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leva [86]
3 years ago
13

Sam started with savings of $2,500 and adds $40 to his savings each week. Marshall started with $7,300 in savings and spends $60

of his savings each week. After how many weeks will both Sam and Marshall have the same amount of savings in their accounts?
Mathematics
1 answer:
Oliga [24]3 years ago
5 0

The number of weeks it will take both Sam and Marshall to have the same amount of savings in their accounts is 48 weeks

Given:

<em>Amount with Sam </em>= $2,500

<em>Additional savings each week</em> = $40

<em>Amount with Marshal</em> = $7,300

<em>Amount spent each week</em> = $60

let

x = <em>number of weeks</em>

<em>Sam equation:</em>

2500 + 40x

<em>Marshall's equation:</em>

7300 - 60x

Equate both

2500 + 40x = 7300 - 60x

<em>collect like terms</em>

2500 - 7300 = - 60x - 40x

-4800 = -100x

divide both sides by - 100

x = -4800 / -100

x = 48 weeks

Therefore, the number of weeks it will take both Sam and Marshall to have the same amount of savings in their accounts is 48 weeks

Read more:

brainly.com/question/8957185

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There are n questions on a multiple choice exam, and for each question, there are four choices. To pass the exam, one must corre
irinina [24]

Answer:

a) 0.352% probability of the student passing

b) 5

c) 1.875

Step-by-step explanation:

For each question, there are only two possible outcomes. Either the student answers it correctly, or he answers it wrong. The probability of correctly answering a question is independent from the probability of correctly answering other questions. 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.

The expected value of the binomial distribution is:

E(X) = np

The variance of the binomial distribution is:

V(X) = np(1-p)

For each question, there are four choices.

One choice is correct, so p = \frac{1}{4} = 0.25

a. Find the probability of the student passing for n = 10.

0.7*10 = 7

This is P(X \geq 7) when n = 10

P(X \geq 7) = P(X = 7) + P(X = 8) + P(X = 9) + P(X = 10)

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

P(X = 7) = C_{10,7}.(0.25)^{7}.(0.75)^{3} = 0.0031

P(X = 8) = C_{10,8}.(0.25)^{8}.(0.75)^{2} = 0.0004

P(X = 9) = C_{10,9}.(0.25)^{9}.(0.75)^{1} = 0.00002

P(X = 10) = C_{10,10}.(0.25)^{10}.(0.75)^{0} \cong 0

P(X \geq 7) = P(X = 7) + P(X = 8) + P(X = 9) + P(X = 10) = 0.0031 + 0.0004 + 0.00002 = 0.00352

0.352% probability of the student passing

b. Find the expected number of questions answered correctly for n = 20.

Here we have n = 20. So

E(X) = np = 20*0.25 = 5

c. Find the variance for the number of questions answered correctly for n = 10.

Here we have n = 10.

V(X) = np(1-p) = 10*0.25*0.75 = 1.875

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4 years ago
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QveST [7]
Hi,
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Solution: 200,000/year year=12 months
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Step-by-step explanation:

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