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soldier1979 [14.2K]
3 years ago
7

A caterer prepares three times as many pizzas as she usually prepares for a large party.

Mathematics
2 answers:
VMariaS [17]3 years ago
8 0

Answer: There are  45 party guests who will be served pizzas.

Step-by-step explanation:

Since we have given that

Number of pizzas usually prepare by the caterer = 5

Part of pizza will eat by each party guest is given by

\frac{1}{3}

As we have given that a caterer prepares 3 times as many pizzas as she usually prepares for a large party,

So, it becomes

\text{Number of pizzas prepare by carterer this time}=5\times 3=15\\

Let the number of party guests be x

According to question, it becomes,

Number of party guests will get the pizza serve is given by

\frac{1}{3}\times x=15\\\\x=15\times 3\\\\x=45

Hence, there are  45 party guests who will be served pizzas.

anastassius [24]3 years ago
8 0
3 x 5 pizzas = 15 pizzas
1/3 of pizzas  x  people  =  15 pizzas
people  = 15 pizzas / (1/3)  =  45
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two similar cones have surface areas in the ration 4:9. find the ration of their lengths and their volumes
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This exercise refers to a standard deck of playing cards. Assume that 5 cards are randomly chosen from the deck. How many hands
pogonyaev

Answer:

Total number of cards in hand contain exactly two 3s and two 7s is 56.

Step-by-step explanation:

Given : This exercise refers to a standard deck of playing cards. Assume that 5 cards are randomly chosen from the deck.

To find : How many hands contain exactly two 3s and two 7s?

Solution :

In a standard deck of card has 52 cards distributed in 4 sets.

Each set has {2,3,4,5,6,7,8,9,10,J,Q,K,A} has 13 cards.

Number of 3s in 52 cards are 4.

Number of 7s in 52 cards are 4.

So, Getting exactly two 3s out of 4 is ^4C_2

Getting exactly two 7s out of 4 is ^4C_2

The fifth card is from remaining cards i.e. 52-8=44

Getting fifth card is ^{44}C_1

Total number of cards in hand contain exactly two 3s and two 7s is

T=^4C_2+^4C_2+^{44}C_1

T=\frac{4!}{2!(4-2)!}+\frac{4!}{2!(4-2)!}+\frac{44!}{1!(44-1)!}

T=\frac{4\times3\times2!}{2!\times 2}+\frac{4\times3\times2!}{2!\times 2}+\frac{44\times43!}{1\times 43!}

T=6+6+44

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Therefore, Total number of cards in hand contain exactly two 3s and two 7s is 56.

8 0
3 years ago
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