Ask question

Ask question

All categories

3 years ago

14

The value of a car is $30.000and depreciates at a rate of 6.5% each year.

1 answer:

zalisa [80]3 years ago

7 0

find interest and subtract it from 30,000

You might be interested in

Manipulate the triangle so angle A measures 14° and angle B measures 62°. Which statement is true about angle C? The angle is ac

zavuch27 [327]

Answer:

Its an obtuse

Step-by-step explanation:

6 0

4 years ago

Read 2 more answers

<img src="https://tex.z-dn.net/?f=%20%5Csqrt%7Bx%20%5Csqrt%7Bx%20%5Csqrt%7Bx%20%5Csqrt%7Bx....%7D%20%7D%20%7D%20%7D%20%20%3D%20%

andrey2020 [161]

First observe that if $a+b>0$ ,

$(a + b)^2 = a^2 + 2ab + b^2 \\\\ \implies a + b = \sqrt{a^2 + 2ab + b^2} = \sqrt{a^2 + ab + b(a + b)} \\\\ \implies a + b = \sqrt{a^2 + ab + b \sqrt{a^2 + ab + b(a+b)}} \\\\ \implies a + b = \sqrt{a^2 + ab + b \sqrt{a^2 + ab + b \sqrt{a^2 + ab + b(a+b)}}} \\\\ \implies a + b = \sqrt{a^2 + ab + b \sqrt{a^2 + ab + b \sqrt{a^2 + ab + b \sqrt{\cdots}}}}$

Let $a=0$ and $b=x$ . It follows that

$a+b = x = \sqrt{x \sqrt{x \sqrt{x \sqrt{\cdots}}}}$

Now let $b=1$ , so $a^2+a=4x$ . Solving for $a$ ,

$a^2 + a - 4x = 0 \implies a = \dfrac{-1 + \sqrt{1+16x}}2$

which means

$a+b = \dfrac{1 + \sqrt{1+16x}}2 = \sqrt{4x + \sqrt{4x + \sqrt{4x + \sqrt{\cdots}}}}$

Now solve for $x$ .

$x = \dfrac{1 + \sqrt{1 + 16x}}2 \\\\ 2x = 1 + \sqrt{1 + 16x} \\\\ 2x - 1 = \sqrt{1 + 16x} \\\\ (2x-1)^2 = \left(\sqrt{1 + 16x}\right)^2$

(note that we assume $2x-1\ge0$ )

$4x^2 - 4x + 1 = 1 + 16x \\\\ 4x^2 - 20x = 0 \\\\ 4x (x - 5) = 0 \\\\ 4x = 0 \text{ or } x - 5 = 0 \\\\ \implies x = 0 \text{ or } \boxed{x = 5}$

(we omit $x=0$ since $2\cdot0-1=-1\ge0$ is not true)

3 0

2 years ago

Find the following from the table:<br> m = _______<br> b = _______<br> Equation: _______

siniylev [52]

Answer: m= -2

b= 14

y= -2x+14

Step-by-step explanation:

First, you estimate to find the m. After you get -2, you plug the numbers in, and see what you add to get x.

3 0

4 years ago

An electrical engineer wishes to compare the mean lifetimes of two types of transistors in an application involving high-tempera

Olin [163]

Answer:

(115.2642, 222.7358).

Step-by-step explanation:

Given data:

type A: n_1=60, xbar_1=1827, s_1=168

type B: n_2=180, xbar_2=1658, s_2=225

n_1 = sample size 1, n_2= sample size 2

xbar_1, xbar_2 are mean life of sample 1 and 2 respectively. Similarly, s_1 and s_2 are standard deviation of 1,2.

a=0.05, |Z(0.025)|=1.96 (from the standard normal table)

So 95% CI is

(xbar_1 -xbar_2) ± Z×√[s1^2/n1 + s2^2/n2]

=(1827-1658) ± 1.96×sqrt(168^2/60 + 225^2/180)

= (115.2642, 222.7358).

8 0

3 years ago

In what year will the two quantities be equal?

Galina-37 [17]

Its the second because i did this already

3 0

4 years ago