A is -40
B is Barry
hope this helps
I'm assuming

(a) <em>f(x)</em> is a valid probability density function if its integral over the support is 1:

Compute the integral:

So we have
<em>k</em> / 6 = 1 → <em>k</em> = 6
(b) By definition of conditional probability,
P(<em>Y</em> ≤ 0.4 | <em>Y</em> ≤ 0.8) = P(<em>Y</em> ≤ 0.4 and <em>Y</em> ≤ 0.8) / P(<em>Y</em> ≤ 0.8)
P(<em>Y</em> ≤ 0.4 | <em>Y</em> ≤ 0.8) = P(<em>Y</em> ≤ 0.4) / P(<em>Y</em> ≤ 0.8)
It makes sense to derive the cumulative distribution function (CDF) for the rest of the problem, since <em>F(y)</em> = P(<em>Y</em> ≤ <em>y</em>).
We have

Then
P(<em>Y</em> ≤ 0.4) = <em>F</em> (0.4) = 0.352
P(<em>Y</em> ≤ 0.8) = <em>F</em> (0.8) = 0.896
and so
P(<em>Y</em> ≤ 0.4 | <em>Y</em> ≤ 0.8) = 0.352 / 0.896 ≈ 0.393
(c) The 0.95 quantile is the value <em>φ</em> such that
P(<em>Y</em> ≤ <em>φ</em>) = 0.95
In terms of the integral definition of the CDF, we have solve for <em>φ</em> such that

We have

which reduces to the cubic
3<em>φ</em>² - 2<em>φ</em>³ = 0.95
Use a calculator to solve this and find that <em>φ</em> ≈ 0.865.
Answer:
Plant B
Step-by-step explanation:
The only measure of variation we really need to check is the mean, or average, of tomatoes that each plant produced. This is because if Kathryn was a farmer, she would want a plant that on average, produces more tomatoes. For that reason, in this context, the mode, median and range are irrelevant. To find the mean of a data set, you simply divide the sum of all of the data by the number of data in the set. Therefore:
Mean of Plant A: (4 + 6 + 7 + 3 + 6 + 2 + 1 + 3 + 6 + 5) / 10 = 43 / 10 = 4.3
Mean of Plant B: (5 + 6 + 7 + 6 + 8 + 9 + 6 + 7 + 8 + 9) / 10 = 71 / 10 = 7.1
As you can see, the mean tomatoes Plant B produces is larger than that of Plant A, therefore, Kathryn should choose Plant B if she was a farmer.
Answer:
x is 5 cm as the entire base is 10 cm, so 10/2 = 5
Answer:
-24 + 12d = 2(d - 3) + 22
-24 + 12d = 2d - 6 + 22
-24 + 12d = 2d + 16
12d = 2d + 40
10d = 40
d = 4
Step-by-step explanation: