Assume that women's heights are normally distributed with a mean given by mu equals 63.4 in, and a standard deviation given by sigma equals 2.7 in. (a) if 1 woman is randomly selected, find the probability that her height is less than 64 in. (b) if 48 women are randomly selected, find the probability that they have a mean height less than 64 in.

0.9895

Step-by-step explanation:

From the mean to the value of interest:  

63.0

−

62.4

=

0.6

inches

So the count of standard deviations beyond the  

mean of 63 is:  

0.6

2.8

σ

So the all encompassing probability from the left for  

μ

+

6

26

σ

=

0.9895

Used 'Elementary Statistics Tables by Henry R Neave

(if you like, you can add your values to this website http://davidmlane.com/hyperstat/z_table.html)

problem:

f(x) = - 2/3x + 5; f(5/2)

substitute 5/2 in for x:

f(5/2) = -2/3(5/2) + 5

simplify the right side:

f(5/2) = -5/3 + 5

f(5/2) = 10/3

f(5/2) = 10/3

i hope this !

the answer is b: 18

step-by-step explanation:

use the pythagorean theorem to .

a^2+b^2=c^2

plug 18 for "r"

the 2 legs are 18 and 24

18^2+24^2=c^2

324+576=c^2

900=c^2

c=30

30-12=18

18 is the radius

: )