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Nutka1998 [239]
3 years ago
7

Jack drove to work in the morning at an average speed of 45 miles per hour. He returned home in the evening along the same route

and averaged 30 miles per hour. If Jack spent a total of one hour commuting to and from work, how many miles did he drive to work in the morning?
Mathematics
1 answer:
Dmitry [639]3 years ago
8 0

Answer:

Step-by-step explanation:

Let's say that the time it took him to get to work in the morning is  t hours. Then the time it took him to get home in the afternoon must be 1 - t hours. We know that for any trip, distance equals rate times time or d = rt . That means that the distance he drove to work is given by , but we also know that the distance he drove to get home must be the same distance, because he took the same route (and, presumably, no one picked up him house and moved it while he was at work) so for the trip home we can say  d = 30 × (1 - t) and since the distances are equal, we can say: 

45t = 30 × (1 - t)

45t = 30 - 30t

45t + 30t = 30

75t = 30

t = 30/75

t = 2 /5 hour to drive to work at 45mph

Since ,  d = rt

d = 45 ×(2/5) = 18miles

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3 years ago
In triangle STU, u2 = s2 + t2. Triangle STU has sides s, t, u opposite to the corresponding vertices S, T, U Which equation is t
wlad13 [49]
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<span>The measure of angle SUT is equal to 90 degrees
</span>
Explanation:
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6 0
4 years ago
Read 2 more answers
Assume that a sample is used to estimate a population proportion p. Find the 98% confidence interval for a
Vikki [24]

Answer:

0.81 - 2.326 \sqrt{\frac{0.81(1-0.81)}{131}}=0.730

0.81 + 2.326 \sqrt{\frac{0.81(1-0.81)}{131}}=0.890

And the confidence interval would be:

0.730 \leq \p \leq 0.890

Step-by-step explanation:

Information given:

n=131 represent the sample size

\hat p=0.81 represent the estimated proportion

The confidence interval would be given by this formula

\hat p \pm z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}

For the 98% confidence interval the value of \alpha=1-0.98=0.02 and \alpha/2=0.01, with that value we can find the quantile required for the interval in the normal standard distribution.

z_{\alpha/2}=2.326

And replacing into the confidence interval formula we got:

0.81 - 2.326 \sqrt{\frac{0.81(1-0.81)}{131}}=0.730

0.81 + 2.326 \sqrt{\frac{0.81(1-0.81)}{131}}=0.890

And the confidence interval would be:

0.730 \leq \p \leq 0.890

6 0
4 years ago
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