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11111nata11111 [884]
3 years ago
15

What is the scale factor of this dilation? [Not drawn to scale]

Mathematics
2 answers:
Margaret [11]3 years ago
8 0
How many times bigger is the right triangle than the left triangle?
A side on the right triangle is 15. The same side on the left triangle is 10. 15/10=1.5.
No matter what side you choose on the right triangle and divide it by the corresponding side of the left triangle, you get the same answer: 1.5.
So, the right triangle is 1.5 times larger than the left triangle. This is the scale factor.
What is 1.5 in fraction form?
The answer is left as an exercise for the reader.
schepotkina [342]3 years ago
6 0
<h2>Answer:</h2>

The scale factor of the dilation is:

                           1\dfrac{1}{2}

<h2>Step-by-step explanation:</h2>

<u>Scale factor--</u>

It is a fixed amount by which the each  of the dimension of the original figure is multiplied in order to obtain the dilated image of the figure.

Here we see that there is a enlargement dilation.

( since the side of the image increases after the dilation)

Let the scale factor be k.

From the figure we see that:

The side of length 6 units is transformed to get a side of length 9 units.

i.e.

6\times k=9

i.e.

k=\dfrac{9}{6}\\\\i.e.\\\\k=\dfrac{3}{2}\\\\i.e.\\\\k=1\dfrac{1}{2}

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The state education commission wants to estimate the fraction of tenth grade students that have reading skills at or below the e
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Answer:

(a) The proportion of tenth graders reading at or below the eighth grade level is 0.1673.

(b) The 95% confidence interval for the population proportion of tenth graders reading at or below the eighth grade level is (0.198, 0.260).

Step-by-step explanation:

Let <em>X</em> = number of students who read above the eighth grade level.

(a)

A sample of <em>n</em> = 269 students are selected. Of these 269 students, <em>X</em> = 224 students who can read above the eighth grade level.

Compute the proportion of students who can read above the eighth grade level as follows:

\hat p=\frac{X}{n}=\frac{224}{269}=0.8327

The proportion of students who can read above the eighth grade level is 0.8327.

Compute the proportion of tenth graders reading at or below the eighth grade level as follows:

1-\hat p=1-0.8327

        =0.1673

Thus, the proportion of tenth graders reading at or below the eighth grade level is 0.1673.

(b)

the information provided is:

<em>n</em> = 709

<em>X</em> = 546

Compute the sample proportion of tenth graders reading at or below the eighth grade level as follows:

\hat q=1-\hat p

  =1-\frac{X}{n}

  =1-\frac{546}{709}

  =0.2299\\\approx 0.229

The critical value of <em>z</em> for 95% confidence interval is:

z_{\alpha/2}=z_{0.05/2}=z_{0.025}=1.96

Compute the 95% confidence interval for the population proportion as follows:

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

     =0.229\pm 1.96\times \sqrt{\frac{0.229(1-0.229)}{709}}\\=0.229\pm 0.03136\\=(0.19764, 0.26036)\\\approx (0.198, 0.260)

Thus, the 95% confidence interval for the population proportion of tenth graders reading at or below the eighth grade level is (0.198, 0.260).

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