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pychu [463]
3 years ago
12

See picture. Please show work

Mathematics
2 answers:
gtnhenbr [62]3 years ago
8 0

Answer:

option b) is not similar but they have ( mid point theorem) EF parallel to vu

and half of UV

lapo4ka [179]3 years ago
5 0

Answer:

its the last one i had this question like 5 days ago i am sure

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3 years ago
The computers of six faculty members in a certain department are to be replaced. Two of the faculty members have selected laptop
Anettt [7]

Answer:

a. \frac{1}{15}

b. \frac{2}{5}

c. \frac{14}{15}

d. \frac{8}{15}

Step-by-step explanation:

Given that there are two laptop machines and four desktop machines.

On a day, 2 computers to be set up.

To find:

a. probability that both selected setups are for laptop computers?

b. probability that both selected setups are desktop machines?

c. probability that at least one selected setup is for a desktop computer?

d. probability that at least one computer of each type is chosen for setup?

Solution:

Formula for probability of an event E can be observed as:

P(E) = \dfrac{\text{Number of favorable cases}}{\text {Total number of cases}}

a. Favorable cases for Both the laptops to be selected = _2C_2 = 1

Total number of cases = 15

Required probability is \frac{1}{15}.

b. Favorable cases for both the desktop machines selected = _4C_2=6

Total number of cases = 15

Required probability is \frac{6}{15} = \frac{2}{5}.

c. At least one desktop:

Two cases:

1. 1 desktop and 1 laptop:

Favorable cases = _2C_1\times _4C_1 = 8

2. Both desktop:

Favorable cases = _4C_2=6

Total number of favorable cases = 8 + 6 = 14

Required probability is \frac{14}{15}.

d. 1 desktop and 1 laptop:

Favorable cases = _2C_1\times _4C_1 = 8

Total number of cases = 15

Required probability is \frac{8}{15}.

8 0
4 years ago
Number 7. The answer is (b). Can you explain how?
alina1380 [7]

u have to multiply all the sides and

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3 years ago
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