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3 years ago

Janet drove 300 miles in 4.5 hours. Write and equation to find the rate at which she was traveling.

1 answer:

7 0

If you would like to know the rate at which she was travelling, you can calculate this using the following steps:300 miles ... 4.5 hoursx miles = ? ... 1 hour300 * 1 = 4.5 * x300 = 4.5 * xx = 300 / 4.5x = 200 / 3x = 66.67 miles per hour Janet was travelling at the rate 66.67 miles per hour.

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Step-by-step explanation:

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Estimate the product by rounding 3x5,186

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2 years ago

A ferry traveled 1/6 of the distance between two ports in 3/7 hour at this rate what fraction of the distance between the two po

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First, we determine the speed of the ferry by dividing the distance by the time it took to cover that certain distance.
speed = (1/6) / (3/7)
speed = 7/18
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3 years ago

Read 2 more answers

The average amount of time that students use computers at a university computer center is 36 minutes with a standard deviation o

frosja888 [35]

Answer:

2119 students use the computer for more than 40 minutes. This number is higher than the threshold estabilished of 2000, so yes, the computer center should purchase the new computers.

Step-by-step explanation:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

$\mu = 36, \sigma = 5$

The first step to solve this question is finding the proportion of students which use the computer more than 40 minutes, which is 1 subtracted by the pvalue of Z when X = 40. So

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{40 - 36}{5}$

$Z = 0.8$

$Z = 0.8$ has a pvalue of 0.7881.

1 - 0.7881 = 0.2119

So 21.19% of the students use the computer for longer than 40 minutes.

Out of 10000

0.2119*10000 = 2119

2119 students use the computer for more than 40 minutes. This number is higher than the threshold estabilished of 2000, so yes, the computer center should purchase the new computers.

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2 years ago