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Marrrta [24]
3 years ago
6

Coby sells apples and bananas at his fruit store. He charges $6 for a bunch of bananas, and $8 for a pound of apples. If a custo

mer buys a total of 13 pieces of fruit and is charged $92, how many bunches of bananas did he buy? Also, how many pounds of apples did he get?
Mathematics
1 answer:
Over [174]3 years ago
6 0

Answer:

  • 6 bunches of bananas
  • 7 pounds of apples

Step-by-step explanation:

We have to assume that a "piece of fruit" is either a bunch of bananas or a pound of apples. Without that assumption, there is insufficient information to work the problem.

Let B represent the number of bunches of bananas. Then 13-B is the number of pounds of apples. The total cost is ...

  6B +8(13 -B) = 92

  -2B + 104 = 92 . . . . . eliminate parentheses

  B = -12/-2 = 6 . . . . . . subtract 104, then divide by the coefficient of B

  13-B = 7 . . . . . . . . . . . the number of pounds of apples

The customer bought 6 bunches of bananas and 7 pounds of apples.

_____

<em>Comment on the solution</em>

You will note that finding the value of the variable involved arithmetic with negative numbers. If you want the numbers to stay positive, then you can choose the variable to represent <em>the most expensive</em> of the items: the number of pounds of apples.

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A center for medical services reported that there were 295,000 appeals for hospitalization and other services. For this group, 4
pshichka [43]

Answer:

a) 0.0025 = 0.25% probability that none of the appeals will be successful.

b) 0.0207 = 2.07% probability that exactly one of the appeals will be successful.

c) 0.9768 = 97.68% probability that at least two of the appeals will be successful

Step-by-step explanation:

For each appeal, there are only two possible outcomes. Either it is succesful, or it is not. The probability of an appeal being succesful is independent of other appeals, so we use the binomial probability distribution to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

In which C_{n,x} is the number of different combinations of x objects from a set of n elements, given by the following formula.

C_{n,x} = \frac{n!}{x!(n-x)!}

And p is the probability of X happening.

45% of first-round appeals were successful.

This means that p = 0.45

Suppose 10 first-round appeals have just been received by a Medicare appeals office.

This means that n = 10

(a) Compute the probability that none of the appeals will be successful.

This is P(X = 0).

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

P(X = 0) = C_{10,0}.(0.45)^{0}.(0.55)^{10} = 0.0025

0.0025 = 0.25% probability that none of the appeals will be successful.

(b) Compute the probability that exactly one of the appeals will be successful.

This is P(X = 1).

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

P(X = 1) = C_{10,1}.(0.45)^{1}.(0.55)^{9} = 0.0207

0.0207 = 2.07% probability that exactly one of the appeals will be successful.

(c) What is the probability that at least two of the appeals will be successful

This is P(X \geq 2)

Either less than two appeals are succesful, or at least two are. The sum of the probabilities of these events is decimal 1. So

P(X < 2) + P(X \geq 2) = 1

P(X < 2) = P(X = 0) + P(X = 1) = 0.0025 + 0.0207 = 0.0232

P(X \geq 2) = 1 - P(X < 2) = 1 - 0.0232 = 0.9768

0.9768 = 97.68% probability that at least two of the appeals will be successful

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