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bonufazy [111]
3 years ago
8

Devon purchased a new car valued at $16,000 that depreciated continuously at a rate of 35%. Its current value is $2,000. The equ

ation mc025-1.jpg represents the situation, where t is the age of the car in years and r is the rate of depreciation. About how old is Devon’s car? Use a calculator and round your answer to the nearest whole number.
Mathematics
2 answers:
PilotLPTM [1.2K]3 years ago
7 0

Cost =16000
rate of depreciation = 35%
written down value = 2000
depreciation for one year =16000*35% = 5600
5600*2 = 11200+5600*6\12 = 2000
the car is 2.5 years old. It has been depreciated for 2 whole years and half years.

Andrej [43]3 years ago
7 0

Answer:

5 years

Step-by-step explanation:

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Answer:

a) 0.7291 is the probability that more than half out of 10 vehicles carry just 1 person.

b) 0.996 is the probability that more than half of the vehicles  carry just one person.    

Step-by-step explanation:

We are given the following information:

A) Binomial distribution

We treat vehicle on road with one passenger as a success.

P(success) = 64% = 0.64

Then the number of vehicles follows a binomial distribution, where

P(X=x) = \binom{n}{x}.p^x.(1-p)^{n-x}

where n is the total number of observations, x is the number of success, p is the probability of success.

Now, we are given n = 10

We have to evaluate:

P(x \geq 6) = P(x =6) +...+ P(x = 10) \\= \binom{10}{6}(0.64)^6(1-0.64)^4 +...+ \binom{10}{10}(0.64)^{10}(1-0.79)^0\\=0.7291

0.7291 is the probability that more than half out of 10 vehicles carry just 1 person.

B) By normal approximation

Sample size, n = 92

p = 0.64

\mu = np = 92(0.64) = 58.88

\sigma = \sqrt{np(1-p)} = \sqrt{92(0.64)(1-0.64)} = 4.60

We have to evaluate the probability that more than 47 cars carry just one person.

P(x \geq 47)

After continuity correction, we will evaluate

P( x \geq 46.5) = P( z > \displaystyle\frac{46.5 - 58.88}{4.60}) = P(z > -2.6913)

= 1 - P(z \leq -2.6913)

Calculation the value from standard normal z table, we have,  

P(x > 46.5) = 1 - 0.004 = 0.996 = 99.6\%

0.996 is the probability that more than half out of 92 vehicles carry just one person.

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NemiM [27]
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