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madam [21]
3 years ago
7

HELP!!! Estimate: 32% of 50 =

Mathematics
2 answers:
nirvana33 [79]3 years ago
8 0

Answer:

32% of 50 = 16% of 100

So, 32% of 50 is equal to 16.

Let me know if this helps!

natali 33 [55]3 years ago
6 0

15 is the answer

if its estimate, its basically asking 30% of 50

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since the "b" in y=mx+b ends up being 0, that means that the y intercept is 0, which also means that the x intercept is 0.

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the painted lines that separate parking spaces are parallel. The measure of amgle 1 is 60. what us the measure of angle 2
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A rectangle has a width, w, equal to 6 less than the length, l. Which expressions represent ways to find the perimeter of the re
Irina-Kira [14]

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3 years ago
How many 2/3 are in 1?
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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
3 years ago
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