Ask question

Ask question

All categories

3 years ago

7

Mitchell purchased a car in 2008 for $25,980. The function y 25,980(0.88y represents the depreciated value of the car after t ye

ars since 2008. What is the approximate value of the car in 2012?

1 answer:

Tasya [4]3 years ago

7 0

Answer:

$\$15,580.09$

Step-by-step explanation:

we know that

The exponential decay function that represent the depreciated value of the car is given by

$y=25,980(0.88^t)$

where

y is the depreciated value of the car

t is the number of years since 2008

What is the approximate value of the car in 2012?

Find the value of t

$t=2012-2008=4\ years$

substitute in the equation

$y=25,980(0.88^4)=\$15,580.09$

You might be interested in

Matt is a software engineer writing a script involving 6 tasks. Each must be done one after the other. Let ti be the time for th

Masteriza [31]

Answer:

Let $t_i$ be the time for the $i$ th task.

We know these times have a certain structure:

Any 3 adjacent tasks will take half as long as the next two tasks.

In the form of an equations we have

$t_1+t_2+t_3=\frac{1}{2}t_4+\frac{1}{2}t_5 \\\\t_2+t_3+t_4=\frac{1}{2}t_5+\frac{1}{2}t_6$

The second task takes 1 second $t_2=1$

The fourth task takes 10 seconds $t_4=10$

So, we have the following system of equations:

$t_1+t_2+t_3-\frac{1}{2}t_4-\frac{1}{2}t_5=0 \\\\t_2+t_3+t_4-\frac{1}{2}t_5-\frac{1}{2}t_6=0\\\\t_2=1\\\\t_4=10$

a) An augmented matrix for a system of equations is a matrix of numbers in which each row represents the constants from one equation (both the coefficients and the constant on the other side of the equal sign) and each column represents all the coefficients for a single variable.

Here is the augmented matrix for this system.

$\left[ \begin{array}{cccccc|c} 1 & 1 & 1 & - \frac{1}{2} & - \frac{1}{2} & 0 & 0 \\\\ 0 & 1 & 1 & 1 & - \frac{1}{2} & - \frac{1}{2} & 0 \\\\ 0 & 1 & 0 & 0 & 0 & 0 & 1 \\\\ 0 & 0 & 0 & 1 & 0 & 0 & 10 \end{array} \right]$

b) To reduce this augmented matrix to reduced echelon form, you must use these row operations.

Subtract row 2 from row 1 $\left(R_1=R_1-R_2\right)$ .
Subtract row 2 from row 3 $\left(R_3=R_3-R_2\right)$ .
Add row 3 to row 2 $\left(R_2=R_2+R_3\right)$ .
Multiply row 3 by −1 $\left({R}_{{3}}=-{1}\cdot{R}_{{3}}\right)$ .
Add row 4 multiplied by $\frac{3}{2}$ to row 1 $\left(R_1=R_1+\left(\frac{3}{2}\right)R_4\right)$ .
Subtract row 4 from row 3 $\left(R_3=R_3-R_4\right)$ .

Here is the reduced echelon form for the augmented matrix.

$\left[ \begin{array}{ccccccc} 1 & 0 & 0 & 0 & 0 & \frac{1}{2} & 15 \\\\ 0 & 1 & 0 & 0 & 0 & 0 & 1 \\\\ 0 & 0 & 1 & 0 & - \frac{1}{2} & - \frac{1}{2} & -11 \\\\ 0 & 0 & 0 & 1 & 0 & 0 & 10 \end{array} \right]$

c) The additional rows are

$\begin{array}{ccccccc} 0 & 0 & 0 & 0 & 0 & 1 & 20 \\\\ 1 & 1 & 1 & 0 & 0 & 0 & 50 \end{array} \right$

and the augmented matrix is

$\left[ \begin{array}{ccccccc} 1 & 0 & 0 & 0 & 0 & \frac{1}{2} & 15 \\\\ 0 & 1 & 0 & 0 & 0 & 0 & 1 \\\\ 0 & 0 & 1 & 0 & - \frac{1}{2} & - \frac{1}{2} & -11 \\\\ 0 & 0 & 0 & 1 & 0 & 0 & 10 \\\\ 0 & 0 & 0 & 0 & 0 & 1 & 20 \\\\ 1 & 1 & 1 & 0 & 0 & 0 & 50 \end{array} \right]$

d) To solve the system you must use these row operations.

