Answer:
1/5^8
Step-by-step explanation:
We know that a^b^c = a^(b*c)
(5^-2)^4
5^(2*-4)
5^-8
We know that a^-b = 1/a^b
1/5^8
Question:
If a sample of 2 hammer is selected
(a) find the probability that all in the sample are defective.
(b) find the probability that none in the sample are defective.
Answer:
a 
b 
Step-by-step explanation:
Given
--- hammers
--- selection
This will be treated as selection without replacement. So, 1 will be subtracted from subsequent probabilities
Solving (a): Probability that both selection are defective.
For two selections, the probability that all are defective is:




Solving (b): Probability that none are defective.
The probability that a selection is not defective is:

For two selections, the probability that all are not defective is:




All you have to do is follow PEMDAS (if you don't know what that is, search it up).
-2-5(4n-6)=-132
Distribute the 5 to everything in the parentheses.
-2-20n+30= -132
Combine like terms.
-20n+28 = -132
Subtract 28 from both sides.
-20n = -160
Divide by -20 on both sides.
n = 8
To the nearest tenth: 3,333,330
To the nearest hundred: 3,333,300
To the nearest thousand: 3,333,000
To the nearest ten thousand: 3,330,000
To the nearest hundred thousand: 3,300,000
To the nearest million: 3,000,000
Hope that helps :)