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son4ous [18]
3 years ago
6

Caleb went to the grocery store to purchase hamburger meat and hamburger buns for an upcoming barbecue with his friends. At the

grocery store, hamburger meat sells for $1.49 a pound and an 8-pack of hamburger buns sells for $2.99. He plans on making 4 hamburgers from each pound of hamburger meat and for each person, p, to eat 1 hamburger. He used the expression below to determine the cost per person of the hamburger meat and hamburger buns.
Mathematics
2 answers:
ikadub [295]3 years ago
8 0

Answer:

The cost per hamburger patty.

Step-by-step explanation:

Well if it's 1.49 a pound and a pound feeds 4 people then 1.49/4 is the cost per patty.

Eva8 [605]3 years ago
6 0
About $0.75 per person.



First, let’s find the cost for each portion of hamburger meat and the cost for each bun.

We can do this for the meat by doing 1.49/4 to get 0.3725.

Now we can do the same with the buns by doing 2.99/8 to get 0.37375.

Now just add these together. 0.3725+0.37375 is 0.74625, which can be rounded to $0.75.

The cost per person is approximately $0.75.



Please consider marking this answer as Brainliest to help me advance.
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Five cards are drawn from a standard 52-card playing deck. A gambler has been dealt five cards—two aces, one king, one 3, and on
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Answer:

The probability that he ends up with a full house is 0.0083.

Step-by-step explanation:

We are given that a gambler has been dealt five cards—two aces, one king, one 3, and one 6. He discards the 3 and the 6 and is dealt two more cards.

We have to find the probability that he ends up with a full house (3 cards of one kind, 2 cards of another kind).

We know that gambler will end up with a full house in two different ways (knowing that he has given two more cards);

  • If he is given with two kings.
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Only in these two situations, he will end up with a full house.

Now, there are three kings and two aces left which means at the time of drawing cards from the deck, the available cards will be 47.

So, the ways in which we can draw two kings from available three kings is given by =  \frac{^{3}C_2 }{^{47}C_2}   {∵ one king is already there}

              =  \frac{3!}{2! \times 1!}\times \frac{2! \times 45!}{47!}           {∵ ^{n}C_r = \frac{n!}{r! \times (n-r)!} }

              =  \frac{3}{1081}  =  0.0028

Similarly, the ways in which one king and one ace can be drawn from available 3 kings and 2 aces is given by =  \frac{^{3}C_1 \times ^{2}C_1 }{^{47}C_2}

                                                                   =  \frac{3!}{1! \times 2!}\times \frac{2!}{1! \times 1!} \times \frac{2! \times 45!}{47!}

                                                                   =  \frac{6}{1081}  =  0.0055

Now, probability that he ends up with a full house = \frac{3}{1081} + \frac{6}{1081}

                                                                                    =  \frac{9}{1081} = <u>0.0083</u>.

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Answer:

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