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polet [3.4K]
3 years ago
11

Graph a triangle on the coordinate plane. choose 1 transformation (reflection, rotation, translation or dilation) and apply that

transformation to the triangle on the coordinate plane to create the second image

Mathematics
1 answer:
Marina86 [1]3 years ago
7 0

Answer:

The answer in the procedure

Step-by-step explanation:

Let

A(0,0),B(0,3),C(4,0) ----> the coordinates of the pre-image

The transformation will be a translation

The rule of the translation is

(x,y)-------> (x+2, y-5)

That means----> The translation is 2 units to the right and 5 units down

Find the coordinates of the image

Applying the rule

A(0,0)-------> A'(0+2, 0-5) ------> A'(2, -5)

B(0,3)-------> B'(0+2, 3-5) ------> B'(2, -2)

C(4,0)-------> C'(4+2, 0-5) ------> C'(6, -5)

using a graphing tool

see the attached figure

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scZoUnD [109]

Step-by-step explanation:

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3 years ago
A sample of 900 college freshmen were randomly selected for a national survey. Among the survey participants, 372 students were
Rufina [12.5K]

Answer:

a) ME=2.58 \sqrt{\frac{0.413(1-0.413)}{900}}=0.0423

b) 0.413 - 2.58 \sqrt{\frac{0.413(1-0.413)}{900}}=0.371

Step-by-step explanation:

1) Data given and notation  

n=900 represent the random sample taken    

X=372 represent the students were pursuing liberal arts degrees

\hat p=\frac{372}{900}=0.413 estimated proportion of students were pursuing liberal arts degrees

\alpha=0.01 represent the significance level

z would represent the statistic (variable of interest)    

p_v represent the p value (variable of interest)    

p= population proportion of students were pursuing liberal arts degrees

2) Solution to the problem

The confidence interval would be given by this formula

\hat p \pm z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}

For the 99% confidence interval the value of \alpha=1-0.99=0.01 and \alpha/2=0.005, with that value we can find the quantile required for the interval in the normal standard distribution.

z_{\alpha/2}=2.58

The margin of error is given by:

ME=z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}

If we replace we have:

ME=2.58 \sqrt{\frac{0.413(1-0.413)}{900}}=0.0423

And replacing into the confidence interval formula we got:

0.413 - 2.58 \sqrt{\frac{0.413(1-0.413)}{900}}=0.371

0.413 + 2.58 \sqrt{\frac{0.413(1-0.413)}{900}}=0.455

And the 99% confidence interval would be given (0.371;0.455).

We are confident (99%) that about 37.1% to 45.5% of students were pursuing liberal arts degrees.

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3 years ago
Please and thank you
vredina [299]

Answer:

C.255

Step-by-step explanation:

To find the answer to this question, we can 14 by 5 and subtract it from the total of 325.

14*5=70

325-70=255

<u>The correct answer is C. 255</u>

Hope this helps!

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