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n200080 [17]
3 years ago
15

Mary wants to buy one large pizza, one medium pizza, and three drinks. The drinks cost $p$ dollars each, the medium pizza costs

two times as much as one drink, and the large pizza costs three times as much as one drink. If Mary started with $30$ dollars, how much money would she have left after making all of her purchases?
Mathematics
1 answer:
Korolek [52]3 years ago
4 0

Answer:

30-8p

Step-by-step explanation:

Let p represent cost of each drink.

We have been given that the medium pizza costs two times as much as one drink, so the cost of medium pizza would be 2p.

We are also told that the large pizza costs three times as much as one drink,  so the cost of la pizza would be 3p.

We have been given that Mary wants to buy one large pizza, one medium pizza, and three drinks.

Since the cost of one drink is p, so cost of 3 three drinks would be 3p.

The total cost of one large pizza, one medium pizza, and three drinks would be 2p+3p+3p=8p

Mary started with $30, so amount left after all of her purchases would be 30 minus total cost of all purchases.

30-8p

Therefore, Mary will have 30-8p dollars left after making all of her purchases.

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a committee has eleven members. there are 3 members that currently serve as the boards chairman, ranking members, and treasurer.
miss Akunina [59]

Answer:

\frac{1}{990}

Step-by-step explanation:

<u>The full question:</u>

<em>"A committee has eleven members. there are 3 members that currently serve as the boards chairman, ranking members, and treasurer. each member is equally likely to serve in any of the positions. Three members are randomly selected and assigned to be the new chairman, ranking member, and treasurer. What is the probability of randomly selecting the three members who currently hold the positions of chairman, ranking member, and treasurer and reassigning them to their current​ positions?"</em>

<em />

<em />

The permutation of choosing 3 members from a group of 11 would be:

P(n,r) = \frac{n!}{(n-r)!}

Where n would be the total [in this case n is 11] & r would be 3

Which is:

P(11,3) = \frac{11!}{(11-3)!}=\frac{11!}{8!}=11*10*9=990

So there are total of 990 possible way and there is ONLY ONE WAY for them to be reassigned. Hence the probability would be:

1/990

8 0
4 years ago
Match each expression with an equivalent expression.
jolli1 [7]

1=6x-4x

2= 6x-4

3=6-4x

4=6x-4y

6 0
3 years ago
The graph shows that f(x)=(one-third) Superscript x is translated horizontally and vertically to get the function (one-third) Su
harina [27]

It is the number 3.              

6 0
3 years ago
Write a expression that is equivalent of 4+3(5+x)
Serggg [28]

Answer:19+3x

Step-by-step explanation:

4+3(5+x)

Open bracket

4+15+3x

19+3x

3 0
4 years ago
If 9 were subtracted from three times a number, the result would be the same as if 4 were added to twice the number. what is the
amid [387]
3x-9=2x+4
3x-9+9=2x+4+9
3x=2x+13
3x-2x=2x-2x+13
x=13
Check:
3(13)-9=2(13)+4
39-9=26+4
30=30
4 0
4 years ago
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