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Crazy boy [7]
3 years ago
8

The distribution of heights of adult American men is approximately Normal with mean 69 inches and standard deviation 2.5 inches.

What percent of men are between 64 and 66.5 inches tall?
answer choices
A. 50%B. 2.5%C. 13.5%D. 34%
Mathematics
1 answer:
nalin [4]3 years ago
5 0

Answer:

option C 13.5%

Step-by-step explanation:

As the heights of adults is normally distributed with mean=69 and standard deviation=2.5 so, the percent of men that are between 64 and 66.5 inches tall can be calculated as

P(64<X<66.5)=P[ (64-69)/2.5<(X-μ)/σ<(66.5-69)/2.5]

P(64<X<66.5)=P(-2<Z<-1)

P(64<X<66.5)=P(-2<Z<0)-P(-1<Z<0)

P(64<X<66.5)=0.4772-0.3413=0.1359

Thus, the percent of men are between 64 and 66.5 inches tall is 13.59%.

If we round the resultant quantity then it will be rounded to 13.6% but considering the given options, option C is most appropriate.

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Please help quick Solve for x.​
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Answer:

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Step-by-step explanation:

So, lets start with what we know.

Line = 180 degrees

Triangle = 180 degrees, a + b + c = 180

So, we can form the equations,

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-17x + 1                -17x + 1

             <CDB = 181 - 17x

and

84 +7x + 5 + <CDB = 180

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-89                              -89 + 7x

                     <CDB = 91 + 7x

Here, we are given 2 values of <CDB, but they must be the same, so let's substitute them into each other

181 - 17x = 91 + 7x

-91 +17x   -91  +17x

90         =        24x

90 / 24  =        24x / 24

3.75       =           x

x            =           3.75

There, you have it, x is 3.75

If you have any questions, please let me know in the comments section of this answer! If you could mark this answer as the brainliest, I would greatly appreciate it!

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