HELP PLSS ITS DATA HANDLING CLASS 8

The vertices of a triangle are A(2,5), B(1, 2), and C(3, 1). Find the coordinates of the image after a dilation with respect

Dmitry_Shevchenko [17]

The coordinates of the image after a dilation with respect to the origin is A'' (2,10), B" (0,4), C" (4,2) .

<h3>How does dilation work?</h3>

Dilation of a figure will leave its sides get scaled (multiplied) by same number. That number is called the scale factor of that dilation.

Its also called scaling of a figure, but due to the involvement of coordinates, it involves a center of a dilation, which is like a pinned point that stays at the same place after dilation.

Its like we enlarge or shorten the size of the figure (or keep it same, when scale factor = 1).

The vertices of a triangle are A(2,5), B(1, 2), and C(3, 1).

A' = 2(2.5) = (4,10)

B' = 2(1,2) = (2,4)

C' = 2(3,1) = (6,2)

Translation of 2 units left means change in x

A'' = (4-2,10) = (2,10)

B" = (2-2,4) = (0,4)

C" = (6-2,2) = (4,2)

Translation of 1 units up means change in y

Learn more about vertices ;

brainly.com/question/27886655

#SPJ1

7 0

2 years ago

In right triangle KLM, KL =14 and angle L=30° and angle M=90°. Find KM. Leave your answer in the simplest radical form.

salantis [7]

Answer:

KM = 7 Units

Step-by-step explanation:

In the given structure ΔKLM

∠KLM = 30° and side KL = 14 and ∠KML = 90°

and we have to find the measurement of KM.

From right angle triangle KLM

Side Sin30° = KM/KL = KM/14

KM = 14×sin30° = 14×1/2 = 7

So the answer is KM = 7 Units

5 0

3 years ago

‘write an equation of the line that passes through each pair of points’ (1,2) and (4,8)

Levart [38]

Hi there!

Step-by-step explanation:

$Slope=\frac{Y_2-Y_1}{X_2-X_1}=\frac{RISE}{RUN}$

$\frac{8-2}{4-1}=\frac{6}{3}=2$

So therefore, the slope is 2.

Final answer is 2.

*The answer must have a positive sign.*

I hope this helps you!

Have a nice day! :)

-Charlie

:D

8 0

3 years ago

jason wants to choose 9 players for his track team. there are 12 players to choose from. how many different teams can jason make

gtnhenbr [62]

the answer is 21 if that's a choice

3 0

3 years ago

Many high school students take the AP tests in different subject areas. In 2007, of the 144,796 students who took the biology ex

Nataly [62]

Answer:

$(0.582-0.485) - 1.64 \sqrt{\frac{0.582(1-0.582)}{144796} +\frac{0.485(1-0.485)}{211693}}=0.0942$

$(0.582-0.485) + 1.64 \sqrt{\frac{0.582(1-0.582)}{144796} +\frac{0.485(1-0.485)}{211693}}=0.09978$

And the 90% confidence interval would be given (0.0942;0.09978).

We are confident at 90% that the difference between the two proportions is between $0.0942 \leq p_A -p_B \leq 0.09978$

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

$p_A$ represent the real population proportion female for Biology

$\hat p_A =\frac{84199}{144796}=0.582$ represent the estimated proportion female for biology

$n_A=144796$ is the sample size for A

$p_B$ represent the real population proportion female for calculus AB

$\hat p_B =\frac{102598}{211693}=0.485$ represent the estimated proportion female for Calculus AB

$n_B=211693$ is the sample size required for B

$z$ represent the critical value for the margin of error

The population proportion have the following distribution

$p \sim N(p,\sqrt{\frac{p(1-p)}{n}})$

The confidence interval for the difference of two proportions would be given by this formula

$(\hat p_A -\hat p_B) \pm z_{\alpha/2} \sqrt{\frac{\hat p_A(1-\hat p_A)}{n_A} +\frac{\hat p_B (1-\hat p_B)}{n_B}}$

For the 90% confidence interval the value of $\alpha=1-0.90=0.1$ and $\alpha/2=0.05$ , with that value we can find the quantile required for the interval in the normal standard distribution.

$z_{\alpha/2}=1.64$

And replacing into the confidence interval formula we got:

$(0.582-0.485) - 1.64 \sqrt{\frac{0.582(1-0.582)}{144796} +\frac{0.485(1-0.485)}{211693}}=0.0942$

$(0.582-0.485) + 1.64 \sqrt{\frac{0.582(1-0.582)}{144796} +\frac{0.485(1-0.485)}{211693}}=0.09978$

And the 90% confidence interval would be given (0.0942;0.09978).

We are confident at 90% that the difference between the two proportions is between $0.0942 \leq p_A -p_B \leq 0.09978$

5 0

3 years ago