V=hpir^2
r=2
h=5
pi≈3.141592
v=5*3.141592*2^2
v=5*3.141592*4
v=20*3.141592
v=62.83185307179586476925286766559
round to tenth
62.8 cubic units
Well what I would do is split the 40% into 20% so from 40% to 20% that is /2. So 32/2=16 so 20% of a number is 16, we know there is 5, 20% in 100% so multiply 16 and 5 which gives you 80. Now 25% of 80 is the same as 80*.25 or 80/4 which is 20
Your Answer 20
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Well, as you can see from the rectangle RT and SW should have equal lengths. So to find the value of x, we need to do.....
4x + 10 = 5x - 20
-x + 10 = -20 (Subtraction property of equality)
-x = -30 (Subtraction property of equality)
x = 30 (Division property of equality)
To check our work:
4(30)+10 = 130
5(30)-20 = 130
So, the value of x is 30!
(a) 0.059582148 probability of exactly 3 defective out of 20
(b) 0.98598125 probability that at least 5 need to be tested to find 2 defective.
(a) For exactly 3 defective computers, we need to find the calculate the probability of 3 defective computers with 17 good computers, and then multiply by the number of ways we could arrange those computers. So
0.05^3 * (1 - 0.05)^(20-3) * 20! / (3!(20-3)!)
= 0.05^3 * 0.95^17 * 20! / (3!17!)
= 0.05^3 * 0.95^17 * 20*19*18*17! / (3!17!)
= 0.05^3 * 0.95^17 * 20*19*18 / (1*2*3)
= 0.05^3 * 0.95^17 * 20*19*(2*3*3) / (2*3)
= 0.05^3 * 0.95^17 * 20*19*3
= 0.000125* 0.418120335 * 1140
= 0.059582148
(b) For this problem, let's recast the problem into "What's the probability of having only 0 or 1 defective computers out of 4?" After all, if at most 1 defective computers have been found, then a fifth computer would need to be tested in order to attempt to find another defective computer. So the probability of getting 0 defective computers out of 4 is (1-0.05)^4 = 0.95^4 = 0.81450625.
The probability of getting exactly 1 defective computer out of 4 is 0.05*(1-0.05)^3*4!/(1!(4-1)!)
= 0.05*0.95^3*24/(1!3!)
= 0.05*0.857375*24/6
= 0.171475
So the probability of getting only 0 or 1 defective computers out of the 1st 4 is 0.81450625 + 0.171475 = 0.98598125 which is also the probability that at least 5 computers need to be tested.