Subtract row 1 from row 6 $\left(R_6=R_6-R_1\right)$ .
Subtract row 2 from row 6 $\left(R_6=R_6-R_2\right)$ .
Subtract row 3 from row 6 $\left(R_6=R_6-R_3\right)$ .
Swap rows 5 and 6.
Add row 5 to row 3 $\left(R_3=R_3+R_5\right)$ .
Multiply row 5 by 2 $\left(R_5=\left(2\right)R_5\right)$ .
Subtract row 6 multiplied by 1/2 from row 1 $\left(R_1=R_1-\left(\frac{1}{2}\right)R_6\right)$ .
Add row 6 multiplied by 1/2 to row 3 $\left(R_3=R_3+\left(\frac{1}{2}\right)R_6\right)$ .

$\left[ \begin{array}{ccccccc} 1 & 0 & 0 & 0 & 0 & 0 & 5 \\\\ 0 & 1 & 0 & 0 & 0 & 0 & 1 \\\\ 0 & 0 & 1 & 0 & 0 & 0 & 44 \\\\ 0 & 0 & 0 & 1 & 0 & 0 & 10 \\\\ 0 & 0 & 0 & 0 & 1 & 0 & 90 \\\\ 0 & 0 & 0 & 0 & 0 & 1 & 20 \end{array} \right]$

The solutions are: $(t_1,...,t_6)=(5,1,44,10,90,20)$ .

5 0

3 years ago

PLS IM SO CONFUSLEZZZD <br> 2+2

cluponka [151]

The answer is 4 hope this helps kindergartner

7 0

3 years ago

Apply the distributive property to factor out the greatest common factor.<br> 90+27=

solniwko [45]

Answer:

9 (10+3)

Step-by-step explanation:

90 = 9*10 = 9*5*2

27 = 9*3

The greatest common factor is 9

9 (10+3)

6 0

3 years ago

Read 2 more answers

What is the square root of -16?<br> е в<br> -87<br> 0 ООО<br> 8

Monica [59]

$4i$

$i=\sqrt{-1}$

$\sqrt{-16}=\sqrt{-1}*4$

$\sqrt{-16}=i*4$

$4i$

Hope this helps.

頑張って!

8 0

3 years ago

When I sit and watch my students take exams, I often think to myself "I wonder if students with bright calculators are impacted

hodyreva [135]

Answer:

There is a difference between the two means.

Step-by-step explanation:

The hypothesis can be defined as:

<em>H₀</em>: The mean exam scores of my SAT 215 students with colorful calculators are same as the mean scores of my STA 215 students with plain black calculators, i.e. <em>μ</em>₁ - <em>μ</em>₂ = 0.

<em>Hₐ</em>: The mean exam scores of my SAT 215 students with colorful calculators are different than the mean scores of my STA 215 students with plain black calculators, i.e. <em>μ</em>₁ - <em>μ</em>₂ ≠ 0.

Assume that the significance level of the test is, <em>α</em> = 0.05. Also assuming that the population variances are equal.

The decision rule:

A 95% confidence interval for mean difference can be used to determine the result of the hypothesis test. If the 95% confidence interval contains the null hypothesis value, i.e. 0 then the null hypothesis will not be rejected.

The 95% confidence interval for mean difference is:

$CI=\bar x_{1}-\bar x_{2}\pm t_{\alpha/2, (n_{1}+n_{2}-2)}\times S_{p}\times \sqrt{\frac{1}{n_{1}}+\frac{1}{n_{2}}}$

Compute the pooled standard deviation as follows:

$S_{p}=\sqrt{\frac{(n_{1}-1)s_{1}^{2}+(n_{2}-1)s_{2}^{2}} {n_{1}+n_{2}-2}}}=\sqrt{\frac{(49-1)(4.7)^{2}+(38-1)(5.7)^{2}}{49+38-2}}=5.16$

The critical value of <em>t</em> is:

$t_{\alpha/2, (n_{1}+n_{2}-2)}=t_{0.05/2, (49+38-2)}=t_{0.025, 85}=1.984$

*Use a <em>t</em>-table.

Compute the 95% confidence interval for mean difference as follows:

$CI=\bar x_{1}-\bar x_{2}\pm t_{\alpha/2, (n_{1}+n_{2}-2)}\times S_{p}\times \sqrt{\frac{1}{n_{1}}+\frac{1}{n_{2}}}$

$=(84-87)\pm 1.984\times 5.16\times \sqrt{\frac{1}{49}+\frac{1}{38}}$

$=-3\pm 2.133\\=(-5.133, -0.867)\\\approx(-5.13, -0.87)$

The 95% confidence interval for mean difference is (-5.13, -0.87).

The confidence interval does not contains the value 0. This implies that the null hypothesis will be rejected at 5% level of significance.

Hence, concluding that the mean exam scores of my STA 215 students with colorful calculators are different than the mean scores of my STA 215 students with plain black calculators.

7 0

4 years ago