1) Find the x and y intercept of the line.

Solve the quadratic equation numerically (using tables of x- and y- values). -4x=-x2+12

Levart [38]

-4x=-x2+12 should be written <span>-4x=-x^2+12, with " ^ " representing exponentiation.

You are to invent a table of x values and find the difference between -4x and -x^ 2+12. The smaller the difference, the closer you are to finding the correct root, x.

To get you started:

x -4x x^2+12 difference
--- ------------ ---------------- --------------
2 -8 16 24 (pretty bad!)
-4 16 28 12 (better)
-10 40 112 72 (much worse)

To answer this problem correctly, you must present such a table, even tho' it'd be faster to use the quadratic formula (or some other method) to find x.

</span>

5 0

3 years ago

Valerie has 10 CDs in her music case.Seven of the CDs are pop music CDs.What is the amount written as a decimal?

kolezko [41]

0.7 is the correct answer

5 0

3 years ago

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25 divided by 9 what is the remainder?

Grace [21]

9 * 2 = 18 under 25
9 * 3 = 27 over 25

9 * 2 + 7 = 25

25/9 = 2, remainder 7

6 0

3 years ago

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James Desreumaux, VP of Human Resources of American First Banks (AFB), is reviewing the employee training programs of AFB banks.

BartSMP [9]

Answer: (24.02,25.98)

Step-by-step explanation:

Answer:

Step-by-step explanation:

Given : Sample size : n= 100

Sample mean : $\overline{x}=25\text{ hours}$

Standard deviation : $\sigma=5\text{ hours}$

Significance level : $\alpha: 1-0.95=0.05$

Critical value : $z_{\alpha/2}=1.96$

The confidence interval for population mean is given by :-

$\overline{x}\pm\ z_{\alpha/2}\dfrac{\sigma}{\sqrt{n}}\\\\=25\pm(1.96)\dfrac{5}{\sqrt{100}}\\\\=25\pm0.98\\\\=(25-0.98,\ 25+0.98)=(24.02,25.98)$

Hence, the 95% confidence interval for the population mean of training times is (24.02, 25.98).

5 0

4 years ago

The value of an investment A (in dollars) after t years is given by the function A(t) = A0ekt. If it takes 10 years for an inves

Ludmilka [50]

Answer:

20 years.

Step-by-step explanation:

We have been given a formula $A(t)=A_0\cdot e^{kt}$ , which represents the value of an investment A (in dollars) after t years.

Substitute the given values:

$\$3,000=\$1,000\cdot e^{k*10}$

Let us solve for k.

$\frac{\$3,000}{\$1,000}=\frac{\$1,000\cdot e^{k*10}}{\$1,000}$

$3=e^{k*10}$

Take natural log of both sides:

$\text{ln}(3)=\text{ln}(e^{k*10})$

Using property $\text{ln}(a^b)=b\cdot \text{ln}(a)$ , we will get:

$\text{ln}(3)=10k\cdot\text{ln}(e)$

We know that $\text{ln}(e)=1$ , so

$\text{ln}(3)=10k\cdot 1$

$\text{ln}(3)=10k$

$\frac{\text{ln}(3)}{10}=\frac{10k}{10}$

$\frac{\text{ln}(3)}{10}=k$

$\$9,000=\$1,000\cdot e^{\frac{\text{ln}(3)}{10}*t}$

Dividing both sides by 1000, we will get:

$9=e^{\frac{\text{ln}(3)}{10}*t}$

Take natural log of both sides:

$\text{ln}(9)=\text{ln}(e^{\frac{\text{ln}(3)}{10}*t)$

$\text{ln}(9)=\frac{\text{ln}(3)}{10}*t\cdot\text{ln}(e)$

$\text{ln}(9)=\frac{\text{ln}(3)}{10}*t\cdot1$

$10*\text{ln}(9)=10*\frac{\text{ln}(3)}{10}*t$

$10\text{ln}(9)=\text{ln}(3)*t$

$10\text{ln}(3^2)=\text{ln}(3)*t$

$2\cdot 10\text{ln}(3)=\text{ln}(3)*t$

$20\text{ln}(3)=\text{ln}(3)*t$

Divide both sides by $\text{ln}(3)$ :

$\frac{20\text{ln}(3)}{\text{ln}(3)}=\frac{\text{ln}(3)*t}{\text{ln}(3)}$

$20=t$

Therefore, it will take 20 years for the investment to be $9,000.

5 0

3 years ago