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3 years ago

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Mathematics

1 answer:

mina [271]3 years ago

4 0

Answer:

The unit rate for green pencils is $0.32

The green pencils are priced less per pencil

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The area of the rectangle shown is 24 square inches. how much longer is its length than its width?

Lina20 [59]

If one length is 6in, the other would be 4, since 6×4 is 24, which is the area of the rectangle. The length is 2 longer than the width, 6-4=2.

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3 years ago

Read 2 more answers

Factored completely the expression 2y^2 + 12y- 54 is equivalent to

MrMuchimi

2y^2 + 12y- 54
=2(y^2 + 6y- 27)
=2 (y - 3)(y+9)

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3 years ago

How do u slove this for y intercept? (4, 1); slope=3/2

Valentin [98]

Answer:

y-intercept=-5

Step-by-step explanation:

y=mx+b m/slope=3/2 the point is (x,y)/(4,1)

1=3/2(4)+b

1=12/2+b

1=6+b

-6-6

-5=b

8 0

3 years ago

Ms. Green tells you that a right triangle has a hypotenuse of 13 and a leg of 5. Find the other leg of the triangle

melomori [17]

A2 +b2= c2
a2 +25= 169
-25 -25
a2= 144
divide by the square root and
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4 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