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PilotLPTM [1.2K]
2 years ago
14

Can someone help me with these 6 problems I need them within 9 mins. I’m currently in class it says to (Find the slope of each l

ine)

Mathematics
1 answer:
balandron [24]2 years ago
7 0

See below for the calculations of the slopes of the graphs

<h3>How to determine the slope?</h3>

The slope of a line is calculated using:

m= (y2 - y1)/(x2 - x1)

Using the above formula, we have:

<u>Line 1: (-2, 0) and (3, 2)</u>

Slope, m= (2 - 0)/(3 + 2)

m= 2/5

Hence, the slope is 2/5

<u>Line 2: (2, -3) and (-2, 4)</u>

Slope, m= (4 + 3)/(-2 - 2)

m= -7/4

Hence, the slope is -7/4

<u>Line 3: (-4, -2) and (1, 3)</u>

Slope, m= (3 + 2)/(1 + 4)

m= 1

Hence, the slope is 1

<u>Line 4: (0, 0) and (-3, -4)</u>

Slope, m= (-4 - 0)/(-3 - 0)

m= 4/3

Hence, the slope is 4/3

<u>Line 5: (0, 0) and (-1, -4)</u>

Slope, m= (-4 - 0)/(-1 - 0)

m= 4

Hence, the slope is 4

<u>Line 6: (-2, 1) and (3, 2)</u>

Slope, m= (2 - 1)/(3 + 2)

m= 1/5

Hence, the slope is 1/5

<u>Line 7: (-2, -1) and (2, -4)</u>

Slope, m= (-4 + 1)/(2 + 2)

m= -3/4

Hence, the slope is -3/4

<u>Line 8: (-2, -3) and (-1, -4)</u>

Slope, m= (-4 + 3)/(-1 + 2)

m= -1

Hence, the slope is -1

Read more about slopes at:

brainly.com/question/3493733

#SPJ1

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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.

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