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expeople1 [14]
3 years ago
8

There were 75 roasted turkeys at Ally's Eats. Bob's Goods had 4/5 as many roasted turkeys as Ally's Eats. If Ally's Eats had 5/6

as many roasted turkeys as Charlotte's Market what was the average number of roasted turkeys owned by each of the three shops? Please help! My teacher thinks ya girl is a genius!
Mathematics
1 answer:
Sever21 [200]3 years ago
7 0

Answer:

75 roasted turkeys.

Step-by-step explanation:

Ally's Eats had 75 roasted turkeys.

Bob's Goods had 4/5 as many roasted turkeys as Ally's Eats, hence, the number of roasted turkeys Bob's Goods had is:

4/5 * 75 = 60 roasted turkeys

Ally's Eats had 5/6 as many roasted turkeys as Charlotte's Market. In other words, Charlotte's market had 6/5 as many roasted turkeys as Ally's Eats:

6/5 * 75 = 90 roasted turkeys.

The average is given as the sum of all toasted turkeys in the three shops divided by 3. Hence:

A = (75 + 60 + 90) / 3

A = 225 / 3 = 75 roasted turkeys

The average number of roasted turkeys in the three shops is 75.

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Let x be total hours of battery life. We have been given that battery life of Vera's phone is down to 2/5. So remaining battery life,

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We have been also given that it drains another 1/9 every other hour. Now let us substitute our given information into an equation.

\frac{1}{9}\cdot x=\frac{3}{5}

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x=\frac{3\cdot 9}{5}=\frac{27}{5}

x=(5+\frac{2}{5})hours

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Answer:

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Step-by-step explanation:

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Answer:

Step-by-step explanation:

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-----------

check

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