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Ivahew [28]
4 years ago
12

Jeri finds a pile of money with at least $\$200$. If she puts $\$50$ of the pile in her left pocket, gives away $\frac23$ of the

rest of the pile, and then puts the rest in her right pocket, she'll have more money than if she instead gave away $\$200$ of the original pile and kept the rest. What are the possible values of the number of dollars in the original pile of money? (Give your answer as an interval.)
Mathematics
1 answer:
Marizza181 [45]4 years ago
6 0

Answer:

The possible values of the number of dollars in the original pile of money is ≥ $200 but < $350

Step-by-step explanation:

Here we have, pile of money ≥ $200

Amount in put the left pocket = $50

Fraction given away = 2/3 of rest of pile ≥ 2/3×150 ≥ $100

Amount put in right pocket = ≥ $150 - $100 ≥ $50

Total amount remaining with Jeri = $50 +≥ $50 ≥ $100

Also original pile - $200 < $100

Therefore where maximum amount given away to have more money = $200 we have

2/3× (original pile - 50) = $200

Maximum amount for original pile = $350

Therefore the possible values of the number of dollars in the original pile of money is ≥ $200 but < $350.

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a cone has a volume of 1230.88 units cubed and a diameter of 14 units what is the height of the cone use 3.14 for pi
ANEK [815]

Answer:

h= 24

Step-by-step explanation:

v = \pi {r}^{2}  \frac{h}{3}

r =  \frac{d}{2}

r =  \frac{14}{2}

r= 7

1230.88=3.14×7^2×h/3

1230.88= 153.86h/3

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3 years ago
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ella [17]
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E
Evgesh-ka [11]

Answer:

He bought 20 shirts

Step-by-step explanation:

shirts -- x

pants -- 3/4 of x

Therefore,

He bought --- 28(3/4 of x) pants and 40(x) shirts.

So, 28(3/4 of x) + 40(x) = 1220

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3 years ago
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sertanlavr [38]
<u>→ Chapter : Fractions ←</u>
<u><em>≡ We know that:</em></u>
⇔ (\frac{a}{b}) \div (\frac{c}{d})=(\frac{a}{b}) \times (\frac{d}{c})

<u><em>≡ Solution:</em></u>
⇒ (\frac{3}{7}) \times (d)=\frac{9}{14}
⇒ d=(\frac{9}{14}) \div (\frac{3}{7})
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