OOOOH my favorite LONG DIVISION but anyways the answer is 164.4
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:
B. y = 2x
Step-by-step explanation:
every y value is the value of x multiplied by 2
so that means y=2x
hope this helps luv <3
Answer:
Angle of A = 90 degree-62 degree = 28 degree.
Step-by-step explanation:
tan (62) = opposite / adjacent = 10 / a ---> a = 10/tan (62) = 10/ 1.88 = 5.319
cos (62) = a/c --->c = a/cos (62) = 5.319 / 0.47 = 5.3 / 0.47 = 11.3
or another way.
sin (62) = 10 /c ---> c = 10/ sin (62) = 10 / 0.88 = 11.3
So Angle of A = 28 degree.
a = 5. 3
c = 11.3.
please give me brainliest. I hope it help.
