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1 year ago

Written as a simplified polynomial in standard form, what is the result when (x+7)^2 is subtracted from7

Mathematics

1 answer:

Oxana [17]1 year ago

8 0

7-(x^2+14x+49)

7-x^2-14x-49

-x^2-14x-42

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The following information was obtained from independent random samples taken of two populations. Assume normally distributed pop

Volgvan

Answer:

1. The 95% confidence interval for the difference between means is (-5.34, 11.34).

2. The standard error of (x-bar1)-(x-bar2) is 4.

$s_{M_d}=\sqrt{\dfrac{\sigma_1^2}{n_1}+\dfrac{\sigma_2^2}{n_2}}=\sqrt{\dfrac{9.2195^2}{10}+\dfrac{9.4868^2}{12}}\\\\\\s_{M_d}=\sqrt{8.5+7.5}=\sqrt{16}=4$

Step-by-step explanation:

We have to calculate a 95% confidence interval for the difference between means.

The sample 1, of size n1=10 has a mean of 45 and a standard deviation of √85=9.2195.

The sample 2, of size n2=12 has a mean of 42 and a standard deviation of √90=9.4868.

The difference between sample means is Md=3.

$M_d=M_1-M_2=45-42=3$

The estimated standard error of the difference between means is computed using the formula:

$s_{M_d}=\sqrt{\dfrac{\sigma_1^2}{n_1}+\dfrac{\sigma_2^2}{n_2}}=\sqrt{\dfrac{9.2195^2}{10}+\dfrac{9.4868^2}{12}}\\\\\\s_{M_d}=\sqrt{8.5+7.5}=\sqrt{16}=4$

The critical t-value for a 95% confidence interval is t=2.086.

The margin of error (MOE) can be calculated as:

$MOE=t\cdot s_{M_d}=2.086 \cdot 4=8.34$

Then, the lower and upper bounds of the confidence interval are:

$LL=M_d-t \cdot s_{M_d} = 3-8.34=-5.34\\\\UL=M_d+t \cdot s_{M_d} = 3+8.34=11.34$

The 95% confidence interval for the difference between means is (-5.34, 11.34).

6 0

4 years ago

Solve the system of equations:x + 3y - z = -4 2x - y + 2z = 13 3x - 2y - z = -9

tatiyna

Answer:

The solution to the system of equations is

$\begin{gathered} x=\frac{179}{13} \\ \\ y=-\frac{279}{39} \\ \\ z=-\frac{48}{13} \end{gathered}$

Explanation:

Giving the system of equations:

$\begin{gathered} x+3y-z=-4\ldots\ldots\ldots\ldots\ldots\ldots..........\ldots\ldots\ldots\ldots.\ldots\text{.}\mathrm{}(1) \\ 2x-y+2z=13\ldots\ldots...\ldots\ldots\ldots\ldots..\ldots..\ldots\ldots\ldots\ldots\ldots.(2) \\ 3x-2y-z=-9\ldots\ldots\ldots.\ldots\ldots\ldots\ldots....\ldots\ldots.\ldots\ldots\ldots\text{.}\mathrm{}(3) \end{gathered}$

To solve this, we need to first of all eliminate one variable from any two of the equations.

Subtracting (2) from twice of (1), we have:

$5y-4z=-21\ldots\ldots\ldots\ldots\ldots.\ldots.\ldots..\ldots..\ldots\ldots.\ldots..\ldots\text{...}\mathrm{}(4)$

Subtracting (3) from 3 times (1), we have

$3y-5z=-3\ldots\ldots...\ldots\ldots..\ldots\ldots\ldots\ldots\ldots.\ldots\ldots\ldots\ldots\ldots..\ldots\ldots(5)$

From (4) and (5), we can solve for y and z.

Subtract 5 times (5) from 3 times (4)

$\begin{gathered} 13z=-48 \\ \\ z=-\frac{48}{13} \end{gathered}$

Using the value of z obtained in (5), we have

$\begin{gathered} 3y-5(-\frac{48}{13})=-3 \\ \\ 3y+\frac{240}{13}=-3 \\ \\ 3y=-3-\frac{240}{13} \\ \\ 3y=-\frac{279}{13} \\ \\ y=-\frac{279}{39} \end{gathered}$

Using the values obtained for y and z in (1), we have

$\begin{gathered} x+3(-\frac{279}{39})-(-\frac{48}{13})=-4 \\ \\ x-\frac{279}{13}+\frac{48}{13}=-4 \\ \\ x-\frac{231}{13}=-4 \\ \\ x=-4+\frac{231}{13} \\ \\ x=\frac{179}{13} \end{gathered}$

8 0

1 year ago

Dave bought a soft drink for $4 and 8 candy bars. He spent a total of $36. How much did each candy bar cost?

Natali5045456 [20]

Each candy bar cost $4.

Step by step:

(Total spent) $36 - (soft drink) $4 = $32 (left for candy bars)

$32 divided by 8 (candy bars) = $4

Therefore, each candy bar was $4.

3 0

4 years ago

How do you write numbers shown as a fraction greater than one?

baherus [9]

40/10 >1 YAYAY!!!!!!!!!

4 0

3 years ago

What is the image of the point (3,1) translated 8 units left and 3 units up?

tamaranim1 [39]

Answer:(-5,4)

Step-by-step explanation:

3 0

4 years ago