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Goshia [24]
3 years ago
11

The environmental club at Paul sells gum and candy bars as a fundraiser. One afternoon, they sold $200 worth of candy. They need

to report how many of each item they sold for their club records. Maura knows that they sold 295 items that day. Candy bars sell for 75 cents, and gum sells for 50 cents. How many candy bars and gums did Paul sell?
Mathematics
1 answer:
Marysya12 [62]3 years ago
8 0

Answer:

210 candy bars

85 gum

Step-by-step explanation:

$200=0.75cb+0.50gum

cb+gum=295

gum=295-cb

substitute

$200=0.75cb+0.50(295-cb)

$200=0.75cb+147.5-0.50cb

$200=0.25cb+147.5

0.25cb=52.5

cb=210

gum=295-210=85

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Answer:

a) No.

b) Yes.

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a) It's not proportional. Although the graph represents a straight line, it does not pass through the origin (0,0).

b) It is proportional. It is a straight line that passes through the origin.

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kati45 [8]

Answer:

15 = a_1 r^4 (1)

1 = a_1 r^5 (2)

If we divide equations (2) and (1) we got:

\frac{r^5}{r^4}= \frac{1}{15}

And then r= \frac{1}{15}

And then we can find the value a_1 and we got from equation (1)

a_1 = \frac{15}{r^4} = \frac{15}{(\frac{1}{15})^4} =759375

And then the general term for the sequence would be given by:

a_n = 759375 (\frac{1}{15})^n-1 , n=1,2,3,4,...

And the best option would be:

C) a1=759,375; an=an−1⋅(1/15)

Step-by-step explanation:

the general formula for a geometric sequence is given by:

a_n = a_1 r^{n-1}

For this case we know that a_5 = 15, a_6 = 1

Then we have the following conditions:

15 = a_1 r^4 (1)

1 = a_1 r^5 (2)

If we divide equations (2) and (1) we got:

\frac{r^5}{r^4}= \frac{1}{15}

And then r= \frac{1}{15}

And then we can find the value a_1 and we got from equation (1)

a_1 = \frac{15}{r^4} = \frac{15}{(\frac{1}{15})^4} =759375

And then the general term for the sequence would be given by:

a_n = 759375 (\frac{1}{15})^n-1 , n=1,2,3,4,...

And the best option would be:

C) a1=759,375; an=an−1⋅(1/15)

4 0
3 years ago
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