13·[18÷6]
13·3
39
the answer is 39
note that first we start with the problem that is the most inside of the operation
Answer:
95.69%
Step-by-step explanation:
We have X is the number of parts produced up to (and including) the first slightly defective part. So, X is Geometric (2/92), which would be the following:
P (X => 3) = Summation i = 3, up to infinity of {[(90/92)^(i-1)] * (2/92)}
We replace and solve and we are left with:
P (X => 3) = (2/92) * (90/92)^(3-1) * 1/(1 - 90/92)
P (X => 3) = 0.9569
Which means that the probability that at least 3 parts must be produced until there is a slightly defective part produced is 95.69%
5
First you can add the two functions together, since that is what (h+k) is asking us to do.
x^2 + 1 + x - 2 ---> then simplify like terms
x^2 + x -1 ----> then plug in 2 for all of the x's in the equation
2^2 + 2 - 1 ----> and solve
4 + 2 - 1 = 5
The answer is C 15 because 37+15=52, 52+15=67, 67+15=82