(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.
Answer:
Option (a) is correct.
The system of equation becomes

Step-by-step explanation:
Given : Equation 
We have to construct a system of equations that can be used to find the roots of the equation 
Consider the given equation 
To construct a system of equation put both sides of the given equation equal to a same variable.
Let the variable be "y", Then the equation 
becomes,
Thus, The system of equation becomes

Option (a) is correct.
Answer:
The probability is 0.31
Step-by-step explanation:
To find the probability, we will consider the following approach. Given a particular outcome, and considering that each outcome is equally likely, we can calculate the probability by simply counting the number of ways we get the desired outcome and divide it by the total number of outcomes.
In this case, the event of interest is choosing 3 laser printers and 3 inkjets. At first, we have a total of 25 printers and we will be choosing 6 printers at random. The total number of ways in which we can choose 6 elements out of 25 is
, where
. We have that 
Now, we will calculate the number of ways to which we obtain the desired event. We will be choosing 3 laser printers and 3 inkjets. So the total number of ways this can happen is the multiplication of the number of ways we can choose 3 printers out of 10 (for the laser printers) times the number of ways of choosing 3 printers out of 15 (for the inkjets). So, in this case, the event can be obtained in 
So the probability of having 3 laser printers and 3 inkjets is given by

Answer:
(3/4)a
Step-by-step explanation:
The angle at K is 120°, so the angle at L is its supplement: 60°. That makes triangle FKL an equilateral triangle with a base of FL = a. The vertex at K is centered over the base, so is a/2 from G.
The midsegement length is the average of GK and FL, so is ...
midsegment = (GK +FL)/2 = (a/2 +a)/2
midsegment = (3/4)a