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Andrews [41]
3 years ago
6

The amount of fun P. Art Eman has at a party is inversely proportional to the number of people at the party. So, if he is having

twice as much fun at a birthday party as he had at last week's holiday party, what is the relationship in the number of people at the two parties?
a. Twice as many at the second party
b. Four times as many at the second party
c. Half as many at the second party
d. The numbers are the same.
Mathematics
1 answer:
Andrej [43]3 years ago
3 0

Answer:

There are half as many at the birthday party.

Step-by-step explanation:

The amount of fun that Eman has at a party is inversely proportional to the number of people at the party.

If F represents the amount of fun and N represents the number of people in the party, then F ∝ 1/ N

So, comparing last week's holiday party where fun was F_{h} and number of people was N_{h} and the birthday party where fun is F_{b} and the number of people is N_{b}, we can write

F_{h}N_{h} = F_{b}N_{b}

⇒ N_{b} = \frac{F_{h} }{F_{b}} \times N_{h} = \frac{N_{h}}{2}

{Since, Eman is having twice as much fun at a birthday party as he had at last week's holiday party}

Therefore, there are half as many people at the birthday party. (Answer)

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At a carry-out pizza restaurant, an order of 3 slices of pizza, 4 breadsticks, and 2 juice drinks costs $13.35. A second order o
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Answer: Pizza = $2.95, Breadstick = $0.50, Juice = $1.25

<u>Step-by-step explanation:</u>

Let P represent the cost of a slice of pizza

and B represent the cost of breadstick

and J represent the cost of a juice drink.

EQ1: 3P + 4B + 2J = 13.35

EQ2: 5P + 2B + 3J = 19.50

EQ3: 4B + J = P + 0.30     -->     P - 4B - J = -0.30

Let's eliminate B from EQ1 and EQ2 to form EQ4:

3P + 4B + 2J = 13.35   →  1(3P + 4B + 2J = 13.35)   →   3P + 4B + 2J =  13.35

5P + 2B + 3J = 19.50  → -2(5P + 2B + 3J = 19.50)  →  <u>-10P -  4B -  6J</u> = <u>-39.00</u>

                                                                        EQ4:     -7P          -4J    = -25.65

And eliminate B from EQ1 and EQ3 to form EQ5:

               3P + 4B + 2J = 13.35  

             <u>    P -  4B  - J </u> =<u> -0.30</u>

EQ5:       4P           + J  = 13.05    

Now, eliminate J from EQ4 and EQ5 to solve for P:

-7P  - 4J = -25.65  →    1(-7P  - 4J = -25.65)   →   -7P  - 4J = -25.65

 4P  +  J = 13.05   →    4(4P  +   J =  13.05)    →  <u>  16P  +4J </u>=  <u>52.20</u>

                                                                             9P          = 26.55

                                                                           <u>÷9   </u>           <u>÷9     </u>

                                                                                    P = 2.95

Plug in P = 2.95 into EQ4 or EQ5 to solve for J:

EQ5: 4P + J = 13.05

        4(2.95) + J = 13.05

        11.80 + J = 13.05

                   J = 1.25

Plug in P = 2.95 and J = 1.25 into either EQ1 or EQ2 or EQ3 to solve for B:

EQ3: 4B + J = P + 0.30

        4B + 1.25 = 2.95 + 0.30

        4B + 1.25 = 3.25

                   4B = 2.00

                     B = 0.50

Check (since we used EQ3 to find B, use either EQ1 or EQ2):

EQ2: 5P + 2B + 3J = 19.50

       5(2.95) + 2(0.50) + 3(1.25) = 19.50

       14.75 + 1.00 + 3.75 = 19.50

                            19.50 = 19.50     \checkmark

6 0
3 years ago
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