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:
(15,18)
Step-by-step explanation:
write two equations with the information provided.
2x + 3y = 84
x + 4y = 87
use the property of substitution to answer.
x + 4y = 87, x = 87 - 4y
2(87 - 4y) + 3y = 84
174 - 8y + 3y = 84
174 - 84 = 8y - 3y
90 = 5y
90/5 = y
y = 18
Add the value of Y to an original equation. Solve for X
2x + 3(18) = 84
2x + 54 = 84
2x = 84 - 54
2x = 30
x = 30/2
x = 15
Answer:
Idk how to explain but I think its C
Please don't delete my answer unless its wrong to those who love deleteing answers
Step-by-step explanation:
