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)
D=51/77
First set up the equation equal to zero(0), the distribute from there.
Answer:
The coordinates of the new vertices are:
K'(4,4), L'(0,- 2), and M'(-6,4).
Step-by-step explanation:
In order to dilate a shape by 2, you can simply multiply the coordinates of the orignal triangle each by 2 to find the new coordinates of the new vertices.
K(2,2), L(0,- 1), and M(-3,2)
K'(2*2,2*2), L'(0*2,- 1*2), and M'(-3*2,2*2).
Now become...
K'(4,4), L'(0,- 2), and M'(-6,4).
You can now graph ∆K'L'M' using these coordinates.
Answer:
33.5 km
Step-by-step explanation:
If 1cm = 5km, then you multiply 6.7 by 5 to get distance, which would be 33.5 km.
(hope this helps :P)
Answer:
A) 0 and 13
Step-by-step explanation:
64 - 37 = 27
27/2 = 13.5
13 teachers can teach 3 classes at max
