Answer:
I think the answer is orange 17
Step-by-step explanation:
Answer:
The open interval of convergence is 1
Step-by-step explanation:
The detailed explanation is given in the attached document.
Answer:
c)No, P is the distance from A to B, and Q is the distance from B to A.
Step-by-step explanation:
Point P partitions the directed segment from A to B into a 1:3 ratio.
Ratio is 1:3
So AP is 1 and PB is 3
Q partitions the directed segment from B to A into a 1:3 ratio.
Ratio is 1:3
So BQ is 1 and QA is 3.
That is AQ= 3 and BQ= 1
The ratio of Q and P varies
AP =1 and AQ=3
So P and Q are not at the same point.
Because P is the distance from A to B and Q is the distance from B to A
There are 10³×26³ = 17,576,000 possible different plates. If any of those is randomly chosen, the probability of picking one in particular is
... 1/17576000 ≈ 0.0000000569
(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.
