Answer:
We are given that According to government data, 75% of employed women have never been married.
So, Probability of success = 0.75
So, Probability of failure = 1-0.75 = 0.25
If 15 employed women are randomly selected:
a. What is the probability that exactly 2 of them have never been married?
We will use binomial
Formula : 
At x = 2
b. That at most 2 of them have never been married?
At most two means at x = 0 ,1 , 2
So, 
c. That at least 13 of them have been married?
P(x=13)+P(x=14)+P(x=15)
Okay so, togo flies 30 miles in 3/5 of an hour with the wind.togo flies 30 miles in 5/6 of an hour against the wind. his rate of speed with the wind was 30 / (3/5) = 30*5/3 = 150/3 = 50 mph.his rate of speed against the wind was 30 / (5/6) = 30*6/5 = 6*6 = 36 mph.3/5 of an hour times 50 mph = 30 miles.
5/6 of an hour times 36 mph = 30 miles.mph looks good.
going he was with the wind.
his rate of speed then was the airplane speed plus the wind speed.coming back he was against the wind.his rate of speed then was the airplane speed minus the wind speed.-----50 mph = A + W36 mph = A - W-----looks like A + W - (A - W) should equal 50 - 36.removing parentheses, equation becomesA + W - A + W = 14.
combining like terms, the equation becomes2*W = 14.dividing both sides by 2, equation becomesW = 7.if W = 7, then A can be found as follows:50 = A + 7A = 43.likewise,36 = A - 7A = 43.since the airplane speed is the same, it looks like the answer is good.solving in the original equations, we get3/5 * (43 + 7) = 303/5 * 50- = 3030 = 30good.-----5/6 * (43 - 7) = 305/6 * 36 = 3030 = 30good.
answer is W = 7 and A = 43.
10t = b - 4
12b+8t = $348
This is a system of equations. I’ll be solving through substitution.
In the first equation. solving for b (the easier variable to isolate) gives you:
b = 10t + 4
Substitute this into the second equation:
12(10t+4) +8t = 348
120t+48+8t = 348
128t = 300
t = 2.34375 —> round it to the nearest cent to get 2.34 dollars
b = 10t+4
b = 10(2.34)+4
b = 27.4 dollars
The area of an equilateral triangle of side "s" is s^2*sqrt(3)/4. So the volume of the slices in your problem is
(x - x^2)^2 * sqrt(3)/4.
Integrating from x = 0 to x = 1, we have
[(1/3)x^3 - (1/2)x^4 + (1/5)x^5]*sqrt(3)/4
= (1/30)*sqrt(3)/4 = sqrt(3)/120 = about 0.0144.
Since this seems quite small, it makes sense to ask what the base area might be...integral from 0 to 1 of (x - x^2) dx = (1/2) - (1/3) = 1/6. Yes, OK, the max height of the triangles occurs where x - x^2 = 1/4, and most of the triangles are quite a bit shorter...
