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Lera25 [3.4K]
2 years ago
11

Jill's bowling scores are approximately normally distributed with mean 170 and standard deviation 25, while jack's scores are ap

proximately normally distributed with mean 160 and standard deviation 10. if jack and jill each bowl one game, then assuming their scores are independent random variables, find the probability that
Mathematics
1 answer:
olya-2409 [2.1K]2 years ago
5 0
Jill's bowling scores<span> are </span>approximately normally distributed<span> with </span>mean 170<span> and </span>standard deviation<span> 20,</span>while Jack's scores<span> are ... </span>If Jack<span> and </span>Jill each bowl one game<span>, </span>then assuming<span> that </span>their scores<span> are ... The difference of </span>means<span> of </span>Jack<span> and </span>Jill<span> follows a </span>normal distribution<span> with </span>mean<span> -</span>10<span> and </span>standard deviation<span> is: </span>25<span>.</span>
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50% of 24.98 = $12.49

24.98 + 12.49 = 37.47

187.34 / 37.47 = 5

5 x 2 hats = 10 hats

He bought 10 hats.

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Which pairs of angles are alternate interior angles? Select all that apply.
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9 and 10 i dont know why they have us doing this confusing  stuff

Step-by-step explanation:

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Which graph represents 9x2 – 16y2 – 54x + 64y – 127 = 0?
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Right now I'm wondering what kind of class you are taking to get such a weird graph, but I graphed it for you on the screenie.

Step-by-step explanation:

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Jensen Tire &amp; Auto is in the process of deciding whether to purchase a maintenance contract for its new computer wheel align
sasho [114]

Answer:

Step-by-step explanation:

Hello!

The given data corresponds to the variables

Y:  Annual Maintenance  Expense ($100s)

X: Weekly Usage  (hours)

n= 10

∑X= 253; ∑X²= 7347; \frac{}{X}= ∑X/n= 253/10= 25.3 Hours

∑Y= 346.50; ∑Y²= 13010.75; \frac{}{Y}= ∑Y/n= 346.50/10= 34.65 $100s

∑XY= 9668.5

a)

To estimate the slope and y-intercept you have to apply the following formulas:

b= \frac{sumXY-\frac{(sumX)(sumY)}{n} }{sumX^2-\frac{(sumX)^2}{n} } = \frac{9668.5-\frac{253*346.5}{10} }{7347-\frac{(253)^2}{10} }= 0.95

a= \frac{}{Y} -b\frac{}{X} = 34.65-0.95*25.3= 10.53

^Y= a + bX

^Y= 10.53 + 0.95X

b)

H₀: β = 0

H₁: β ≠ 0

α:0.05

F= \frac{MS_{Reg}}{MS_{Error}} ~~F_{Df_{Reg}; Df_{Error}}

F= 47.62

p-value: 0.0001

To decide using the p-value you have to compare it against the level of significance:

If p-value ≤ α, reject the null hypothesis.

If p-value > α, do not reject the null hypothesis.

The decision is to reject the null hypothesis.

At a 5% significance level you can conclude that the average annual maintenance expense of the computer wheel alignment and balancing machine is modified when the weekly usage increases one hour.

b= 0.95 $100s/hours is the variation of the estimated average annual maintenance expense of the computer wheel alignment and balancing machine is modified when the weekly usage increases one hour.

a= 10.53 $ 100s is the value of the average annual maintenance expense of the computer wheel alignment and balancing machine when the weekly usage is zero.

c)

The value that determines the % of the variability of the dependent variable that is explained by the response variable is the coefficient of determination. You can calculate it manually using the formula:

R^2 = \frac{b^2[sumX^2-\frac{(sumX)^2}{n} ]}{[sumY^2-\frac{(sumY)^2}{n} ]} = \frac{0.95^2[7347-\frac{(253)^2}{10} ]}{[13010.75-\frac{(346.50)^2}{10} ]} = 0.86

This means that 86% of the variability of the annual maintenance expense of the computer wheel alignment and balancing machine is explained by the weekly usage under the estimated model ^Y= 10.53 + 0.95X

d)

Without usage, you'd expect the annual maintenance expense to be $1053

If used 100 hours weekly the expected maintenance expense will be 10.53+0.95*100= 105.53 $100s⇒ $10553

If used 1000 hours weekly the expected maintenance expense will be $96053

It is recommendable to purchase the contract only if the weekly usage of the computer is greater than 100 hours weekly.

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