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agasfer [191]
3 years ago
14

For the 405 highway that car pass through a checkpoint, assume the speeds are normally distributed such that μ= 61 miles per hou

r and δ=4 miles per hour. Calculate the Z value for the next car that passes through the checkpoint will be traveling slower than 65 miles per hour.
Mathematics
1 answer:
Strike441 [17]3 years ago
4 0

Answer:

Z = 1

Step-by-step explanation:

Z - score

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 = 61, \sigma = 4

Calculate the Z value for the next car that passes through the checkpoint will be traveling slower than 65 miles per hour.

This is Z when X = 65. So

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

Z = \frac{65 - 61}{4}

Z = 1

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Step-by-step explanation:

8 0
3 years ago
Dose someone know how to figure this out can someone pleas help its a test
riadik2000 [5.3K]

Complete question is;

The levels of mercury in two different bodies of water are rising. In one body of water the initial measure of mercury is 0.05 parts per billion (ppb) and is rising at a rate of 0.1 ppb each year. In the second body of water the initial measure is 0.12 ppb and the rate of increase is 0.06 ppb each year.

Which equation can be used to find y, the year in which both bodies of water have the same amount of mercury?

A) 0.05 – 0.1y = 0.12 – 0.06y

B) 0.05y + 0.1 = 0.12y + 0.06

C) 0.05 + 0.1y = 0.12 + 0.06y

D) 0.05y – 0.1 = 0.12y – 0.06

Answer:

Option C: 0.05 + 0.1y = 0.12 + 0.06y

Step-by-step explanation:

In the first body, the initial measure of mercury is 0.05 parts per billion (ppb) while it's rising at a rate of 0.1 ppb each year. We are told to use y for the number of years.

Thus, amount of mercury for y years in this first body is;

A1 = 0.05 + 0.1y

Now, for the second body, we are told that;the initial measure is 0.12 ppb and the rate of increase is 0.06 ppb each year.

Thus, amount of mercury for y years in this second body is;

A2 = 0.12 + 0.06y

Since we want to find the year in which both bodies of water have the same amount of mercury. Thus, it means;

A1 = A2

Thus;

0.05 + 0.1y = 0.12 + 0.06y

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Step-by-step explanation:

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