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3 years ago

10% more than a number of 132

Mathematics

2 answers:

sergij07 [2.7K]3 years ago

8 0

142. Because 132+10 is 142 I think that’s it

timurjin [86]3 years ago

4 0

145.2 because 10% of 132 is 13.2.
13.2+132 leaves you with 145.2

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yes they can

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3 years ago

Question

kondor19780726 [428]

Answer: r = 11

Step-by-step explanation:

We know that the point (-2, r) lies on the graph of:

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6 0

3 years ago

What’s this answer ? will give ton of points!!!!!!!!!!

Ann [662]

Answer:

268 cm^2

Step-by-step explanation:

Pink prism:

top: 2 cm * 2 cm = 4 cm^2

right, left, back, front: 4 * 2 cm * 6 cm = 48 cm^2

bottom: completely covered in connection to green prism

Total surface area of pink: 52 cm^2

Green prism:

bottom: 10 cm * 5 cm = 50 cm^2

front, back: 2 * 10 cm * 4 cm = 80 cm^2

right, left: 2 * 5 cm * 4 cm = 40 cm^2

top: 10 cm * 5 cm - 2 cm * 2 cm = 46 cm^2

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3 0

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