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3 years ago

Michael has a free throw record of .625. If he has to shoot 50 baskets, how many can we expect him to make?

Mathematics

1 answer:

ziro4ka [17]3 years ago

3 0

Answer:

The number of free shoot to be make is 80 .

Step-by-step explanation:

Given as :

The free throw record of Michael = x = 0.625

The number of basket to be shoot = n = 50 baskets

Let The number of free shoot to be make = y

<u>Now, according to question</u>

The number of free shoot be make = $\dfrac{number of basket to be shoot}{\textrm the free throw record}$

i.e y = $\dfrac{n}{x}$

or, y = $\dfrac{50}{025}$

∴ y = 80

So, The number of free shoot to be make = y = 80

Hence,The number of free shoot to be make is 80 . Answer

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3 years ago

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To learn more on rigid transformations: brainly.com/question/28004150

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3 years ago

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b. $\frac{2}{5}$

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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:

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Total number of cases = 15

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

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