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)
Answer:
GFD and jik...a pair of angles on the outer side
Answer:
62.92
Step-by-step explanation:
21 percent of 52$ is 10.92 then add em together hope it helps
The word "times" means you have to multiply.
2 1/5 times 1 3/4 is 3 17/20.
The phrase "how many more" means you have to subtract.
3 17/20 minus 2 1/5 is 1 13/20.
Carson walked her dog 1 13/20 more miles than Claire did.
Sorry for the lazy explanations Imao
