I think the domain is -1 and the range is 2. You may want to double check though.
Answer:
It's 15°
Step-by-step explanation:
BC+60°+85°+200° =360°
BC+345°=360°
BC=360°-345°
Bc=15°
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)
That would be the area of a circle of radius 12 miles
= pi r^2 = 452.4 square miles to nearest tenth.
Plug 4 in for every x
(3*4)-4=8
(4^2)-1=15 [(4^2) is 4*4]
(2*(4^2))-15= 15 [because of PEMDAS the exponent comes first and then multiply by 2 then subtract]
Perimeter is the sum of the lengths so add up all the sides from above
8+15+15=38
