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:




Answer:
The first one
Step-by-step explanation:
If you have any questions about the way I solved it, don't hesitate to ask me in the comments below ;)
Answer:
The concession stands needs to sell 14 or more bags.
Step-by-step explanation:
Given that:
Cost per bag of popcorn = $2.50
Number of bags already sold = 30
Cost of 30 bags = 30*2.50 = $75
Bags needs to be sold to make $110 = x
2.50x + 75 ≥ 110
2.50x ≥ 110-75
2.50x ≥ 35
Dividing both sides by 2.50

Hence,
The concession stands needs to sell 14 or more bags.
Answer:
1256
Step-by-step explanation:
Given the function F(x)=1256(1.24)^x, the initial value occurs at x = 0
Substitute x = 0 into the function;
F(0)=1256(1.24)^0
f(0) = 1256(1) (any value raise to sero is 1)
f(0) = 1256
hence the initial value is 1256
They are equal. 2/5 is 4/10.