Step-by-step explanation:
L=2B
P=2(L+B)
48=2(2B+B)
48=2(3B)
48=6B
B=8yd
L=2*8
=16yd
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:
Step-by-step explanation:
8.3g+2.61=32.49
8.3g=29.88
g=3.6
The question asks per week. So 1.5 cups times 7 days is 10.5 cups.
The pipe leaks 10.5 cups every week.
Answer:
D
Step-by-step explanation:
4 + 12 + 18 + 68= 102/2 51
40 + 28 + 52 + 76= 196/2 98
51:98
