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:
1. y = 11
x = -48
2. a =8
b= -10
3. x= 1
y= 2
Step-by-step explanation:
1 . 6x + 5y = 7
x-y= 3
6x+5y = 7
6x-6y =18
-1y = - 11
6x + 5y = 7
y = 11
6x + 55 = 7
6x = -48
x = -48
Probablity=desiredoutcomes/totalpossibleoutcomes
total possible outcomes=1+2+4+3=10
desired outcomes=4 blue
probablity=4/10=2/5=40%
Answer:
2/3, 12/18 and 18/27
Step-by-step explanation:
Answer:
The answer is 221/2 or 73 ⅔
Step-by-step explanation:
5 ⅔ = (3 × 5 + 2) ÷ 3 = 17/3
Now,
17/3 × 13 = 221/2 = 73 ⅔
Thus, The answer is 221/2 or 73 ⅔
