Ok.
Convert 3/4 to 6/8.

There is your answer. 3/8.
Answer: A. S(t)= 1000 + 0.63t + 0.48t - 17t
B. S(t)= 984.11
Step-by-step explanation:
Answer:A
Step-by-step explanation:
FOIL
Follow the order of operations. PEMDAS. Parenthesis, Exponents, Multiplication, Division, Addition, and Subtraction. Let's start with parenthesis. First we so everything inside the parenthesis. So the first thing we do, is exponents, 2 x 2 = 4. Now the top equation is 3 x 4 - 23. Next thing in the order of operations, is Multiplication. So now we do 3 x 4. Which equals 12. Now the top equation is 10 - 12 - 23. Now we do Subtraction in order. First 10 - 12 = negative 2 (-2) Then we do -2 - 23 which equals negative 25 (-25). Now let's work on the bottom equation. Once again follow the order of operations. We do the exponents in the parenthesis first. 10 x 10 = 100. Then do the rest in the parenthesis, 1 + 100 = 101. Now the exponents in the outside of the parenthesis. - 2 x - 2 = 4. And 5 x 5 = 25. Then we do Multiplication, 4 x 25, which equals 100. Then we do 101 - 100 = 1. Then we reduce the fraction. -25/1 = -25. Therefore the answer to your problem is -25. I apologize for taking so long to write this. Hope this helps though!
-Twixx
Answer:
0.0177 = 1.77% probability that the first defect is caused by the seventh component tested.
The expected number of components tested before a defective component is found is 50, with a variance of 0.0208.
Step-by-step explanation:
Assume that the probability of a defective computer component is 0.02. Components are randomly selected. Find the probability that the first defect is caused by the seventh component tested.
First six not defective, each with 0.98 probability.
7th defective, with 0.02 probability. So

0.0177 = 1.77% probability that the first defect is caused by the seventh component tested.
Find the expected number and variance of the number of components tested before a defective component is found.
Inverse binomial distribution, with 
Expected number before 1 defective(n = 1). So

Variance is:

The expected number of components tested before a defective component is found is 50, with a variance of 0.0208.