Ask question

Ask question

All categories

3 years ago

6

A painting is 16 inches wide. it is placed inside a frame with a border that is x inches wide. The total width of the painting a

nd the frame is 20 inches. Create an equation to model the situation, then solve the equation to find the width of the border.

1 answer:

zloy xaker [14]3 years ago

4 0

16 + 2x = 20
2x = 20 - 16 = 4
x = 4/2 = 2 inches

You might be interested in

If y varies directly as x, find the constant of variation if y = 22 and x = 2.

GarryVolchara [31]

THE ANSWER IS 44
like

8 0

3 years ago

(EASY MULTIPLE CHOICE I BEG YOU)

iogann1982 [59]

1) its number 2. 30x^7y^4

8 0

2 years ago

(xy + 9y + 2) and (xy – 3)

Bezzdna [24]

Answer:

x^2y^2 +9xy^2-xy-27y-6

Step-by-step explanation:

(xy+9y+2) (xy-3)

Each term of second bracket will be multiplied with the terms of first bracket

= xy(xy+9y+2) -3(xy+9y+2)

= x^2y^2+9xy^2+2xy-3xy-27y-6

=x^2y^2+9xy^2-xy-27y-6

4 0

3 years ago

How to round o.175 thousands

Sliva [168]

It has a bunch of zeroes after the five so it is rounded unless you want to go to .18

4 0

3 years ago

A researcher compares the effectiveness of two different instructional methods for teaching electronics. A sample of 138 student

yan [13]

Answer:

The 98% confidence interval for the true difference between testing averages for students using Method 1 and students using Method 2 is (-8.04, 0.84).

Step-by-step explanation:

The (1 - <em>α</em>)% confidence interval for the difference between population means is:

$CI=(\bar x_{1}-\bar x_{2})\pm z_{\alpha/2}\times \sqrt{\frac{\sigma^{2}_{1}}{n_{1}}+\frac{\sigma^{2}_{2}}{n_{2}}}$

The information provided is as follows:

$n_{1}= 138\\n_{2}=156\\\bar x_{1}=61\\\bar x_{2}=64.6\\\sigma_{1}=18.53\\\sigma_{2}=13.43$

The critical value of <em>z</em> for 98% confidence level is,

$z_{\alpha/2}=z_{0.02/2}=2.326$

Compute the 98% confidence interval for the true difference between testing averages for students using Method 1 and students using Method 2 as follows:

$CI=(\bar x_{1}-\bar x_{2})\pm z_{\alpha/2}\times \sqrt{\frac{\sigma^{2}_{1}}{n_{1}}+\frac{\sigma^{2}_{2}}{n_{2}}}$

$=(61-64.6)\pm 2.326\times\sqrt{\frac{(18.53)^{2}}{138}+\frac{(13.43)^{2}}{156}}\\\\=-3.6\pm 4.4404\\\\=(-8.0404, 0.8404)\\\\\approx (-8.04, 0.84)$

Thus, the 98% confidence interval for the true difference between testing averages for students using Method 1 and students using Method 2 is (-8.04, 0.84).

5 0

2 years ago