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3 years ago

If 1/2 gallon of paint covers 1/12 of a wall, then how many quarts of paint are needed for the

Mathematics

1 answer:

Orlov [11]3 years ago

6 0

Answer:

Explanation:

½ gallons × 12 = 6 (gallons of paint needed to cover the whole wall)

6 gallons = 24 quarts

i hope this helps! :D

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Which of the following expressions is written in simplest form

nadya68 [22]

Answer: I x^2 y^3 z

Step-by-step explanation:

This is the one most simplified. I’ll tell you why the others are incorrect.

F) 3^5 x^2 can be simplified. 3^5= 243. The simplified answer would be 243 x^2

G) (5y)^3= 125y^3

H) a^0 b (^0= 1 always) -> ab

Hope this helps!

3 0

3 years ago

A circular sheet was cut to make cone without a base of slant height equal to 7 m. If the net of the cone has a perimeter of 47

Usimov [2.4K]

Answer:

75%

Step-by-step explanation:

5 0

3 years ago

Read 2 more answers

I need help with this one also don’t put a link

Nadya [2.5K]

Answer: 306 cm^2

Step-by-step explanation:

Surface is found by add all the areas of the sides together, including sides, top and bottom. You might need to divide the 3D shape into different shapes in order to find the area of the sides/top/bottoms. Area for rectangle = bxh.

8 0

3 years ago

When v= -7 and C=2 look at picture

Bond [772]

Answer:

Step-by-step explanation:

insert the numbers and divide

-4(-7)= 28

2^2= 4

28/4= 7

5 0

3 years ago

Read 2 more answers

Exercise 1.A committee of five people is chosen randomly from four men and six women. Find the probability that:a) Exactly four

MatroZZZ [7]

Answer with explanation:

Given : Number of men = 4

Number of women = 6

Total people = $6+4=10$

Total number of ways to make a committee of five people from 10 persons :-

$^{10}C_{4}=\dfrac{10!}{4!(10-4)!}=210$

a) Number of ways to make committee that has exactly four women :

$^6C_4\times ^4C_1=\dfrac{6!}{4!(6-4)!}\times4=60$

The probability that committee has exactly four women :

$\dfrac{60}{210}=\dfrac{2}{7}$

b) Number of ways to make committee that has at-least four women :

$^6C_4\times ^4C_1+^6C_5=\dfrac{6!}{4!(6-4)!}\times4+6=66$

The probability that committee at-least four women :

$\dfrac{66}{210}=\dfrac{22}{70}$

c) Number of ways that committee has more than 4 women :-

$^6C_5\times^4C_0=6$

The probability that committee has more than 4 women :-

$\dfrac{6}{210}$

Now, the probability that committee has at most four women :-

$1-\dfrac{6}{210}=\dfrac{204}{210}=\dfrac{102}{105}$

5 0

3 years ago