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Archy [21]
10 months ago
14

The value of a car is depreciating at a rate of 5% per year. In 2010, the car was worth $32,000. Find the value of the car in 20

13.step 1: decide whether it is growing or decaying:step 2: solve for the rate:step 3: solve:
Mathematics
1 answer:
PSYCHO15rus [73]10 months ago
8 0

Data:

Depreciating: The value decrease

Rate: 5% per year

2010: $32,000

2013: $?

Step 1: As depreciating means decrease, the value of the car is decaying

Step 2 and step 3: You use the formula below to find the value of the car after 3 years (as the value is decaying the rate in the formula is substracting)

Final value: F

Principal value: P=32000

rate: r=5

Time in years: n=3 (2013-2010=3)

\begin{gathered} F=P(1-\frac{r}{100})^n \\  \\ F=32000(1-\frac{5}{100})^3 \\  \\ F=32000(0.95)^3 \\  \\ F=27436 \end{gathered}

In 2013 the value of the car is $27,436
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According to a study, 50 % of adult smokers started smoking before 21 years old. 5 smokers 21 years old or older are randomly se
belka [17]

Answer:

a) The probability that at least 2 of them started smoking before 21 years of age is 0.1875 = 18.75%.

b) The probability that at most 4 of them started smoking before 21 years of age is 0.96875 = 96.875%.

c) The probability that exactly 3 of them started smoking before 21 years of age is 0.3125 = 31.25%.

Step-by-step explanation:

For each smoker, there are only two possible outcomes. Either they started smoking before 21 years old, or they did not. Smokers are independent, which means that the binomial probability distribution is used 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 adult smokers started smoking before 21 years old.

This means that p = 0.5

5 smokers 21 years old or older are randomly selected, and the number of smokers who started smoking before 21 is recorded.

This means that n = 5.

a) The probability that at least 2 of them started smoking before 21 years of age is

This is:

P(X \geq 2) = 1 - P(X < 2)

In which

P(X < 2) = P(X = 0) + P(X = 1)

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

P(X = 0) = C_{5,0}.(0.5)^{0}.(0.5)^{5} = 0.03125

P(X = 1) = C_{5,1}.(0.5)^{1}.(0.5)^{4} = 0.15625

P(X < 2) = P(X = 0) + P(X = 1) = 0.03125 + 0.15625 = 0.1875

The probability that at least 2 of them started smoking before 21 years of age is 0.1875 = 18.75%.

b) The probability that at most 4 of them started smoking before 21 years of age is

This is:

P(X \leq 4) = 1 - P(X = 5)

In which

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

P(X = 5) = C_{5,5}.(0.5)^{5}.(0.5)^{0} = 0.03125

P(X \leq 4) = 1 - P(X = 5) = 1 - 0.03125 = 0.96875

The probability that at most 4 of them started smoking before 21 years of age is 0.96875 = 96.875%.

c) The probability that exactly 3 of them started smoking before 21 years of age is

This is P(X = 3). So

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

P(X = 3) = C_{5,3}.(0.5)^{3}.(0.5)^{2} = 0.3125

The probability that exactly 3 of them started smoking before 21 years of age is 0.3125 = 31.25%.

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Which expressions are equivalent to the given expression?
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-51 - 6i AND -3 + 2i - 2(24) - 8i are correct :)

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51%, 0.57, 0.61, 6/10 least to greatest
Evgen [1.6K]

Answer:

least is 51%

Step-by-step explanation:

so 51%, .57,6/10 and .61

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