet). Today, with the price increase to $12 per bouquet, we need to find out how many bouquets were sold to still make $120 in total sales. The calculation is as follows:

Total Sales = Number of Bouquets Sold  imes Price per Bouquet

We know that the Total Sales should still be $120, and the Price per Bouquet is now $12. Therefore:

$120 = Number of Bouquets Sold  imes $12

Number of Bouquets Sold = $120 / $12

Number of Bouquets Sold = 10 bouquets

Thus, the flower stand sold 10 bouquets today to make the same amount of money as yesterday after increasing the price to $12 each.

Answer:

The probability that at least 280 of these students are smokers is 0.9664.

Step-by-step explanation:

Let the random variable X be defined as the number of students at a particular college who are smokers

The random variable X follows a Binomial distribution with parameters n = 500 and p = 0.60.

But the sample selected is too large and the probability of success is close to 0.50.

So a Normal approximation to binomial can be applied to approximate the distribution of X if the following conditions are satisfied:

1. np ≥ 10

2. n(1 - p) ≥ 10

Check the conditions as follows:

 [tex]np=500\times 0.60=300>10\\n(1-p)=500\times(1-0.60)=200>10[/tex]

Thus, a Normal approximation to binomial can be applied.

So,  

[tex]X\sim N(\mu=600, \sigma=\sqrt{120})[/tex]

Compute the probability that at least 280 of these students are smokers as follows:

Apply continuity correction:

P (X ≥ 280) = P (X > 280 + 0.50)

                   = P (X > 280.50)

                   [tex]=P(\frac{X-\mu}{\sigma}>\frac{280-300}{\sqrt{120}}\\=P(Z>-1.83)\\=P(Z<1.83)\\=0.96638\\\approx 0.9664[/tex]

*Use a z-table for the probability.

Thus, the probability that at least 280 of these students are smokers is 0.9664.

The probability of a minimum of 280 students being radio listeners would be:

0.9664

The standard deviation is a statistic that measures the dispersion of a dataset relative to its mean and is calculated as the square root of the variance in statistics.

So, the formula is,

[tex]SD=\sqrt{\sigma}[/tex]

Suppose it is known that 60% of students at a college are smokers.

A sample of 500 students from the college is selected at random.

[tex]mean=np\\=500\times 0.6\\=300[/tex]

The standard deviation is,

[tex]S=\sqrt{np(1-p)}\\ =\sqrt{500\times 0.6\times 0.4}\\ =10.95445[/tex]

So, the probability is,

[tex]P(X > 280)=\frac{P(x-mean)}{s} \\=\frac{280-300}{10.95445} \\=0.9664[/tex]

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Answer:

-121

Step-by-step explanation:

[tex] \frac{t}{ - 11} = 11 \\ t = - 11 \times 11 \\ \huge \red{ \boxed{t = - 121}}[/tex]

Answer:

$420/5

$85 a day is Jesse's daily wage

Step-by-step explanation:

Answer:

$84

Step-by-step explanation:

$420 a week divided by 5 days per week gives $84 per day

Answer:

The value of the coefficient of determination is 0.263 or 26.3%.

Step-by-step explanation:

R-squared is a statistical quantity that measures, just how near the values are to the fitted regression line. It is also known as the coefficient of determination.

A high R² value or an R² value approaching 1.0 would indicate a high degree of explanatory power.

The R-squared value is usually taken as “the percentage of dissimilarity in one variable explained by the other variable,” or “the percentage of dissimilarity shared between the two variables.”

The R² value is the square of the correlation coefficient.

The correlation coefficient between heights​ (in inches) and weights​ (in lb) of 40 randomly selected men is:

r = 0.513.

Compute the value of the coefficient of determination as follows:

[tex]R^{2}=(r)^{2}\\=(0.513)^{2}\\=0.263169\\\approx0.263[/tex]

Thus, the value of the coefficient of determination is 0.263 or 26.3%.

This implies that the percentage of variation in the variable height explained by the variable weight is 26.3%.

The probability of winning no more than 8 times out of 10 plays of the arcade game is approximately 0.6394.

Here, we have,

To calculate the probability of winning no more than 8 times out of 10 plays of the arcade game, we need to find the cumulative probability of winning 0, 1, 2, 3, 4, 5, 6, 7, and 8 times.

We'll use the binomial distribution formula to do this:

The binomial distribution formula for a probability of k successes in n trials is given by:

[tex]P(X = k) = C(n, k) * p^k * (1 - p)^{(n - k)[/tex]

where:

P(X = k) is the probability of exactly k successes,

n is the total number of trials (10 in this case),

k is the number of successes we want to calculate the probability for (0 to 8 in this case),

p is the probability of winning (0.632 in this case),

C(n, k) is the binomial coefficient, calculated as

C(n, k) = n! / (k! * (n - k)!).

