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2 years ago

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Henry buys 6 circus tickets for himself and five friends for a total of $174. Each friend pays Henry back for his or her ticket.

If one of Henry's friends gives him two $20 bills, how much change should Henry return?

1 answer:

Deffense [45]2 years ago

5 0

Answer:

$11

Step-by-step explanation:

According to the statement, you have that 6 tickets for the price of each ticket is equal to $174 and from that, you can determine the price of each ticket:

6*P=174

P= price of each ticket

P= 174/6= 29

Now, as it says that one of Henry's friends gives him two $20 bills, that means that he gives him $40 and you have to subtract the price of the ticket from that:

40-29= 11

The answer is that if one of Henry's friends gives him two $20 bills, Henry has to return $11.

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no. of Classes = x

cans = 220

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Step-by-step explanation:

Given that,

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1 year ago

The Slow Ball Challenge or The Fast Ball Challenge.

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Answer:

Fast ball challenge

Step-by-step explanation:

Given

Slow Ball Challenge

$Pitches = 7$

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$Win = \$60$

$Lost = \$10$

Fast Ball Challenge

$Pitches = 3$

$P(Hit) = 70\%$

$Win = \$60$

$Lost = \$10$

Required

Which should he choose?

To do this, we simply calculate the expected earnings of both.

Considering the slow ball challenge

First, we calculate the binomial probability that he hits all 7 pitches

$P(x) =^nC_x * p^x * (1 - p)^{n - x}$

Where

$n = 7$ --- pitches

$x = 7$ --- all hits

$p = 80\% = 0.80$ --- probability of hit

So, we have:

$P(x) =^nC_x * p^x * (1 - p)^{n - x}$

$P(7) =^7C_7 * 0.80^7 * (1 - 0.80)^{7 - 7}$

$P(7) =1 * 0.80^7 * (1 - 0.80)^0$

$P(7) =1 * 0.80^7 * 0.20^0$

Using a calculator:

$P(7) =0.2097152$ --- This is the probability that he wins

i.e.

$P(Win) =0.2097152$

The probability that he lose is:

$P(Lose) = 1 - P(Win)$ ---- Complement rule

$P(Lose) = 1 -0.2097152$

$P(Lose) = 0.7902848$

The expected value is then calculated as:

$Expected = P(Win) * Win + P(Lose) * Lose$

$Expected = 0.2097152 * \$60 + 0.7902848 * \$10$

Using a calculator, we have:

$Expected = \$20.48576$

Considering the fast ball challenge

First, we calculate the binomial probability that he hits all 3 pitches

$P(x) =^nC_x * p^x * (1 - p)^{n - x}$

Where

$n = 3$ --- pitches

$x = 3$ --- all hits

$p = 70\% = 0.70$ --- probability of hit

So, we have:

$P(3) =^3C_3 * 0.70^3 * (1 - 0.70)^{3 - 3}$

$P(3) =1 * 0.70^3 * (1 - 0.70)^0$

$P(3) =1 * 0.70^3 * 0.30^0$

Using a calculator:

$P(3) =0.343$ --- This is the probability that he wins

i.e.

$P(Win) =0.343$

The probability that he lose is:

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$P(Lose) = 1 - 0.343$

$P(Lose) = 0.657$

The expected value is then calculated as:

$Expected = P(Win) * Win + P(Lose) * Lose$

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Using a calculator, we have:

$Expected = \$27.15$

So, we have:

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