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)
<u>40 is shown 3 times, 41 is shown 3 times, 42 is shown 3 times, and 43 is shown 3 times. </u>
I would put as the answer that <u>all the numbers are equally shown the same amount of times. </u>
Answer:
x = 10.8
Step-by-step explanation:
cos 26° = x/12
0.8988 = x/12
x = 10.785
Answer:
-2x^2+5x+12
Step-by-step explanation:
(4-x)(3+2x)
12-3x+8x-2x^2
12+5x-2x^2
-2x^2+5x+12
