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3 years ago

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The results of a recent survey of the students at Rockdale Middle School show that 17 out of every 22 students enjoy the cafeter

ia’s cheeseburgers, "5" /"6" of the students enjoy the cafeteria’s pizza, and 80.5% enjoy the cafeteria’s salad bar. Based upon this survey, which of these three foods is most enjoyed?

1 answer:

madam [21]3 years ago

5 0

Answer:

Pizza is the most enjoyed food with 83.3%

Step-by-step explanation:

Given

Cheeseburgers = 17 out of 22 students

Pizza = 5/6

Salad Bar = 80.5%

Required

Which of the food is most enjoyed

First, we have to convert all goven parameters to percentage

$Cheeseburgers = \frac{17}{22}$

$Cheeseburgers = 77.3\%$

$Pizza = \frac{5}{6}$

$Pizza = 83.3\%$

Comparing the percentages;

Pizza is the most enjoyed food with 83.3%

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Canadians who visit the United States often buy liquor and cigarettes, which are much cheaper in the United States. However, the

fenix001 [56]

Answer:

(a): Marginal pmf of x

$P(0) = 0.72$

$P(1) = 0.28$

(b): Marginal pmf of y

$P(0) = 0.81$

$P(1) = 0.19$

(c): Mean and Variance of x

$E(x) = 0.28$

$Var(x) = 0.2016$

(d): Mean and Variance of y

$E(y) = 0.19$

$Var(y) = 0.1539$

(e): The covariance and the coefficient of correlation

$Cov(x,y) = 0.0468$

$r \approx 0.2657$

Step-by-step explanation:

Given

<em>x = bottles</em>

<em>y = carton</em>

<em>See attachment for complete question</em>

<em />

Solving (a): Marginal pmf of x

This is calculated as:

$P(x) = \sum\limits^{}_y\ P(x,y)$

So:

$P(0) = P(0,0) + P(0,1)$

$P(0) = 0.63 + 0.09$

$P(0) = 0.72$

$P(1) = P(1,0) + P(1,1)$

$P(1) = 0.18 + 0.10$

$P(1) = 0.28$

Solving (b): Marginal pmf of y

This is calculated as:

$P(y) = \sum\limits^{}_x\ P(x,y)$

So:

$P(0) = P(0,0) + P(1,0)$

$P(0) = 0.63 + 0.18$

$P(0) = 0.81$

$P(1) = P(0,1) + P(1,1)$

$P(1) = 0.09 + 0.10$

$P(1) = 0.19$

Solving (c): Mean and Variance of x

Mean is calculated as:

$E(x) = \sum( x * P(x))$

So, we have:

$E(x) = 0 * P(0) + 1 * P(1)$

$E(x) = 0 * 0.72 + 1 * 0.28$

$E(x) = 0 + 0.28$

$E(x) = 0.28$

Variance is calculated as:

$Var(x) = E(x^2) - (E(x))^2$

Calculate $E(x^2)$

$E(x^2) = \sum( x^2 * P(x))$

$E(x^2) = 0^2 * 0.72 + 1^2 * 0.28$

$E(x^2) = 0 + 0.28$

$E(x^2) = 0.28$

So:

$Var(x) = E(x^2) - (E(x))^2$

$Var(x) = 0.28 - 0.28^2$

$Var(x) = 0.28 - 0.0784$

$Var(x) = 0.2016$

Solving (d): Mean and Variance of y

Mean is calculated as:

$E(y) = \sum(y * P(y))$

So, we have:

$E(y) = 0 * P(0) + 1 * P(1)$

$E(y) = 0 * 0.81 + 1 * 0.19$

$E(y) = 0+0.19$

$E(y) = 0.19$

Variance is calculated as:

$Var(y) = E(y^2) - (E(y))^2$

Calculate $E(y^2)$

$E(y^2) = \sum(y^2 * P(y))$

$E(y^2) = 0^2 * 0.81 + 1^2 * 0.19$

$E(y^2) = 0 + 0.19$

$E(y^2) = 0.19$

So:

$Var(y) = E(y^2) - (E(y))^2$

$Var(y) = 0.19 - 0.19^2$

$Var(y) = 0.19 - 0.0361$

$Var(y) = 0.1539$

Solving (e): The covariance and the coefficient of correlation

Covariance is calculated as:

$COV(x,y) = E(xy) - E(x) * E(y)$

Calculate E(xy)

$E(xy) = \sum (xy * P(xy))$

This gives:

$E(xy) = x_0y_0 * P(0,0) + x_1y_0 * P(1,0) +x_0y_1 * P(0,1) + x_1y_1 * P(1,1)$

$E(xy) = 0*0 * 0.63 + 1*0 * 0.18 +0*1 * 0.09 + 1*1 * 0.1$

$E(xy) = 0+0+0 + 0.1$

$E(xy) = 0.1$

So:

$COV(x,y) = E(xy) - E(x) * E(y)$

$Cov(x,y) = 0.1 - 0.28 * 0.19$

$Cov(x,y) = 0.1 - 0.0532$

$Cov(x,y) = 0.0468$

The coefficient of correlation is then calculated as:

$r = \frac{Cov(x,y)}{\sqrt{Var(x) * Var(y)}}$

$r = \frac{0.0468}{\sqrt{0.2016 * 0.1539}}$

$r = \frac{0.0468}{\sqrt{0.03102624}}$

$r = \frac{0.0468}{0.17614266944}$

$r = 0.26569371378$

$r \approx 0.2657$ --- approximated

8 0

3 years ago

allochka39001 [22]

Answer:

50 + 200h

Step-by-step explanation:

4 0

3 years ago

Write a compound inequality to represent all of the numbers between -4 and 6.

svetlana [45]

-4<x<6 would be the compound inequality

6 0

3 years ago

Please answer right away

Andreas93 [3]

Answer:

20 miles with an error margin of ± 8 miles

Step-by-step explanation:

The margin of error of a result is the range in which an error can vary. To find the margin of error between both distances we have to

28-12 = 16, that is, the variation of the result has a range of 16 miles. So we will look for the midpoint of both distances

(X2-X1)/2+X1=(28-12)/2+12=16/2+12=8+12=20

So from this midpoint the value can vary between 8 points below and 8 points above that would cover the difference of 16 miles that we observed at the beginning

In this way, the correct answer is 20 miles with an error margin of ± 8 miles

Done

8 0

3 years ago

Leah's age is twice her cow old are Leah and her cousin?usin's age. If they add their ages together they get 36. How old are Lea

nignag [31]

Okay so lets call Leah "L" and her cousin "C". We know that L+C=36 ... we also know that Leah is twice her cousins age. Therefore, L=2 times C, or L=2C. This is because Leah's age is equivalent to twice as much as her cousin's.
Now that you know that L=2C, you can plug this back into the equation. This should make it so that's there's only one variable now!
L+C=36
(2C)+C=36 ... here we subbed in L=2C
3C=36 ... we add up the C's
C=12 ... we isolate for C by dividing both sides by 3

So her cousin's age is 12 years old. Leah's age is twice that. Thus, she's 24. If you add the two up: 12+24, you indeed get 36. Hope that helps :))

4 0

4 years ago