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Harman [31]
2 years ago
7

when you two shapes to compare on a coordinate plan,you can determine the scale factor, knowing that the transformation was a di

lation.Generate instructions you would give another student to determine the scale factor.​
Mathematics
1 answer:
iragen [17]2 years ago
8 0

Answer:

jjjjjj

Step-by-step explanation:

You might be interested in
Part A: Clancey and Ethan are starting new books. So far, Clancey has read 1/4 of his book, which has 240 total pages and Ethan
Nonamiya [84]

In order to find who has read more pages, let's find the number of pages read by each one.

The number of pages read by Clancey is:

\frac{1}{4}\cdot240=\frac{240}{4}=60

The number of pages read by Ethan is:

\frac{2}{5}\cdot170=\frac{340}{5}=68

So Ethan read more pages.

The combined number of pages in both books is 240 + 170 = 410.

The combined number of pages read is 60 + 68 = 128

So the fraction is:

\frac{128}{410}=\frac{64}{205}

5 0
1 year ago
Find the common ratio r for the given geometric sequence and find the next three terms.
34kurt

Answer:

The common ratio is -2

Next three terms are 18, -36, and 72

Step-by-step explanation:

The common ratio is -2 since each consecutive term is being multiplied by -2

The next three terms are -9(-2) = 18, 18(-2) = -36, and -36(-2) = 72

4 0
2 years ago
Read 2 more answers
4)
erastovalidia [21]

Answer:

D.) 4x + y

Step-by-step explanation:

3x + x + y

Add  3 x  and  x .

4 x  +  y

7 0
3 years ago
Find 8 and one third percent of 144
madam [21]
\bf 8\frac{1}{3}\implies \cfrac{8\cdot 3+1}{3}\implies \cfrac{25}{3}
\\\\\\
now\qquad \cfrac{\frac{25}{3}}{100}\implies \cfrac{\frac{25}{3}}{\frac{100}{1}}\implies \cfrac{25}{3}\cdot \cfrac{1}{100}\implies \cfrac{1}{3\cdot 4}\implies \cfrac{1}{12}
\\\\\\
8\frac{1}{3}\%\ of\ 144\implies  \left( \cfrac{\frac{25}{3}}{100} \right)\cdot 144\implies \cfrac{1}{12}\cdot 144\implies \cfrac{144}{12}\implies 12
3 0
3 years ago
A bank with a branch located in a commercial district of a city has the business objective of developing an improved process for
tatiyna

Answer:

(a) The test statistic value is -4.123.

(b) The critical values of <em>t</em> are ± 2.052.

Step-by-step explanation:

In this case we need to determine whether there is evidence of a difference in the mean waiting time between the two branches.

The hypothesis can be defined as follows:

<em>H₀</em>: There is no difference in the mean waiting time between the two branches, i.e. <em>μ</em>₁ - <em>μ</em>₂ = 0.

<em>Hₐ</em>: There is a difference in the mean waiting time between the two branches, i.e. <em>μ</em>₁ - <em>μ</em>₂ ≠ 0.

The data collected for 15 randomly selected customers, from bank 1 is:

S = {4.21, 5.55, 3.02, 5.13, 4.77, 2.34, 3.54, 3.20, 4.50, 6.10, 0.38, 5.12, 6.46, 6.19, 3.79}

Compute the sample mean and sample standard deviation for Bank 1 as follows:

\bar x_{1}=\frac{1}{n_{1}}\sum X_{1}=\frac{1}{15}[4.21+5.55+...+3.79]=4.29

s_{1}=\sqrt{\frac{1}{n_{1}-1}\sum (X_{1}-\bar x_{1})^{2}}\\=\sqrt{\frac{1}{15-1}[(4.21-4.29)^{2}+(5.55-4.29)^{2}+...+(3.79-4.29)^{2}]}\\=1.64

The data collected for 15 randomly selected customers, from bank 2 is:

S = {9.66 , 5.90 , 8.02 , 5.79 , 8.73 , 3.82 , 8.01 , 8.35 , 10.49 , 6.68 , 5.64 , 4.08 , 6.17 , 9.91 , 5.47}

Compute the sample mean and sample standard deviation for Bank 2 as follows:

\bar x_{2}=\frac{1}{n_{2}}\sum X_{2}=\frac{1}{15}[9.66+5.90+...+5.47]=7.11

s_{2}=\sqrt{\frac{1}{n_{2}-1}\sum (X_{2}-\bar x_{2})^{2}}\\=\sqrt{\frac{1}{15-1}[(9.66-7.11)^{2}+(5.90-7.11)^{2}+...+(5.47-7.11)^{2}]}\\=2.08

(a)

It is provided that the population variances are not equal. And since the value of population variances are not provided we will use a <em>t</em>-test for two means.

Compute the test statistic value as follows:

t=\frac{\bar x_{1}-\bar x_{2}}{\sqrt{\frac{s_{1}^{2}}{n_{1}}+\frac{s_{2}^{2}}{n_{2}}}}

  =\frac{4.29-7.11}{\sqrt{\frac{1.64^{2}}{15}+\frac{2.08^{2}}{15}}}

  =-4.123

Thus, the test statistic value is -4.123.

(b)

The degrees of freedom of the test is:

m=\frac{[\frac{s_{1}^{2}}{n_{1}}+\frac{s_{2}^{2}}{n_{2}}]^{2}}{\frac{(\frac{s_{1}^{2}}{n_{1}})^{2}}{n_{1}-1}+\frac{(\frac{s_{2}^{2}}{n_{2}})^{2}}{n_{2}-1}}

   =\frac{[\frac{1.64^{2}}{15}+\frac{2.08^{2}}{15}]^{2}}{\frac{(\frac{1.64^{2}}{15})^{2}}{15-1}+\frac{(\frac{2.08^{2}}{15})^{2}}{15-1}}

   =26.55\\\approx 27

Compute the critical value for <em>α</em> = 0.05 as follows:

t_{\alpha/2, m}=t_{0.025, 27}=\pm2.052

*Use a <em>t</em>-table for the values.

Thus, the critical values of <em>t</em> are ± 2.052.

3 0
3 years ago
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