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____ [38]
3 years ago
15

A caterer needs to buy 21 pounds of pasta to cater a wedding. At a local store, 8 pounds of pasta cost $12. How much will the ca

terer pay for the pasta there?
1. Write a ratio for the given information about the cost of pasta



2. Would it be more helpful to write an equivalent ratio with 1 pound of pasta as one of the numbers, or with $1 as one of the numbers? Explain your reasoning, then write that equivalent ratio.




3. Find the answer and explain or show your reasoning.
Mathematics
2 answers:
Serggg [28]3 years ago
7 0

Answer:

1. 8/$12=21/$x then cross multiply.

2. 1/$1.5, we're looking for cost so $1 is more reasonable

3. $31.5

Step-by-step explanation:

12x 21/ 8 = 31.5

zalisa [80]3 years ago
5 0

Answer:

<h2>$31.50 </h2>

Step-by-step explanation:

To solve this problem, we need to use the rule of three, which is based on ratios and proportional relationships.

The ratio that shows the cost of pasta is:

\frac{\$12}{8 lb}

Which means that they charge $12 per 8 pounds of pasta.

Now, it would be help if we use the given ratio to apply the rule of three. Looking for a ratio per pound of pasta would be waste of time, and could confuse a lot. So, we just use the given ratio.

If 8 pounds of pasta costs $12, how much would be 21 pounds of pasta?

x=21 \ lb\frac{\$12}{8 \ lb}=\$ 31.50

Therefore, according to the given ration, it would cost $31.50 to get 21 pounds of pasta.

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There are 8 ounces in 1 cup. The table shows some conversions between ounces and cups. Which equation describes how to convert b
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Z=8C

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Five thousand raffle tickets are sold for $5 each. 50 winnings numbers will be chosen, 40 of which will win $100 each and 10 of
Allisa [31]

Answer:

Step-by-step explanation:

Given

there are 50 winning number out of which 40 get $ 100 each and 10 wins $ 1000 .

let x denotes the winning from a randomly selected ticket

x=100\ with\ Probability\ \frac{40}{5000}

x=1000\ with\ Probability\ \frac{10}{5000}

Expected value is given by

E(x)=100\times \frac{40}{5000}+1000\times \frac{10}{5000}

E(x)=0.8+2=2.8

and Expected return is given by

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3 years ago
For the following telescoping series, find a formula for the nth term of the sequence of partial sums
gtnhenbr [62]

I'm guessing the sum is supposed to be

\displaystyle\sum_{k=1}^\infty\frac{10}{(5k-1)(5k+4)}

Split the summand into partial fractions:

\dfrac1{(5k-1)(5k+4)}=\dfrac a{5k-1}+\dfrac b{5k+4}

1=a(5k+4)+b(5k-1)

If k=-\frac45, then

1=b(-4-1)\implies b=-\frac15

If k=\frac15, then

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This means

\dfrac{10}{(5k-1)(5k+4)}=\dfrac2{5k-1}-\dfrac2{5k+4}

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The sum telescopes so that

S_n=\dfrac2{14}-\dfrac2{5n+4}

and as n\to\infty, the second term vanishes and leaves us with

\displaystyle\sum_{k=1}^\infty\frac{10}{(5k-1)(5k+4)}=\lim_{n\to\infty}S_n=\frac17

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3 years ago
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