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3 years ago

11

errence buys a new car for $20,000. The value of the car depreciates by 15% each year. If f(x) represents the value of the car a

fter x years, which function represents the car’s value?

1 answer:

s344n2d4d5 [400]3 years ago

3 0

Hi there

The function is
F (x)=20000 (1-0.15)^x
F (x)=20000 (0.85)^x
Where x is the number of years

Hope it helps

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1 PERFORMANCE - Essay

Amiraneli [1.4K]

Answer:

Thus the length = 230 and width = 90.

Step-by-step explanation:

Farmer Fernanda wants to put a fence around her rectangular field. The length of the field is 40 feet less than 3 times the width.

She has 320 feet of fencing.

Let the length of the rectangular fence be x and the width of the fence be y.

Then the relation between them is

x = 3y - 40

Also x + y = 320

Solving these equations,

320 - y = 3y - 40

4y = 360

y = 90

x = 230

Thus the length = 230 and width = 90.

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3 years ago

The table shows the time Monique worked and the amount of money she earned during four different weeks.

spayn [35]

Answer:

330

Step-by-step explanation:

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3 years ago

The following results come from two independent random samples taken of two populations.

photoshop1234 [79]

Answer:

$(a)\ \bar x_1 - \bar x_2 = 2.0$

$(b)\ CI =(1.0542,2.9458)$

$(c)\ CI = (0.8730,2.1270)$

Step-by-step explanation:

Given

$n_1 = 60$ $n_2 = 35$

$\bar x_1 = 13.6$ $\bar x_2 = 11.6$

$\sigma_1 = 2.1$ $\sigma_2 = 3$

Solving (a): Point estimate of difference of mean

This is calculated as: $\bar x_1 - \bar x_2$

$\bar x_1 - \bar x_2 = 13.6 - 11.6$

$\bar x_1 - \bar x_2 = 2.0$

Solving (b): 90% confidence interval

We have:

$c = 90\%$

$c = 0.90$

Confidence level is: $1 - \alpha$

$1 - \alpha = c$

$1 - \alpha = 0.90$

$\alpha = 0.10$

Calculate $z_{\alpha/2}$

$z_{\alpha/2} = z_{0.10/2}$

$z_{\alpha/2} = z_{0.05}$

The z score is:

$z_{\alpha/2} = z_{0.05} =1.645$

The endpoints of the confidence level is:

$(\bar x_1 - \bar x_2) \± z_{\alpha/2} * \sqrt{\frac{\sigma_1^2}{n_1}+\frac{\sigma_2^2}{n_2}}$

$2.0 \± 1.645 * \sqrt{\frac{2.1^2}{60}+\frac{3^2}{35}}$

$2.0 \± 1.645 * \sqrt{\frac{4.41}{60}+\frac{9}{35}}$

$2.0 \± 1.645 * \sqrt{0.0735+0.2571}$

$2.0 \± 1.645 * \sqrt{0.3306}$

$2.0 \± 0.9458$

Split

$(2.0 - 0.9458) \to (2.0 + 0.9458)$

$(1.0542) \to (2.9458)$

Hence, the 90% confidence interval is:

$CI =(1.0542,2.9458)$

Solving (c): 95% confidence interval

We have:

$c = 95\%$

$c = 0.95$

Confidence level is: $1 - \alpha$

$1 - \alpha = c$

$1 - \alpha = 0.95$

$\alpha = 0.05$

Calculate $z_{\alpha/2}$

$z_{\alpha/2} = z_{0.05/2}$

$z_{\alpha/2} = z_{0.025}$

The z score is:

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

The endpoints of the confidence level is:

$(\bar x_1 - \bar x_2) \± z_{\alpha/2} * \sqrt{\frac{\sigma_1^2}{n_1}+\frac{\sigma_2^2}{n_2}}$

$2.0 \± 1.96 * \sqrt{\frac{2.1^2}{60}+\frac{3^2}{35}}$

$2.0 \± 1.96* \sqrt{\frac{4.41}{60}+\frac{9}{35}}$

$2.0 \± 1.96 * \sqrt{0.0735+0.2571}$

$2.0 \± 1.96* \sqrt{0.3306}$

$2.0 \± 1.1270$

Split

$(2.0 - 1.1270) \to (2.0 + 1.1270)$

$(0.8730) \to (2.1270)$

Hence, the 95% confidence interval is:

$CI = (0.8730,2.1270)$

8 0

3 years ago

Ty’s experimental probability of winning a certain game is 1/3. Jana’s expiremental probability of winning the same game is 3/4.

Dafna1 [17]

Answer:

Jana is expected to win 40 more games over Ty.

Step-by-step explanation:

If Ty plays the game 120 times, he is expected to win in 1/3 part of 120 games, which is ( 1/3 * 120 =) 40 games.

If Jana plays the game 120 times, she is expected to win in 2/3 part of 120 games, which is ( 2/3 * 120 =) 80 games.

If Jana is expected to win 80 times an Ty is expected to 40 times, then Jana is expected to win ( 80 - 40 =) 40 more games over Ty.

4 0

3 years ago

Need some help it'd be appreciated

Natalka [10]

Answer:

C

Step-by-step explanation:

The parallel postulate is one of Euclid main axioms. Since it was accepted as self evident.

A is wrong, any two points can be drawn a straight line through. And straight lines exist in plane geometry.

B and D is wrong, the angle measures of a triangle must add up to 180 degrees in plane geometry.

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3 years ago