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pantera1 [17]
2 years ago
15

Brandon earns a rebate each month on his meals and movie tickets. He earns $0.50 for each movie ticket and $2 for each meal. Las

t month Brandon purchased 22 movie tickets and earned $25.
Part A: Create an equation that will determine the number of meals he purchased. (3 points)

Part B: Solve this equation justifying each step with an algebraic property of equality. (6 points)
Mathematics
1 answer:
AURORKA [14]2 years ago
7 0

Let ticket be x and meal be y

So $0.50 for each movie ticket = 0.50x

$2 for each meal = 2y


He purchased 22 tickets, so x = 22

0.50 * 22 = $ 11


Now the total value is $25 (with both ticket and meal)

total = t

0.50x + 2y = t

you know x = 22 and t = 25

11 + 2y = 25


Part A)

2y = 25 -11

y = 14/2

y = 7

Number of meal is 7


Part B)

(don't know about part B)

0.50x + 2y = 25

0.50(22) + 2(7) = 25

11 + 14 = 25

25 = 25

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Step-by-step explanation:

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P(X = x) = 12Cx . 0.5^x . (1 - 0.5)^(12- x)

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since we are interested in the probability of the number of times an even number occurs

it can occur either as P(X = 0), P(X =1), P(X =2), P(X =3), P(X =4), P(X =5), P(X =6), P(X =7), P(X =8), P(X =9), P(X =10), P(X =11), and P(X =12)

For no even number in 12 rolls,

P(X = 0) = 12C0 . 0.5^0 . 0.5^(12- 0) = 0.000244

For one even number in 12 rolls,

P(X = 1) = 12C1 . 0.5^1 . 0.5^(12- 1) = 0.002930

For two even number in 12 rolls,

P(X = 2) = 12C2 . 0.5^2 . 0.5^(12- 2) = 0.016113  

For three even number in 12 rolls,

P(X = 3) = 12C3 . 0.5^3 . 0.5^(12- 3) = 0.053711  

For four even number in 12 rolls,

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For five even number in 12 rolls,

P(X = 5) = 12C5 . 0.5^5 . 0.5^(12- 5) = 0.193359

For six even number in 12 rolls,

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For seven even number in 12 rolls,

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For eight even number in 12 rolls,

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i.e.

P(X = 0) + P(X =1) + P(X =2) + P(X =3) + P(X =4) + P(X =5) + P(X =6) + P(X =7) + P(X =8) + P(X =9) + P(X =10) + P(X =11) + P(X =12) = 1

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the probability value stands

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