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kompoz [17]
3 years ago
8

Emma decided to buy some paint to paint some of the rooms in his house. She found out that one room required 1 1/2 cans of paint

. If Emma buys 9 cans of paint, how many rooms can she paint?
6
27/3
13 1/2
5
​
Mathematics
2 answers:
erastova [34]3 years ago
4 0

Answer:

I believe the answer would be six rooms

Step-by-step explanation: This may seem a little difficult, but really it is quite simple. All you have to do is divide the amount of paint she has (9) by how much it takes to paint one room (1 1/2). The equation would be,            (9 / 1 1/2) which equals 6. I hope this helps.

evablogger [386]3 years ago
3 0

Answer:

6

Step-by-step explanation:

To solve this, divide 9 by 1 1/2.

To make this easier to divide, convert 1 1/2 into an improper fraction.

1 1/2 = 3/2

Dividing 9 by 3/2 is the same as multiplying 9 by 2/3

Multiply the numerators and denominators

[9 is the same as 9/1]

9 x 2 = 18

1 x 3 = 3

18/3 = 6

So, Emma can paint 6 rooms.

I hope this helps!

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A friend of mine is giving a dinner party. His current wine supply includes 9 bottles of zinfandel, 10 of merlot, and 12 of cabe
e-lub [12.9K]

Answer:

Step-by-step explanation:

From the given information:

The total number of wine = 9 + 10 + 12 = 31

(1)

The number of distinct sequences used for serving any five wines can be estimated by using the permutation of the number of total wines with the number of wines served.

i.e

= ^{31}P_5

=\dfrac{31!}{(31-5)!}

=\dfrac{31!}{(26)!}

=\dfrac{31\times 30\times 29\times 28\times 27\times 26!}{(26)!}

= 20389320

(2)

If the first two wines served = zinfandel and the last three is either merlot or cabernet;

Then, the no of ways we can achieve this is:

= ^9P_2\times ^{22}P_3

= \dfrac{9!}{(9-2)!}\times \dfrac{22!}{(22-3)!}

= \dfrac{9!}{(7)!}\times \dfrac{22!}{(19)!}

= \dfrac{9*8*7!}{(7)!}\times \dfrac{22*21*20*19!}{(19)!}

= 665280

(3)

The probability that no zinfandel is served is computed as follows:

Total wines (with zinfandel exclusion) = 31 - 9 = 22

Now;

the required probability is:

= \dfrac{^{22}P_5 }{^{31}P_5}

= \dfrac{\dfrac{22!}{(22-5)!} } {\dfrac{31!}{(31-5)!} }

= \dfrac{\dfrac{22!}{17!} } {\dfrac{31!}{(26)! }}

= \dfrac{(\dfrac{22*21*20*19*18*17!}{17!})} {(\dfrac{31*30*29*28*27*26!}{(26)! })}

= 0.1549

≅ 0.155

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3 years ago
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nirvana33 [79]

Answer:

Step-by-step explanation:

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