Now, let's calculate the probabilities for each value of k and then sum them up to find the cumulative probability:

P(X ≤ 8) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6) + P(X = 7) + P(X = 8)

now, we have,

[tex]P(X = 0) = C(10, 0) * (0.632)^0 * (1 - 0.632)^{(10 - 0)}\\P(X = 1) = C(10, 1) * (0.632)^1 * (1 - 0.632)^{(10 - 1)}\\P(X = 2) = C(10, 2) * (0.632)^2 * (1 - 0.632)^{(10 - 2)}\\P(X = 3) = C(10, 3) * (0.632)^3 * (1 - 0.632)^{(10 - 3)}\\P(X = 4) = C(10, 4) * (0.632)^4 * (1 - 0.632)^{(10 - 4)}\\[/tex]

[tex]P(X = 5) = C(10, 5) * (0.632)^5 * (1 - 0.632)^{(10 - 5)}\\P(X = 6) = C(10, 6) * (0.632)^6 * (1 - 0.632)^{(10 - 6)}\\P(X = 7) = C(10, 7) * (0.632)^7 * (1 - 0.632)^{(10 - 7)}\\P(X = 8) = C(10, 8) * (0.632)^8 * (1 - 0.632)^{(10 - 8)}[/tex]

Now, let's calculate each of these probabilities:

[tex]P(X = 0) = 1 * 1 * (0.368)^{10} = 0.0010434068\\P(X = 1) = 10 * 0.632 * (0.368)^9 = 0.0091045617\\P(X = 2) = 45 * 0.632^2 * (0.368)^8 = 0.0333533058\\P(X = 3) = 120 * 0.632^3 * (0.368)^7 = 0.0789130084\\P(X = 4) = 210 * 0.632^4 * (0.368)^6 = 0.1312387057\\P(X = 5) = 252 * 0.632^5 * (0.368)^5 = 0.1552134487\\P(X = 6) = 210 * 0.632^6 * (0.368)^4 = 0.1320219902\\P(X = 7) = 120 * 0.632^7 * (0.368)^3 = 0.0702045343\\P(X = 8) = 45 * 0.632^8 * (0.368)^2 = 0.0272837154\\[/tex]

Now, let's sum up these probabilities:

P(X ≤ 8) ≈ 0.0010434068 + 0.0091045617 + 0.0333533058 + 0.0789130084 + 0.1312387057 + 0.1552134487 + 0.1320219902 + 0.0702045343 + 0.0272837154

≈ 0.6393716762

So, the probability of winning no more than 8 times out of 10 plays of the arcade game is approximately 0.6394.

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Finally, we find:

P(X ≤ 8) ≈ P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6) + P(X = 7) + P(X = 8)

To calculate the probability of winning no more than 8 times out of 10 plays in the arcade game, we can use the binomial distribution.

The binomial distribution calculates the probability of having a certain number of successes (in this case, winning) in a fixed number of independent trials (10 plays) when the probability of success (p) remains constant.

Probability of winning (p) = 0.632

Probability of losing (q, since p + q = 1) = 0.368

Number of trials (n) = 10

To find the probability of winning no more than 8 times, we need to sum the probabilities of winning 0, 1, 2, 3, 4, 5, 6, 7, and 8 times.

P(X ≤ 8) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6) + P(X = 7) + P(X = 8)

Using the binomial probability formula:

P(X = k) = (n choose k) * p^k * q^(n-k)

where (n choose k) is the binomial coefficient, given by:

(n choose k) = n! / (k! * (n-k)!)

Calculating the probabilities for each individual case:

P(X = 0) = (10 choose 0) * (0.632^0) * (0.368^(10-0))

P(X = 1) = (10 choose 1) * (0.632^1) * (0.368^(10-1))

P(X = 2) = (10 choose 2) * (0.632^2) * (0.368^(10-2))

...

P(X = 8) = (10 choose 8) * (0.632^8) * (0.368^(10-8))

After calculating the probabilities for each individual case, we can sum them up to find P(X ≤ 8).

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Answer:

Null Hypothesis, [tex]H_0[/tex] : [tex]\mu[/tex] [tex]\geq[/tex] $13.04  

Alternate Hypothesis, [tex]H_A[/tex] : [tex]\mu[/tex] < $13.04

Step-by-step explanation:

We are given that the mean cost of a flank steak, broccoli, and rice bought at the grocery store is $13.04.

A sample of 100 neighborhood restaurants showed a mean price of $12.65 and a standard deviation of $2 for a comparable restaurant meal.

Let [tex]\mu[/tex] = mean cost of a restaurant meal

So, Null Hypothesis, [tex]H_0[/tex] : [tex]\mu[/tex] [tex]\geq[/tex] $13.04  

Alternate Hypothesis, [tex]H_A[/tex] : [tex]\mu[/tex] < $13.04

Here, null hypothesis states that the mean cost of a restaurant meal is more than or equal to fixing a comparable meal at home.

On the other hand, alternate hypothesis states that the mean cost of a restaurant meal is less than fixing a comparable meal at home.

The test statistics that we can use for conducting this hypothesis would be t test statistics as we don't know about population standard deviation;

                            T.S. = [tex]\frac{\bar X -\mu}{\frac{s}{\sqrt{n} } }[/tex]  ~ [tex]t_n_-_1[/tex]

where, [tex]\bar X[/tex] = sample mean price of a restaurant meal $12.65

             s = sample standard deviation = $2

             n = sample of neighborhood restaurants = 100

The null hypothesis (H0) is that the cost of a restaurant meal is equal to or more than dining in. The alternative hypothesis (Ha) is that the restaurant meal is cheaper. If the test statistic lies in the critical region, we would then reject the null hypothesis in favor of the alternative.

To determine whether the mean cost of a restaurant meal is cheaper than fixing a similar meal at home, we have to consider hypothesis testing. In this case, the null hypothesis (H0) would be that the restaurant meal is equally or more expensive than dining in, and the alternative hypothesis (Ha) would be that the meal at the restaurant is cheaper.

H0: μ ≥ $13.04

Ha: μ < $13.04

Where μ stands for the mean cost of a restaurant meal. This hypotheses are based on a sample of 100 neighborhood restaurants with a mean price of $12.65 and a standard deviation of $2. If we get a test statistic that lies in the critical region, we could then reject the null hypothesis in favor of the alternative, thereby concluding that it is cheaper to eat at a restaurant.

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