Answer:
The answer is 72
Step-by-step explanation:
Hope this helps :))
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)
The complete question in the attached figure
we know that
sin40° = opposite side / hypotenuse
opposite side =AC
hypotenuse = AB-------> 10 in
so
sin 40=AC/AB---------> AC=AB*sin 40-------> AC=10*sin 40
the answer is
AC=10*sin 40
Answer:
x=−7/5and y=−6/5
Step-by-step explanation:
Step: Substitute−2x−4foryiny=3x+3:
y=3x+3
−2x−4=3x+3
−2x−4+−3x=3x+3+−3x(Add -3x to both sides)
−5x−4=3
−5x−4+4=3+4(Add 4 to both sides)
−5x=7
−5x
−5
=
7
−5
(Divide both sides by -5)
x=−7/5
