In which number is the value of the first 8 ten times as great as the value of the second 8?

A sandwich store charges a delivery fee to bring lunch to an office building. One office pays $33 for 4 turkey sandwiches. Anoth

Lorico [155]

First setup the system of equations for the given case.
Let sandwich=x
and delivery fee =y
Equations are:
4x+y=33 Equation 1
8x+y=61 Equation 2
Subtract equation 1 from 2.
4x=28
x=7
Now substitute value of x in equation 1.
4(7)+y=33
28+y=33
y=5

So cost of single sandwich=$7 and Delivery fee=$5

Now cost of 12 sandwiches:
12x+y=85
Put y=5
12x+5=85
12x=80
x=6.67 which is less than 7. So, They are not able to get 12 sandwiches for $85.

6 0

3 years ago

Will give brainliest

givi [52]

Answer:

69 and 29

Step-by-step explanation:

if you add 90 by 69 and 29 you will get the answer of 180.

5 0

3 years ago

Which statement is true regarding the diagram of circle P?

masya89 [10]

Answer:

The difference of z and y must be 2x

Step-by-step explanation:

6 0

3 years ago

Read 2 more answers

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

Find the areas of at least two of the triangles below. Show your reasoning.<br> help plzz

scoray [572]

Answer:
A. Area = 8
B. Area = 15.5

Explanation:

Formula for Area of Triangles: b•h / 2

A.
height = 2.
base = 8.
2•8 = 16/2 = 8
Area = 8

B.
height = 3
base = 7
3•7 = 21/2 = 15.5
Area = 15.5

5 0

3 years ago