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Montano1993 [528]
3 years ago
13

Al, Pablo, and Marsha shared the driving on a 1,500-mile trip. Which of the three drove the greatest distance on the trip?(1) Al

drove 1 hour longer than Pablo but at an average rate of 5 miles per hour slower than Pablo.(2) Marsha drove 9 hours and averaged 50 miles per hour.
Mathematics
1 answer:
Svetach [21]3 years ago
6 0

Answer:

Pablo drove the greatest distance of 527 miles.

Missing Content:

Since information on Pablo is not given, it is impossible to solve this problem. Let us assume some scenario for Pablo. Say:

(3)Pablo averaged 52.7 miles per hour and drove for 10 hours.

Step-by-step explanation:

Using this information, we can find out the data on Al,

Al drove 1 hour longer than Pablo so,

Time Al drove= 10+1 =11 hours

Average rate of Al=Pablo's average rate-5= 52.7-5 = 47.7 miles per hour.

We know that,

Average Speed = \frac{Distance}{Time}

Distance =Average Speed x Time

Distance travelled by Al = 47.7 x 11 =524.7 miles

Distance Travelled by Marsha = 50 x 9=450 miles

Distance Travelled by Pablo= 52.7 x 10= 527 miles

<u>As we can see that Pablo drove the greatest distance which is 527 miles.</u>

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Step-by-step explanation:

Information provided

n = 10 sample size

s= 1.186 the sample deviation

\sigma_o =1.34 the value that we want to test

p_v represent the p value for the test

t represent the statistic  (chi square test)

\alpha=0.01 significance level

Part a

On this case we want to test if the true deviation is 1,34 or no, so the system of hypothesis are:

H0: \sigma = 1.34

H1: \sigma \neq 1.34

The statistic is given by:

t=(n-1) [\frac{s}{\sigma_o}]^2

Part b

The degrees of freedom are given by:

df = n-1= 10-1=9

And the critical values with \alpha/2=0.005 on each tail are:

\chi_{\alpha/2}= 1.735, \chi_{1-\alpha/2}= 23.589

Part c

Replacing the info we got:

t=(10-1) [\frac{1.186}{1.34}]^2 =7.05

Part d

For this case since the critical value is not higher or lower than the critical values we have enough evidence to FAIL to reject the null hypothesis and we can conclude that the true deviation is not significantly different from 1.34

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