What is the volume of Alina’s model in cubic centimeters

For integers a, b and k, we know that a >12, b > 20 and a < b. If b=7k, what is the value of k ?

Doss [256]

This question is incomplete, the complete question is;

For integers a, b and k, we know that a > 12, b < 20 and a < b. If b=7k, what is the value of k ?

Answer: the value of k = 2

Step-by-step explanation:

Given that;

a > 12

b < 20

a < b

If b = 7k

Now if k = 1 {b = 7k = 7}

b would be equal to 7 but b has to be greater than 20

IT CANT BE

if k = 2 { b = 7k = 14}

b would be equal to 14, a is greater than 12, b has to be less than b; 14 < 20,

a has to be less than than b ( 12 < 14 )

IT IS

if k = 3 {b = 7k = 21}

b would be equal to 21, so b is greater than 20 and a is less than 21

IT CANT BE

Therefore the value of k = 2

5 0

3 years ago

Karen purchased a prepaid phone card for $30. Long distance calls cost 16 cents a minute using this card. Karen used her card on

Sedbober [7]

30-24.08=5.92, 5.92/.16=37 minutes

3 0

3 years ago

Solve for y you will get brainliest

77julia77 [94]

y = 3p+85, because they are vertical angles.

x = 2p ‐ 10, because they are vertical angles too.

4 0

3 years ago

Find the least number which must be added to 6203 to obtain a perfect square.

andrey2020 [161]

Answer:

79

Step-by-step explanation:

78

---------

7 | 62 03

49

----------

148 | 13 03

11 84

-----------

19

-----------

78^2 = 6084.

We observe that 78^2 < 6203.

79^2 = 6241.

We observe that 79^2 > 6203.

Hence the number to be added to 6203 is 6241 - 6203 = 38.

6203 + 38 = 3241

= 79 * 79

= 79.

Therefore 38 should be added to 6203 to obtain a perfect square.

5 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