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4 years ago

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A new car is purchased for 17000 dollars. The value of the car depreciates at 13.5% per year. What will the value of the car be,

to the nearest cent, after 12 years?

1 answer:

brilliants [131]4 years ago

3 0

The value of car after 12 years is $ 2982.92

<em><u>Solution:</u></em>

Given that,

A new car is purchased for 17000 dollars

The value of the car depreciates at 13.5% per year

<em><u>To find: Value of car after 12 years</u></em>

<em><u>The decreasing function is given as:</u></em>

$y = a(1-r)^t$

Where,

y is the future value

a is the initial value

r is the decreasing rate in decimal

t is the number of years

From given,

$a = 17000\\\\t = 12\\\\r = 13.5 \% = \frac{13.5}{100} = 0.135$

<em><u>Substituting the values we get,</u></em>

$y = 17000(1-0.135)^{12}\\\\y = 17000(0.865)^{12}\\\\y = 17000 \times 0.1754661\\\\y = 2982.9241 \approx 2982.92$

Thus value of car after 12 years is $ 2982.92

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Answer:

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Step-by-step explanation:

4a(3b+5)+2b

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Order of operations (PEMDAS)

4(6)[21+5]+2(7)

4(6)+26+2(7)

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Hope this helps. Have a nice day. May it be filled with joy and happiness. Please, be safe and wash your hands. Happy Holidays.

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Answer:

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Step-by-step explanation:

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Let height equal h & base equal b

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Solve,

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Answer:

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Step-by-step explanation:

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In which $L_{1}, L_{2}$ and $L_{3}$ are the number of miligrams of levonorgestrel taken in each phase.

The total milligrams of ethinyl estradiol is:

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In which $E_{1},E_{2}$ and $E_{3}$ are the number of miligrams of ethinyl estradiol taken in each phase.

We can solve this by phase.

Phase 1: 6 tablets, each containing 0.050 mg of levonorgestrel and 0.030 mg ethinyl estradiol

In this phase,

$6*0.050mg = 0.30mg$ of levonorgestrel and $6*0.030mg = 0.18mg$ of ethinyl estradiol are taken.

So $L_{1} = 0.30$ and $E_{1} = 0.18$

Phase 2: 5 tablets, each containing 0.075 mg of levonorgestrel and 0.040 mg ethinyl estradiol

In this phase,

$5*0.075mg = 0.375mg$ of levonorgestrel and $5*0.040mg = 0.20mg$ of ethinyl estradiol are taken.

So $L_{2} = 0.375$ and $E_{2} = 0.20$

Phase 3: 10 tablets, each containing 0.125 mg of levonorgestrel and 0.030 mg ethinyl estradiol

In this phase,

$10*0.125mg = 1.25mg$ of levonorgestrel and $10*0.030mg = 0.30mg$ of ethinyl estradiol are taken.

So $L_{3} = 1.25$ and $E_{3} = 0.30$

The total milligrams of levonorgestrel is:

$L = L_{1} + L_{2} + L_{3} = 0.30 + 0.375 + 1.25 = 1.925$

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$E = E_{1} + E_{2} + E_{3} = 0.18 + 0.20 + 0.30 = 0.68$

1.925mg of levonorgestrel and 0.68mg of ethinyl estradiol are taken during the 28-day period.

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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

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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.

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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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