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3 years ago

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The distance from New York City to Los Angeles is 4090 kilometers. a. [3 pts]What is the distance in miles? (You must use unit f

ractions. Round to the nearest mile and be sure to include units.) b. [3 pts]If your car averages 31 miles per gallon, how many gallons of gas can you expect to use driving from New York to Los Angeles? (You must use unit fractions. Round to one decimal place and be sure to include units.)

1 answer:

sergejj [24]3 years ago

6 0

Answer: a) 2556.25 miles

b) 82.5 gallons

Step-by-step explanation:

Given : The distance from New York City to Los Angeles is 4090 kilometers.

$=4090\times\dfrac{5}{8}=2556.25\text{ miles}$ [∵ 1 km= $\dfrac{5}{8}$ miles]

Hence, the distance in miles= 2556.25

If your car averages 31 miles per gallon, then the number of gallons of gas required to use driving from New York to Los Angeles = $\dfrac{\text{Distance}}{31}$

$=\dfrac{2556.25}{31}=82.4596774194\approx82.5\text{gallons}$

Hence, the required number of gallons = 82.5 gallons

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Answer:

(A) The mean for the differences is 2.0.

(B) The test statistic is 1.617.

(C) At 90% confidence the null hypothesis should not be rejected.

Step-by-step explanation:

We are given that a random sample of eight cars from each manufacturer is selected, and eight drivers are selected to drive each automobile for a specified distance.

The following data (in miles per gallon) show the results of the test;

Driver Manufacturer A Manufacturer B

1 32 28

2 27 22

3 26 27

4 26 24

5 25 24

6 29 25

7 31 28

8 25 27

Let $\mu_1$ = mean MPG for the fuel efficiency of Manufacturer A brand

$\mu_2$ = mean MPG for the fuel efficiency of Manufacturer B brand

SO, Null Hypothesis, $H_0$ : $\mu_1-\mu_2=0$ or $\mu_1= \mu_2$ {means that there is a not any significant difference in the mean MPG (miles per gallon) when testing for the fuel efficiency of these two brands of automobiles}

Alternate Hypothesis, $H_A$ : $\mu_1-\mu_2\neq 0$ or $\mu_1\neq \mu_2$ {means that there is a significant difference in the mean MPG (miles per gallon) for the fuel efficiency of these two brands of automobiles}

The test statistics that will be used here is <u>Two-sample t test statistics</u> as we don't know about the population standard deviations;

T.S. = $\frac{(\bar X_1-\bar X_2)-(\mu_1-\mu_2)}{s_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2} } }$ ~ $t__n__1+_n__2-2$

where, $\bar X_1$ = sample mean MPG for manufacturer A = $\frac{\sum X_A}{n_A}$ = 27.625

$\bar X_2$ = sample mean MPG for manufacturer B = $\frac{\sum X_B}{n_B}$ = 25.625

$s_1$ = sample standard deviation for manufacturer A = $\sqrt{\frac{\sum (X_A-\bar X_A)^{2} }{n_A-1} }$ = 2.72

$s_2$ = sample standard deviation manufacturer B = $\sqrt{\frac{\sum (X_B-\bar X_B)^{2} }{n_B-1} }$ = 2.20

$n_1$ = sample of cars selected from manufacturer A = 8

$n_2$ = sample of cars selected from manufacturer B = 8

Also, $s_p=\sqrt{\frac{(n_1-1)s_1^{2}+(n_2-1)s_2^{2} }{n_1+n_2-2} }$ = $\sqrt{\frac{(8-1)\times 2.72^{2}+(8-1)\times 2.20^{2} }{8+8-2} }$ = 2.474

(A) The mean for the differences is = 27.625 - 25.625 = 2

(B) <u><em>The test statistics</em></u> = $\frac{(27.625-25.625)-(0)}{2.474 \times \sqrt{\frac{1}{8}+\frac{1}{8} } }$ ~ $t_1_4$

= 1.617

(C) Now at 10% significance level, the t table gives critical values between -1.761 and 1.761 at 14 degree of freedom for two-tailed test. Since our test statistics lies within the range of critical values of t, so we have insufficient evidence to reject our null hypothesis as it will not fall in the rejection region due to which <u>we fail to reject our null hypothesis</u>.

Therefore, we conclude that there is a not any significant difference in the mean MPG (miles per gallon) when testing for the fuel efficiency of these two brands of automobiles.

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Answer:

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From the diagram we can see that the length ad the width of the box are found by using the following expressions:

L=18-2x

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$12(x^{2}-10x+18)=0$

which yields:

x^{2}-10x+18=0

This one can only be solved by using the quadratic formula:

$x=\frac{-b\pm \sqrt{-b^{2}-4ac}}{2a}$

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$x=\frac{-(-10)\pm \sqrt{-(-10)^{2}-4(1)(18)}}{2(1)}$

which yields:

x=7.65in or x=2.35in

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Height=x=2.35

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