Employees in an office wear eyeglasses. 23 have

Ответ:

PrisonKing9334

18.01.2022

26.60% probability that both of them wear bifocals.

22.43% probability that both have single-vision correction.

Step-by-step explanation:

A probability is the number of desired outcomes divided by the number of total outcomes.

In this problem, the order of the employees is not important. For example, John and Elisa, or Elisa and John are the same outcomes.

This is why we use the combinations formula:

$C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

Total outcomes:

23+25 = 48 employees

Two selected. So

$T = C_{48,2} = \frac{48!}{2!(48-2)!} = 1128$

What is the probability that both of them wear bifocals?

25 wear bifocals.

Two selected. So

$D = C_{25,2} = \frac{25!}{2!(25-2)!} = 300$

$P = \frac{D}{T} = \frac{300}{1128} = 0.266$

26.60% probability that both of them wear bifocals.

What is the probability that both have single-vision correction?

23 have single vision correction.

Two selected. So

$D = C_{23,2} = \frac{23!}{2!(23-2)!} = 253$

$P = \frac{D}{T} = \frac{253}{1128} = 0.2243$

22.43% probability that both have single-vision correction.

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mpateltx

22.11.2021

y = 7/3x + 2/3

Step-by-step explanation:

-7x + 3y = -10

3y = 7x - 10

y = 7/3x - 10/3

-4 = 7/3(-2) + b

-4 = -14/3 + b

2/3 = b

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