Answer:
The expected number of defective batteries to be pulled out is 0.9, which rounded to the nearest integer gives a total of 1, that is, 1 of the 3 batteries is expected to be defective.
Step-by-step explanation:
Given that a box contains 3 defective batteries and 7 good ones, and I reach in and pull out three batteries, to determine what is the expected number of defective batteries, the following calculation must be performed:
3 + 7 = 100
3 = X
10 = 100
3 = X
3 x 100/10 = X
300/10 = X
30 = X
3 x 3/10 = X
0.9 = X
Therefore, the expected number of defective batteries to be pulled out is 0.9, which rounded to the nearest integer gives a total of 1, that is, 1 of the 3 batteries is expected to be defective.
Total capacity = sum of the individual production capacities.
Here,
Total capacity = sum of f(m) = (m + 4)^2 + 100 and g(m) = (m + 12)^2 − 50.
Then f(m) + g(m) = (m + 4)^2 + 100 + (m + 12)^2 − 50.
We must expand the binomial squares in order to combine like terms:
m^2 + 8 m + 16 + 100
+m^2 + 24m + 144 - 50
---------------------------------
Then f(m) + g(m) = 2m^2 + 32m + 160 + 50
f(m) + g(m) = 2m^2 + 32m + 210, where m is the number of
minutes during which the two machines operate.
<span> f(x) = –81 (4/3) X-1 f(x) = –81 (-3/4) </span>
Answer:
See the attachment
Step-by-step explanation:
It is convenient to use a spreadsheet or graphing calculator for this.
