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hodyreva [135]
1 year ago
14

Adult men have heights with a mean of 69. 0 inches and a standard deviation of 2. 8 inches. find the z-score of a man who is 71.

9 inches tall. (to 4 decimal places)?
Mathematics
1 answer:
Anarel [89]1 year ago
8 0

The value of the z score is 1.03.

According to the statement

we have given that the value of mean and standard deviation and we have to find the value of the z score.

So, For this purpose we know that the

Z-score indicates how much a given value differs from the standard deviation. The Z-score, or standard score, is the number of standard deviations a given data point lies above or below mean.

And the given values are:

mean value = 69 inches

s.d value = 2.8 inches

And the value of x is 71.9 inches.

So, The Z score is

z = x - mean / standard deviation

substitute the values in it then

z = 71.9 - 69 / 2.8

then

z = 2.9 /2.8

z  = 1.03

So, The value of the z score is 1.03.

Learn more about Z Score here

brainly.com/question/25638875

#SPJ4

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3 years ago
The radius of a cone is decreasing at a constant rate of 7 inches per second, and the volume is decreasing at a rate of 948 cubi
inessss [21]

Answer:

The height of cone is decreasing at a rate of 0.085131 inch per second.        

Step-by-step explanation:

We are given the following information in the question:

The radius of a cone is decreasing at a constant rate.

\displaystyle\frac{dr}{dt} = -7\text{ inch per second}

The volume is decreasing at a constant rate.

\displaystyle\frac{dV}{dt} = -948\text{ cubic inch per second}

Instant radius = 99 inch

Instant Volume = 525 cubic inches

We have to find the rate of change of height with respect to time.

Volume of cone =

V = \displaystyle\frac{1}{3}\pi r^2 h

Instant volume =

525 = \displaystyle\frac{1}{3}\pi r^2h = \frac{1}{3}\pi (99)^2h\\\\\text{Instant heigth} = h = \frac{525\times 3}{\pi(99)^2}

Differentiating with respect to t,

\displaystyle\frac{dV}{dt} = \frac{1}{3}\pi \bigg(2r\frac{dr}{dt}h + r^2\frac{dh}{dt}\bigg)

Putting all the values, we get,

\displaystyle\frac{dV}{dt} = \frac{1}{3}\pi \bigg(2r\frac{dr}{dt}h + r^2\frac{dh}{dt}\bigg)\\\\-948 = \frac{1}{3}\pi\bigg(2(99)(-7)(\frac{525\times 3}{\pi(99)^2}) + (99)(99)\frac{dh}{dt}\bigg)\\\\\frac{-948\times 3}{\pi} + \frac{2\times 7\times 525\times 3}{99\times \pi} = (99)^2\frac{dh}{dt}\\\\\frac{1}{(99)^2}\bigg(\frac{-948\times 3}{\pi} + \frac{2\times 7\times 525\times 3}{99\times \pi}\bigg) = \frac{dh}{dt}\\\\\frac{dh}{dt} = -0.085131

Thus, the height of cone is decreasing at a rate of 0.085131 inch per second.

3 0
3 years ago
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