Step-by-step explanation:
The solution to this problem is very much similar to your previous ones, already answered by Sqdancefan.
Given:
mean, mu = 3550 lbs (hope I read the first five correctly, and it's not a six)
standard deviation, sigma = 870 lbs
weights are normally distributed, and assume large samples.
Probability to be estimated between W1=2800 and W2=4500 lbs.
Solution:
We calculate Z-scores for each of the limits in order to estimate probabilities from tables.
For W1 (lower limit),
Z1=(W1-mu)/sigma = (2800 - 3550)/870 = -.862069
From tables, P(Z<Z1) = 0.194325
For W2 (upper limit):
Z2=(W2-mu)/sigma = (4500-3550)/879 = 1.091954
From tables, P(Z<Z2) = 0.862573
Therefore probability that weight is between W1 and W2 is
P( W1 < W < W2 )
= P(Z1 < Z < Z2)
= P(Z<Z2) - P(Z<Z1)
= 0.862573 - 0.194325
= 0.668248
= 0.67 (to the hundredth)
Remember, sin(|x|)=cos(|90-x|)
so sin(|36|)=cos(|90-36|)=cos(|54|)=cos(54)
answer is C
The chance of getting 5 heads is 50% because there are only two sides to a coin hope this helps
Answer:
add both for a total of 11
then 6/11 = .54 = 54%
we know the diameter of the circle is 54, so the radius of it is half that or 27.
![\textit{area of a circle}\\\\ A=\pi r^2~~ \begin{cases} r=radius\\[-0.5em] \hrulefill\\ r=27 \end{cases}\implies A=\pi (27)^2\implies \stackrel{using~\pi =3.14}{A=2289.06~mi^2}](https://tex.z-dn.net/?f=%5Ctextit%7Barea%20of%20a%20circle%7D%5C%5C%5C%5C%20A%3D%5Cpi%20r%5E2~~%20%5Cbegin%7Bcases%7D%20r%3Dradius%5C%5C%5B-0.5em%5D%20%5Chrulefill%5C%5C%20r%3D27%20%5Cend%7Bcases%7D%5Cimplies%20A%3D%5Cpi%20%2827%29%5E2%5Cimplies%20%5Cstackrel%7Busing~%5Cpi%20%3D3.14%7D%7BA%3D2289.06~mi%5E2%7D)
