Answer:
C) a sample distribution of a sample mean with n = 10
and
Step-by-step explanation:
Here, the random experiment is rolling 10, 6 faced (with faces numbered from 1 to 6) fair dice and recording the average of the numbers which comes up and the experiment is repeated 20 times.So, here sample size, n = 20 .
Let,
= The number which comes up on the ith die on the jth trial.
∀ i = 1(1)10 and j = 1(1)20
Then,
=
= 3.5 ∀ i = 1(1)10 and j = 1(1)20
and,
=
=
=
15.166667
so, =
= 2.91667
and =
Now we get that,
We get that are iid RV's ∀ j = 1(1)20
Let,
So, we get that
= for any i = 1(1)10
= 3.5
and,
Hence, the option which best describes the distribution being simulated is given by,
C) a sample distribution of a sample mean with n = 10
and
The answer is B. x > -22
-5x + 20 < 130
-5x < 130 - 20
-5x < 110
-5x/-5 > 110/-5
x > -22
Answer:
Step-by-step explanation:
<u>We have:</u>
- p = -1/5x + 100 = -0.2x + 100
a) <u>The revenue is:</u>
- R = px
- R = x( -0.2x + 100)
- R= -0.2x² + 100x
b) <u>x = 200, find R:</u>
- R = -0.2(200²) + 100(200) = 12000
c) <u>This is a quadratic function and the maximum value is obtained at vertex.</u>
- x = -100/(-0.2*2) = 250 is the required quantity
d) <u>The max revenue is obtained when -0.2x + 100 at max:</u>
- The maximum possible is p = 100 when x = 0
Answer:
I think it's B
Step-by-step explanation: