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Triss [41]
4 years ago
12

Michael spends an average of $120 per month going out to lunch and dinner. If he can cut that

Mathematics
1 answer:
expeople1 [14]4 years ago
6 0

Answer:

If he spends $120 dollars a month, and he cut's that down to $50 a month, we should create a pattern to visualize the months.

Step-by-step explanation:

120- 50= 70 ($70 saved each month)

Month 1- $70 saved total

Month 2- $140 saved total

Month 3- $210 saved total

Month 4- $280 saved total

Month 5- $350 saved total

Month 6- $420 saved total

It takes 6 months to get 400 saved. We could also divide to get

5.714285, but it would takes 6 months before reaching the goal. :)

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$10.05 divided by 15 equals 0.67 cents.

0.67 x 15 = $10.05

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How much time will have elapsed from 2:37 to 3:19
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Simplify the expression 2x+3(x-6)
Bingel [31]

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3 years ago
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.

6 0
3 years ago
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