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lianna [129]
3 years ago
5

What is the result of a dilation of scale factor 3 centered at the origin of the line 2y + 3x=10?? PLEASE HELP PLEASEEEEEEEEE

Mathematics
1 answer:
maks197457 [2]3 years ago
3 0

Given:

The equation of a line is:

2y+3x=10

The line is dilated by factor 3.

To find:

The result of dilation.

Solution:

The equation of a line is:

2y+3x=10

For x=0,

2y+3(0)=10

2y+0=10

y=\dfrac{10}{2}

y=5

For x=2,

2y+3(2)=10

2y+6=10

2y=10-6

2y=4

Divide both sides by 2.

y=\dfrac{4}{2}

y=2

The given line passes through the two points A(0,5) and B(2,2).

If the line dilated by factor 3 with origin as center of dilation, then

(x,y)\to (3x,3y)

Using this rule, we get

A(0,5)\to A'(3(0),3(5))

A(0,5)\to A'(0,15)

Similarly,

B(2,2)\to B'(3(2),3(2))

B(2,2)\to B'(6,6)

The dilated line passes through the points A'(0,15) and B'(6,6). So, the equation of dilated line is:

y-y_1=\dfrac{y_2-y_1}{x_2-x_1}(x-x_1)

y-15=\dfrac{6-15}{6-0}(x-0)

y-15=\dfrac{-9}{6}(x)

y-15=\dfrac{-3}{2}x

Multiply both sides by 2.

2(y-15)=-3x

2y-30=-3x

2y+3x=30

Therefore, the equation of the line after the dilation is 2y+3x=30.

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A random sample of 12 supermarkets from Region 1 had mean sales of 84 with a standard deviation of 6.6. A random sample of 17 su
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We conclude that there is no difference in potential mean sales per market in Region 1 and 2.

Step-by-step explanation:

We are given that a random sample of 12 supermarkets from Region 1 had mean sales of 84 with a standard deviation of 6.6.

A random sample of 17 supermarkets from Region 2 had a mean sales of 78.3 with a standard deviation of 8.5.

Let \mu_1 = mean sales per market in Region 1.

\mu_2  = mean sales per market in Region 2.

So, Null Hypothesis, H_0 : \mu_1-\mu_2 = 0      {means that there is no difference in potential mean sales per market in Region 1 and 2}

Alternate Hypothesis, H_A : > \mu_1-\mu_2\neq 0      {means that there is a difference in potential mean sales per market in Region 1 and 2}

The test statistics that will be used here is <u>Two-sample t-test statistics</u> because we don't know about population standard deviations;

                            T.S.  =  \frac{(\bar X_1 -\bar X_2)-(\mu_1-\mu_2)}{s_p \times \sqrt{\frac{1}{n_1}+ {\frac{1}{n_2}}} }   ~  t__n_1_+_n_2_-_2

where, \bar X_1 = sample mean sales in Region 1 = 84

\bar X_2 = sample mean sales in Region 2 = 78.3

s_1  = sample standard deviation of sales in Region 1 = 6.6

s_2  = sample standard deviation of sales in Region 2 = 8.5

n_1 = sample of supermarkets from Region 1 = 12

n_2 = sample of supermarkets from Region 2 = 17

Also, s_p=\sqrt{\frac{(n_1-1)\times s_1^{2}+(n_2-1)\times  s_2^{2}  }{n_1+n_2-2} }  = s_p=\sqrt{\frac{(12-1)\times 6.6^{2}+(17-1)\times  8.5^{2}  }{12+17-2} } = 7.782

So, <u><em>the test statistics</em></u> =  \frac{(84-78.3)-(0)}{7.782 \times \sqrt{\frac{1}{12}+ {\frac{1}{17}}} }  ~   t_2_7

                                   =  1.943  

The value of t-test statistics is 1.943.

 

Now, at a 0.02 level of significance, the t table  gives a critical value of -2.472 and 2.473 at 27 degrees of freedom for the two-tailed test.

Since the value of our test statistics lies within the range of critical values of t, so we have<u><em> insufficient evidence to reject our null hypothesis</em></u> as it will not fall in the rejection region.

Therefore, we conclude that there is no difference in potential mean sales per market in Region 1 and 2.

6 0
3 years ago
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