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mars1129 [50]
3 years ago
11

6 people consists of 3 married couples. Each couple wants to sit with older partner on the left.

Mathematics
1 answer:
Lera25 [3.4K]3 years ago
8 0

a. The first part asks for how many ways they can be seated together in a row. Therefore we want the permutations of the set of 6 people, or 6 factorial,

6! = 6 * 5

= 30 * 4

= 360 * 2 = 720 possible ways to order 6 people in a row

b. There are two cases to consider here. If the doctor were to sit in the left - most seat, or the right - most seat. In either case there would be 5 people remaining, and hence 5! possible ways to arrange themselves.

5! = 5 * 4

= 20 * 3

= 120 * 1 = 120 possible ways to arrange themselves if the doctor were to sit in either the left - most or right - most seat.

In either case there are 120 ways, so 120 + 120 = Total of 240 arrangements among the 6 people if the doctor sits in the aisle seat ( leftmost or rightmost seat )

c. With each husband on the left, there are 3 people left, all women, that we have to consider here.

3!  = 3 * 2 6 ways to arrange 3 couples in a row, the husband always to the left

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Answer:

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Step-by-step explanation:

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6 0
3 years ago
One year Roger had the lowest ERA​ (earned-run average, mean number of runs yielded per nine innings​ pitched) of any male p
baherus [9]

Answer:

(a) The <em>z</em>-score of Roger is -1.56 and the <em>z</em>-score of Amber is -1.13.

(b) Roger had a better year relative to his peers.

Step-by-step explanation:

If X follows N (<em>µ, σ</em>₂), then z=\frac{x-\mu}{\sigma}, is a standard normal variate with mean,      E (Z) = 0 and Var (Z) = 1. That is, Z follows N (0, 1).

Let <em>X</em> = ERA of male pitchers and <em>Y</em> = ERA of ERA of female pitchers.

It is provided that the mean and standard deviation of <em>X</em> are:

\mu_{X}=4.371\\\sigma_{X}=0.787

Also, the mean and standard deviation of <em>Y</em> are:

\mu_{Y}=4.363\\\sigma_{Y}=0.869

ER of Roger is 3.14 and ERA of Amber is 3.38.

(1)

Compute the <em>z</em>-score of Roger as follows:

z=\frac{x-\mu_{X}}{\sigma_{X}}=\frac{3.14-4.371}{0.787}=-1.56

Thus, the <em>z</em>-score of Roger is -1.56.

Compute the <em>z</em>-score of Amber as follows:

z=\frac{x-\mu_{Y}}{\sigma_{Y}}=\frac{3.38-4.363}{0.869}=-1.13

Thus, the <em>z</em>-score of Amber is -1.13.

(2)

Compute the probability of ERA's that are greater than Roger's ERA as follows:

P(X>3.14)=P(\frac{X-\mu}{\sigma}>\frac{3.14-4.371}{0.787})\\=P(Z>-1.56)\\=P(Z

This implies that 94% of the other male pitchers had an ERA higher than 3.14.

Compute the probability of ERA's that are greater than Amber's ERA as follows:

P(Y>3.38)=P(\frac{Y-\mu_{Y}}{\sigma_{Y}}>\frac{3.38-4.363}{0.869})\\=P(Z>-1.13)\\=P(Z

This implies that 87% of the other female pitchers had an ERA higher than 3.38.

As it is provided that the lower the ERA the better the pitcher, then it can be concluded that Roger had a better year relative to his peers.

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