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Dima020 [189]
3 years ago
12

During a sale, 20-cent candy bars were sold at 3 for 50 cents. How much is saved on 9 bars?

Mathematics
1 answer:
Mazyrski [523]3 years ago
8 0

Answer:

Step-by-step explanation:

$0.30

Step-by-step explanation:

1 bar of candy = $0.20

3 bars of candy = $0.50

To solve, multiply for both:

If you pay for each candy bar individually, they each cost $0.20. Multiply 9 with 0.20:

9 x 0.20 = $1.80

If you pay for the candy bars by 3's, they cost $0.50 each pack. Divide 9 with 3, then multiply by 0.50:

9/3 = 3

3 x 0.50 = $1.50

Subtract the total cost of the individual from the pack:

$1.80 - $1.50 = $0.30

. $0.30 is your answer.

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16: Like before: you have pick a king with probability 4/52 = 1/13 (there are four kings - one for each suit, out of 52 cards in a standard deck), and you pick "I" from "INCREDIBLE" with probability 2/10 = 1/5 (there are two "I"s out of 10 letters). So, the probability of picking a king and then an "I" is

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Exercises 18 to 20 follow the same scheme: find out the probability of the two events and multiply them.

21-26: Not we DON'T replace the balls, so the second pick will suffer the effects of the first one. Let's dive into ex. 21 for example. For the first pick, we want a 2-digits number. There are 21 of such balls (all balls except balls 1 to 9), so we pick a 2-digits ball with probability 21/30 = 7/10. For the second pick, we want the balls number 4. But we have to assume that we already picked the first ball, and we picked a 2-digits ball. So, for the second pick, we're choosing from a bag with 29 balls, and there is only one ball labeled 4. So, we pick the ball number 4 with probability 1/29. We deduce that the two events happen one after the other with probability

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Exercises 22 to 26 are similar: you find out the probability of the first event, and then you consider the new environment (i.e. you keep track of the first pick) when it comes to the probability distribution for the second pick).

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