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lukranit [14]
3 years ago
14

A group of students bakes 100 cookies to sell at the school bake sale. The students want to ensure that the price of each cookie

offsets the cost of the ingredients. If all the cookies are sold for $0.10 each, the net result will be a loss of $4. If all the cookies are sold for $0.50 each. The students will make a $36 profit. First, write the linear function p(x) that represents the net profit from selling all the cookies, where x is the price of each cookie. Then, determine how much profit the students will make if they sell the cookies for $0.60 each. Explain. Tell how your answer is reasonable.
Mathematics
1 answer:
Goryan [66]3 years ago
7 0

Answer:

46

Step-by-step explanation:

-Let b be the constant in the linear equation.

#The linear equation can be expressed as:

p(x)=100x+b

Substitute the values in the equation to find b:

p(x)=100x+b\\\\-4=100(0.1)+b\\\\b=-14\\\\\#or\\\\36=100(0.5)+b\\\\b=-14

We know have the constant value b=-14, substitute the values of b and x in the p(x) function:

p(x)=100x+b\\\\p(x)=100(0.6)-14\\\\p(x)=60-14\\\\p(x)=46

Hence, the profit when selling price is $0.60 is $46

#From our calculations, it's evident that the cookies production has a very high fixed cost which can only be offset by raisng the selling price or the number of units sold at any given time.

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In a multiple choice quiz there are 5 questions and 4 choices for each question (a, b, c, d). Robin has not studied for the quiz
Ahat [919]

Answer:

a) There is a 18.75% probability that the first question that she gets right is the second question.

b) There is a 65.92% probability that she gets exactly 1 or exactly 2 questions right.

c) There is a 10.35% probability that she gets the majority of the questions right.

Step-by-step explanation:

Each question can have two outcomes. Either it is right, or it is wrong. So, for b) and c), we use the binomial probability distribution to solve this problem.

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}.\pi^{x}.(1-\pi)^{n-x}

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

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

And \pi is the probability of X happening.

In this problem we have that:

Each question has 4 choices. So for each question, Robin has a \frac{1}{4} = 0.25 probability of getting ir right. So \pi = 0.25. There are five questions, so n = 5.

(a) What is the probability that the first question she gets right is the second question?

There is a 75% probability of getting the first question wrong and there is a 25% probability of getting the second question right. These probabilities are independent.

So

P = 0.75(0.25) = 0.1875

There is a 18.75% probability that the first question that she gets right is the second question.

(b) What is the probability that she gets exactly 1 or exactly 2 questions right?

This is: P = P(X = 1) + P(X = 2)

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

P(X = 1) = C_{5,1}.(0.25)^{1}.(0.75)^{4} = 0.3955

P(X = 2) = C_{5,2}.(0.25)^{2}.(0.75)^{3} = 0.2637

P = P(X = 1) + P(X = 2) = 0.3955 + 0.2637 = 0.6592

There is a 65.92% probability that she gets exactly 1 or exactly 2 questions right.

(c) What is the probability that she gets the majority of the questions right?

That is the probability that she gets 3, 4 or 5 questions right.

P = P(X = 3) + P(X = 4) + P(X = 5)

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

P(X = 3) = C_{5,3}.(0.25)^{3}.(0.75)^{2} = 0.0879

P(X = 4) = C_{5,4}.(0.25)^{4}.(0.75)^{1} = 0.0146

P(X = 5) = C_{5,5}.(0.25)^{5}.(0.75)^{0} = 0.001

P = P(X = 3) + P(X = 4) + P(X = 5) = 0.0879 + 0.0146 + 0.001 = 0.1035

There is a 10.35% probability that she gets the majority of the questions right.

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