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Juli2301 [7.4K]

1 year ago

8

Mark bought a brand new car for $35,000 in 2008. If the car depreciates in value approximately 8% each year, write an exponentia

l function to model the situation. Then, find the value of the car in 2015. Is this considered growth or decay?

1 answer:

hram777 [196]1 year ago

3 0

Answer:

$V(t) = 35000(0.92)^{t}$

Decay function

The value of the car in 2015 is $19,525.

Step-by-step explanation:

A exponential value function has the following format:

$V(t) = V(0)(1+r)^{t}$

In which V(t) is the value after t years, V(0) is the initial value and 1+r is the yearly variation rate.

If 1+r>1, the function is a growth function.

If 1-r<1, the function is a decay function.

Mark bought a brand new car for $35,000 in 2008.

This means that $V(0) = 35,000$

If the car depreciates in value approximately 8% each year

Depreciates, then r is negative. So $r = -0.08$

Then

$V(t) = V(0)(1+r)^{t}$

$V(t) = 35000(1-0.08)^{t}$

$V(t) = 35000(0.92)^{t}$

0.92 < 1, so decay function.

Then, find the value of the car in 2015.

2015 is 2015-2008 = 7 years after 2008. So this is V(7).

$V(t) = 35000(0.92)^{t}$

$V(7) = 35000(0.92)^{7}$

$V(7) = 19525$

The value of the car in 2015 is $19,525.

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How can we find how many inches wide the coat box is?

Alex777 [14]

Answer:

First we found the sum of 15.438 and 3.18. The subtract the answer from 32.476.

Option A is correct.

Step-by-step explanation:

The total width of box is = 32.476

Width of box of T-shirts = 15.438

Width of box of Socks = 3.18

We need to find width of coat box.

The width of coat box can be found as:

Total Width = Width of box of T-shirts + Width of box of Socks + width of coat box

32.476 = 15.438 + 3.18 + width of coat box

32.476 = 18.618 + width of coat box

width of coat box= 32.476 - 18.618

width of coat box = 13.858

So, The Width of coat box = 13.858 inches

First we found the sum of 15.438 and 3.18. The subtract the answer from 32.476.

Option A is correct.

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1 year ago

At an ocean-side nuclear power plant, seawater is used as part of the cooling system. This raises the temperature of the water t

grandymaker [24]

Answer:

(a1) The probability that temperature increase will be less than 20°C is 0.667.

(a2) The probability that temperature increase will be between 20°C and 22°C is 0.133.

(b) The probability that at any point of time the temperature increase is potentially dangerous is 0.467.

(c) The expected value of the temperature increase is 17.5°C.

Step-by-step explanation:

Let <em>X</em> = temperature increase.

The random variable <em>X</em> follows a continuous Uniform distribution, distributed over the range [10°C, 25°C].

The probability density function of <em>X</em> is:

$f(X)=\left \{ {{\frac{1}{25-10}=\frac{1}{15};\ x\in [10, 25]} \atop {0;\ otherwise}} \right.$

(a1)

Compute the probability that temperature increase will be less than 20°C as follows:

$P(X$

Thus, the probability that temperature increase will be less than 20°C is 0.667.

(a2)

Compute the probability that temperature increase will be between 20°C and 22°C as follows:

$P(20$

Thus, the probability that temperature increase will be between 20°C and 22°C is 0.133.

(b)

Compute the probability that at any point of time the temperature increase is potentially dangerous as follows:

$P(X>18)=\int\limits^{25}_{18}{\frac{1}{15}}\, dx\\=\frac{1}{15}\int\limits^{25}_{18}{dx}\,\\=\frac{1}{15}[x]^{25}_{18}=\frac{1}{15}[25-18]=\frac{7}{15}\\=0.467$

Thus, the probability that at any point of time the temperature increase is potentially dangerous is 0.467.

(c)

Compute the expected value of the uniform random variable <em>X</em> as follows:

$E(X)=\frac{1}{2}[10+25]=\frac{35}{2}=17.5$

Thus, the expected value of the temperature increase is 17.5°C.

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2 years ago

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2 years ago

Read 2 more answers

The builder receives a 20% contractor's discount, plus and additional 3% for paying in full within 30 days. Tax on building mate

erastova [34]

Answer:

$1340.49

Step-by-step explanation:

From the attached image

Subtotal Cost of Purchased Items

$= (20 \times 16.69)+(2 \times 20.78)+(2 \times 15.58)+(6 \times 21.38)\\+(118 \times 6.60)+(500 \times 0.22)+(10 \times 8.44)+(1 \times 150)\\\\=\$1658$

Since the builder receives a 20% contractor's discount, plus and additional 3% for paying in full within 30 days.

Discount = 23% of $1658 = 0.23 X 1658 =$381.34

Total (Less Discount) = 1658-381.34 = $1276.66

Tax on building materials is 5%.

Therefore:

Tax = 5% of $1276.66=0.05 X 1276.66

Tax=$63.83

Therefore, the total bill:

=$1276.66+63.83

Grand Total =$1340.49

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1 year ago

According to a survey, 10% of Americans are afraid to fly. Suppose 1100 Americans are sampled. a) What is the probability that 1

JulijaS [17]

Answer:

a) 14.46%

b) 0.00%

c) 1.54%

Step-by-step explanation:

According to the survey, 10% of Americans are afraid to fly.

This means $p=0.10$ and $q=1-0.10=0.90$ .

If 1100 Americans are sampled, then the sample size is $n=1100$ .

The mean of the distribution is $\mu=np$ .

This means $\mu=1100\times 0.10=110$

The standard deviation is $\sigma=\sqrt{npq}$

We substitute the values to get:

$\sigma=\sqrt{1100\times 0.1\times 0.9}=9.95$

a) We want to find the probability that 121 or more Americans in the survey are afraid to fly.

We first apply the continuity correction factor to get: $P(X\ge121)=P(X\:>\:121-0.5)\\P(X\ge121)=P(X\:>\:120.5)$

We now convert to Z-scores to get:

$P(X\:>\:120.5)=P(z\:>\:\frac{120.5-110}{9.95})=1.06\\$

From the standard normal distribution table P(z>1.06)=0.1446

As a percentage, the probability is 14.46%

b) We want to find the probability that 165 or more Americans in the survey are afraid to fly.

We apply the CCF to get:

$P(X\ge165)=P(X\:>\:165-0.5)\\P(X\ge165)=P(X\:>\:164.5)$

We convert to z-scores:

$P(X\:>\:164.5)=P(z\:>\:\frac{164.5-110}{9.95})=5.48$

From the normal distribution, P(z>164.5)=0

c) First, 8% of 1100 is 88.

We want to find the probability that 88 or less Americans in the survey are afraid to fly.

We apply the CCF to get:

$P(X\le88)=P(X\:$

We convert to z-scores:

$P(X\:$

From the normal distribution, P(z>\:-2.16)=0.0154

As a percentage, we get 1.54%

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2 years ago