F=182/.8 should be the correct equation to use in this problem since your trying to find the remaining 80% left of the floor
Answer: Expected value of the daily cost of operating the machine is 235.264.
Step-by-step explanation:
Since we have given that
E[x]= 0.96 repairs per day
And Var[x] = 0.96 repairs per day.

![E[c]=160+40E[x^2]\\\\E[c]=160+40(Var[x]+(E[x])^2)\\\\E[c]=160+40(0.96+0.96^2)\\\\E[c]=235.264](https://tex.z-dn.net/?f=E%5Bc%5D%3D160%2B40E%5Bx%5E2%5D%5C%5C%5C%5CE%5Bc%5D%3D160%2B40%28Var%5Bx%5D%2B%28E%5Bx%5D%29%5E2%29%5C%5C%5C%5CE%5Bc%5D%3D160%2B40%280.96%2B0.96%5E2%29%5C%5C%5C%5CE%5Bc%5D%3D235.264)
Hence, Expected value of the daily cost of operating the machine is 235.264.
Answer:
a = 0.519
Step-by-step explanation:
14a + 13a + 8 = 22
27a = 14
a = 0.519
Answer is C
14.2+6.5(p+3q)-4.05q
14.2+6.5p+19.5q-4.05q
14.2+6.5p+15.45q→6.5p+15.45q+14.2
so the answer is C
Answer:
The number of different combinations of three students that are possible is 35.
Step-by-step explanation:
Given that three out of seven students in the cafeteria line are chosen to answer a survey question.
The number of different combinations of three students that are possible is given as:
7C3 (read as 7 Combination 3)
xCy (x Combination y) is defines as
x!/(x-y)!y!
Where x! is read as x - factorial or factorial-x, and is defined as
x(x-1)(x-2)(x-3)...2×1.
Now,
7C3 = 7!/(7 - 3)!3!
= 7!/4!3!
= (7×6×5×4×3×2×1)/(4×3×2×1)(3×2×1)
= (7×6×5)/(3×2×1)
= 7×5
= 35
Therefore, the number of different combinations of three students that are possible is 35.