Answer:
3p*3p*3p*3p*3p*3p
Step-by-step explanation:
Multiply exponents
3p^6
expand
3p*3p*3p*3p*3p*3p = 3p^6
<span>The chance or probability that the lawn mower will hit a piece of glass that is already cracked is calculated by dividing the number of glasses that are cracked by the total number of glasses. In this item, the unknown can be calculated by 3/15. The answer is therefore, 1/5 or 0.20. </span>
Answer:
Step-by-step explanation:
C
We are given
piggy bank has nickels , dimes and quarters
Let's assume
number of nickels =n
every two nickels there are 3 dimes
so, number of dimes are
every two dimes that are 5 quarters there
so, number of quarters are
so, total number of coins = number of nickels + number of dimes +number of quarters
total number of coins =n+1.5n+3.75
there are 500 coins
so, we get
now, we can solve for n
divide both sides by 6.25
so, we get
number of dimes is 1.5n
number of quarters is 3.75n
so,
Number of nickels =80
Number of dimes =120
Number of quarters =300............Answer
(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.
