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Maru [420]
3 years ago
7

The distribution of height of adult american women is approximately normal with a mean of 65.5 inches and a standard deviation o

f 2.5 inches what percentage of women are taller than 70.5 inches?
Mathematics
1 answer:
Blizzard [7]3 years ago
6 0

Answer:

2.275%

Step-by-step explanation:

We start by calculating the z-score

Mathematically;

z-score = (x-mean)SD

here, x = 70.5

mean = 65.5

SD = 2.5

Substituting these values;

z = (70.5-65.5)/2.5

= 5/2.5 = 2

So we want to calculate the probability that;

P( z> 2)

We check the standard normal distribution table for this

That will be

0.02275

In percentage, this is 2.275 %

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Find the volume of the solid generated by revolving the region enclosed by the triangle with vertices (1 comma 0 )​, (3 comma 2
Troyanec [42]

Answer: Volume = \frac{20\pi }{3}

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V = \pi \int\limits^a_b {(R(x))^{2} - (r(x))^{2}} \, dx

For this case, the region generated by the conditions proposed above is shown in the attachment.

Because it is revolting around the y-axis, the formula will be:

V=\pi \int\limits^a_b {(R(y))^{2} - (r(y))^{2}} \, dy

Since it is given points, first find the function for points (3,2) and (1,0):

m = \frac{2-0}{3-1} = 1

y-y_{0} = m(x-x_{0})

y - 0 = 1(x-1)

y = x - 1

As it is rotating around y:

x = y + 1

This is R(y).

r(y) = 1, the lower limit of the region.

The volume will be calculated as:

V = \pi \int\limits^2_0 {[(y+1)^{2} - 1^{2}]} \, dy

V = \pi \int\limits^2_0 {y^{2}+2y+1 - 1} \, dy

V=\pi \int\limits^2_0 {y^{2}+2y} \, dy

V=\pi(\frac{y^{3}}{3}+y^{2} )

V=\pi (\frac{2^{3}}{3}+2^{2} - 0)

V=\frac{20\pi }{3}

The volume of the region bounded by the points is \frac{20\pi }{3}.

6 0
4 years ago
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Answer:

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Step-by-step explanation:

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