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3 years ago

Greg drove 660 miles in 12 hours. At the same rate, how long would it take him to drive 495 miles?

1 answer:

3 0

This one has two parts, first we have to find out the speed to drive 660 miles in 12 hours. Divide 660 by 12 to get the speed of 55 miles per hour, that is your rate for the next part.

Next we need to find out how long it will take to drive 495 miles at the same pace of 55 mph.

Do the same thing 495/55. That will give you your final answer.

495/55= 9

It will take him 9 hours to drive 495 miles at a rate of 55 mph.

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A selective university advertises that 96% of its bachelor’s degree graduates have, on graduation day, a professional job offer

OLEGan [10]

Answer:

The probability is $P( p < 0.9207) = 0.0012556$

Step-by-step explanation:

From the question we are told

The population proportion is $p = 0.96$

The sample size is $n = 227$

The number of graduate who had job is k = 209

Generally given that the sample size is large enough (i.e n > 30) then the mean of this sampling distribution is

$\mu_x = p = 0.96$

Generally the standard deviation of this sampling distribution is

$\sigma = \sqrt{\frac{p (1 - p )}{n} }$

=> $\sigma = \sqrt{\frac{0.96 (1 - 0.96 )}{227} }$

=> $\sigma = 0.0130$

Generally the sample proportion is mathematically represented as

$\^ p = \frac{k}{n}$

=> $\^ p = \frac{209}{227}$

=> $\^ p = 0.9207$

Generally probability of obtaining a sample proportion as low as or lower than this, if the university’s claim is true, is mathematically represented as

$P( p < 0.9207) = P( \frac{\^ p - p }{\sigma } < \frac{0.9207 - 0.96}{0.0130 } )$

$\frac{\^ p - p}{\sigma } = Z (The \ standardized \ value\ of \ \^ p )$

$P( p < 0.9207) = P(Z< -3.022 )$

From the z table the area under the normal curve to the left corresponding to -3.022 is

$P(Z< -3.022 ) = 0.0012556$

=> $P( p < 0.9207) = 0.0012556$

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How do I find the average rate of change?

Alenkasestr [34]

Answer:

add them all up together and then divide by the number of items

Step-by-step explanation:

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