Add (3d+2)+(4d+9) ixl

A has the coordinates (–4, 3) and B has the coordinates (4, 4). If DO,1/2(x, y) is a dilation of △ABC, what is true about the im

loris [4]

Answer:

Coordinates of Vertices of triangle ABC are A (-4,3) , B(4,4) and C(1,1).

As, DO is Dilation of Δ ABC by Scale factor of $\frac{1}{2}$ .

Vertices of A' B'C' are

$A'=(\frac{-4}{2},\frac{3}{2})=(-2,\frac{3}{2}), B'=(\frac{4}{2},\frac{4}{2})=(2,2), C'=(\frac{1}{2},\frac{1}{2})$

So, Image Δ A'B'C' will be smaller than the Pre image Δ ABC.

The two triangles will be congruent.

AO is Dilated by a factor of half , so A'O' will be half of AO.

So, correct Statements are

1. AB is parallel to A'B'.

2.DO,1/2(x, y) =

The distance from A' to the origin is half the distance from A to the origin.

$OA=\sqrt{(-4)^2+3^2}=\sqrt{25}=5\\\\ O'A'=\sqrt{2^2+(\frac{3}{2})^2}=\frac{5}{2}=2.5$

8 0

3 years ago

Read 2 more answers

3-5(3a+6) in standard form

Gre4nikov [31]

Answer:

$3 - 15a - 30 \\ - 15a - 27$

Good luck!

Intelligent Muslim,

From Uzbekistan.

6 0

2 years ago

The stem-and-leaf plot shows the ages of the volunteers at a beach cleanup. How many of the volunteers were less than 20 or olde

Lynna [10]

If you want a answer for this you need to show the graph to us

5 0

2 years ago

(PLEASE HELP) A contractor has 10 lots for sale to build on. 4 lots are greater than two acres. 6 lots are less than two acres.

charle [14.2K]

Probability that the first lot sold will be less than two acres = number of lots less than two acres / total number of lots = 6/10 = 3/5

Probability that the second lot sold will be less than two acres = remaining number of lots less than two acres / remaining total number of lots = 5/9

Required probability = 3/5 x 5/9 = 15/45 = 1/3

8 0

2 years ago

Using the same data, Complete the following

astra-53 [7]

Answer:

a) $weight = 49.995 +0.103height +\epsilon$

$R^2 = 0.67051225196413$

$R^2_{adj} =0.66160717769289$

$F =75.2955260721013$

b) H0: $\beta_1 =0$

H1: $\beta_1 \neq 0$

On this case the test statistc is given by $t=8.67729946884982$

And the p value obtined was: $p_v =1.88977483236692x10^{-10}$ and since the p value is lower than th significance level we can reject the null hypothesis at 5% of significance. The slope is significant.

$SE_{\beta_1}=0.0118255634597743$

Step-by-step explanation:

We assume the following info:

Gender Height Weight

Male 73.84701702 241.8935632

Male 68.78190405 162.3104725

Male 74.11010539 212.7408556

Male 71.7309784 220.0424703

Male 69.88179586 206.3498006

Male 67.25301569 152.2121558

Male 68.78508125 183.9278886

Male 68.34851551 167.9711105

Male 67.01894966 175.9294404

Male 63.45649398 156.3996764

Male 71.19538228 186.6049256

Male 71.64080512 213.7411695

Male 64.76632913 167.1274611

Male 69.2830701 189.4461814

Male 69.24373223 186.434168

Male 67.6456197 172.1869301

Male 72.41831663 196.0285063

Male 63.97432572 172.8834702

Male 69.6400599 185.9839576

Male 67.93600485 182.426648

Male 67.91505019 174.1159291

Male 69.43943987 197.7314216

Male 66.14913196 149.173566

Male 75.20597361 228.7617806

Male 67.89319634 162.0066518

Male 68.1440328 192.3439766

Male 69.08963143 184.4351744

Male 72.80084352 206.8281894

Male 67.42124228 175.2139224

Male 68.49641536 154.3426389

Male 68.61811055 187.5068432

Male 74.03380762 212.9102253

Male 71.52821604 195.0322432

Male 69.1801611 205.1836213

Male 69.57720237 204.1641255

Male 70.4009288 192.9035151

Male 69.07617117 197.4882426

Male 67.19352328 183.8109732

Male 65.80731565 163.8518249

And we can use excel to solve the problem

Part a

On this case we need to click on Data>Data analysis>Regression>Ok

We select the input Y range and th X range and then we click in Ok.

The output obtained is on the first figure attached.

And in order to estimate the regression line we need to look into the column Coefficients, and for this case we got:

$\beta_0 =49.995$

$\beta_1 =0.103$

And the linear model would be given by:

$weight = 49.995 +0.103height +\epsilon$

For the determination coefficient we have:

$R^2 = 0.67051225196413$

And the adjusted $R^2 =0.66160717769289$

And if we analyze the output th F statistic is given by:

$F =75.2955260721013$

Part b

We want to test:

H0: $\beta_1 =0$

H1: $\beta_1 \neq 0$

On this case the test statistc is given by $t=8.67729946884982$

And the p value obtined was: $p_v =1.88977483236692x10^{-10}$ and since the p value is lower than th significance level we can reject the null hypothesis at 5% of significance. The slope is significant.

The standard error for the slope is given in the output:

$SE_{\beta_1}=0.0118255634597743$

$t=8.67729946884982$

$p_v =1.88977483236692x10^{-10}$

We reject the null hypothesis.

5 0

3 years ago