Answer:
put 0 first
Step-by-step explanation:
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)
Since the shirt is 25 percent off your paying for 75 percent of the shirt. 0.75(36) is 27. The shirt will cost 27.00
5/8 = 0.625 = 62.5%
If 62.5% were glazed then you must find 62.5% of 24.
Equation…
x=0.625(24) = 15
if each costs $0.84 cents plus tax then you will multiply by 15.
15 x .84 = 12.6
The total cost was $12.60
