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4 years ago

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chad buys peanut in 2-pound bags.He repackages them into bags that hold 5/6 pound of peanuts.How many 2-pound bags of peanuts sh

ould chad buy so he can fill the 5/6 pund bags without having any peanuts left over.

1 answer:

Nataliya [291]4 years ago

6 0

First, let's find out how many 5/6 pound bags he can fill with on 2 pound bag.

2 divided by 5/6
2 x 6/5
2.4

He can fill 2.4 bags with one 2 bag.
Now we just need to multiply 2.4 by a number that will make it a whole number. You can use the guess and check method starting with 2 to figure your answer out, but 2.4 x 5 = 12 which is a whole number.

This means that Chad needs to buy 5 of the 2 pound bags to fill 12 of the 5/6 pound bags, and he will not have peanuts leftover

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Suppose that 50% of all young adults prefer McDonald's to Burger King when asked to state a preference. A group of 12 young adul

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Answer:

a) 0.194 = 19.4% probability that more than 7 preferred McDonald's

b) 0.787 = 78.7% probability that between 3 and 7 (inclusive) preferred McDonald's

c) 0.787 = 78.7% probability that between 3 and 7 (inclusive) preferred Burger King

Step-by-step explanation:

For each young adult, there are only two possible outcomes. Either they prefer McDonalds, or they prefer burger king. The probability of an adult prefering McDonalds is independent from other adults. 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.

50% of all young adults prefer McDonald's to Burger King when asked to state a preference.

This means that $p = 0.5$

12 young adults were randomly selected

This means that $n = 12$

(a) What is the probability that more than 7 preferred McDonald's?

$P(X > 7) = P(X = 8) + P(X = 9) + P(X = 10) + P(X = 11) + P(X = 12)$

In which

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

$P(X = 8) = C_{12,8}.(0.5)^{8}.(0.5)^{4} = 0.121$

$P(X = 9) = C_{12,9}.(0.5)^{9}.(0.5)^{3} = 0.054$

$P(X = 10) = C_{12,10}.(0.5)^{10}.(0.5)^{2} = 0.016$

$P(X = 11) = C_{12,11}.(0.5)^{11}.(0.5)^{1} = 0.003$

$P(X = 12) = C_{12,12}.(0.5)^{12}.(0.5)^{0} = 0.000$

$P(X > 7) = P(X = 8) + P(X = 9) + P(X = 10) + P(X = 11) + P(X = 12) = 0.121 + 0.054 + 0.016 + 0.003 + 0.000 = 0.194$

0.194 = 19.4% probability that more than 7 preferred McDonald's

(b) What is the probability that between 3 and 7 (inclusive) preferred McDonald's?

$P(3 \leq X \leq 7) = P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6) + P(X = 7)$

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

$P(X = 3) = C_{12,3}.(0.5)^{3}.(0.5)^{9} = 0.054$

$P(X = 4) = C_{12,4}.(0.5)^{4}.(0.5)^{8} = 0.121$

$P(X = 5) = C_{12,5}.(0.5)^{5}.(0.5)^{7} = 0.193$

$P(X = 6) = C_{12,6}.(0.5)^{6}.(0.5)^{6} = 0.226$

$P(X = 7) = C_{12,7}.(0.5)^{7}.(0.5)^{5} = 0.193$

$P(3 \leq X \leq 7) = P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6) + P(X = 7) = 0.054 + 0.121 + 0.193 + 0.226 + 0.193 = 0.787$

0.787 = 78.7% probability that between 3 and 7 (inclusive) preferred McDonald's

(c) What is the probability that between 3 and 7 (inclusive) preferred Burger King?

Since $p = 1-p = 0.5$ , this is the same as b) above.

So

0.787 = 78.7% probability that between 3 and 7 (inclusive) preferred Burger King

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Answer:

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Since it says that (x,2) is a point in the equation, we can substitute it into the equation

2=3/4x-25/4

add 25/4 to both sides

33/4=3/4x

multiply by 4/3

11=x

we can double check by plugging (11,2) into the equation of the line.

2=3/4(11)-25/4

2=33/4-25/4

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it works! :)

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