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Butoxors [25]
2 years ago
9

Nathaniel purchased a new car in 1997 for $20,300. The value of the car has been depreciating exponentially at a constant rate.

If the value of the car was $2,700 in the year 2004, then what would be the predicted value of the car in the year 2007, to the nearest dollar?
Mathematics
1 answer:
Tems11 [23]2 years ago
6 0

Answer:

Step-by-step explanation:

We would apply the formula for exponential decay which is expressed as

A = P(1 - r)^t

Where

A represents the value of the car after t years.

t represents the number of years.

P represents the initial value of the car.

r represents rate of decay.

From the information given,

A = $2700

P = $20300

n = 2004 - 1997 = 7 years

Therefore,

20300 = 2700(1 - r)^7

20300/2700 = (1 - r)^7

7.519 = (1 - r)^7

Taking log of both sides, it becomes

Log 7.519 = 7 log(1 - r)

0.876 = 7 log(1 - r)

Log (1 - r) = 0.876/7 = 0.125

Taking inverse log of both sides, it becomes

10^log1 - r = 10^0.125

1 - r = 1.33

r = 1.33 - 1 = 0.33

The expression would be

A = 20300(1 - 0.33)^t

A = 20300(0.67)^t

Therefore, in 2007,

t = 2007 - 1997 = 10 years

The value would be

A = 20300(0.67)^10

A = $370

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Omari drives a car that gets 18 miles per gallon of gasoline. The car's gasoline tank holds 15 gallons. The distance Omari drive
Art [367]

Answer:

Step-by-step explanation:

I believe 9 miles 15x3= 45 gallons what the car holds but he gets 18 mpg

18x3= 54 gallons 54-45=9

3 0
2 years ago
A certain article indicates that in a sample of 1,000 dog owners, 610 said that they take more pictures of their dog than of the
lbvjy [14]

Answer:

(a) The 90% confidence interval is: (0.60, 0.63) Correct interpretation is (3).

(b) The 95% confidence interval is: (0.42, 0.46) Correct interpretation is (1).

(c) First, the confidence level in part(b) is <u>more than</u> the confidence level in part(a) is, so the critical value of part(b) is <u>more than</u> the critical value of part(a), Second the <u>margin of error</u> in part(b) is <u>more than</u> than in part(a).

Step-by-step explanation:

(a)

Let <em>X</em> = number of dog owners who take more pictures of their dog than of their significant others or friends.

Given:

<em>X</em> = 610

<em>n</em> = 1000

Confidence level = 90%

The (1 - <em>α</em>)% confidence interval for population proportion is:

CI=\hat p\pm z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}

The sample proportion is:

\hat p=\frac{X}{n}=\frac{610}{1000}=0.61

The critical value of <em>z</em> for a 90% confidence level is:

z_{\alpha/2}=z_{0.10/2}=z_[0.05}=1.645

Construct a 90% confidence interval for the population proportion of dog owners who take more pictures of their dog than their significant others or friends as follows:

CI=\hat p\pm z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}\\=0.61\pm 1.645\sqrt{\frac{0.61(1-0.61}{1000}}\\=0.61\pm 0.015\\=(0.595, 0.625)\\\approx(0.60, 0.63)

Thus, the 90% confidence interval for the population proportion of dog owners who take more pictures of their dog than their significant others or friends is (0.60, 0.63).

<u>Interpretation</u>:

There is a 90% chance that the true proportion of dog owners who take more pictures of their dog than of their significant others or friends falls within the interval (0.60, 0.63).

Correct option is (3).

(b)

Let <em>X</em> = number of dog owners who are more likely to complain to their dog than to a friend.

Given:

<em>X</em> = 440

<em>n</em> = 1000

Confidence level = 95%

The (1 - <em>α</em>)% confidence interval for population proportion is:

CI=\hat p\pm z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}

The sample proportion is:

\hat p=\frac{X}{n}=\frac{440}{1000}=0.44

The critical value of <em>z</em> for a 95% confidence level is:

z_{\alpha/2}=z_{0.05/2}=z_{0.025}=1.96

Construct a 95% confidence interval for the population proportion of dog owners who are more likely to complain to their dog than to a friend as follows:

CI=\hat p\pm z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}\\=0.44\pm 1.96\sqrt{\frac{0.44(1-0.44}{1000}}\\=0.44\pm 0.016\\=(0.424, 0.456)\\\approx(0.42, 0.46)

Thus, the 95% confidence interval for the population proportion of dog owners who are more likely to complain to their dog than to a friend  is (0.42, 0.46).

<u>Interpretation</u>:

There is a 95% chance that the true proportion of dog owners who are more likely to complain to their dog than to a friend falls within the interval (0.42, 0.46).

Correct option is (1).

(c)

The confidence interval in part (b) is wider than the confidence interval in part (a).

The width of the interval is affected by:

  1. The confidence level
  2. Sample size
  3. Standard deviation.

The confidence level in part (b) is more than that in part (a).

Because of this the critical value of <em>z</em> in part (b) is more than that in part (a).

Also the margin of error in part (b) is 0.016 which is more than the margin of error in part (a), 0.015.

First, the confidence level in part(b) is <u>more than</u> the confidence level in part(a) is, so the critical value of part(b) is <u>more than</u> the critical value of part(a), Second the <u>margin of error</u> in part(b) is <u>more than</u> than in part(a).

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2 years ago
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konstantin123 [22]

Product means multiplication.

Given w= weight of a trout.

So, we need to convert "product of 16 and weight of trout: w " into an algebraic expression.

So, product of 16 and w can be written as 16*w or simply 16w.

Hence, B is the correct choice.

8 0
2 years ago
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ivolga24 [154]
The answer 456 fe
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