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3 years ago

6

(-5, 18), (19,9) Find the slope of the line through each of pair of points

2 answers:

Rasek [7]3 years ago

7 0

Answer:

slope = -3/8

image attached so its not confusing :D its a negative mixed number, not negative 3 over 8

n200080 [17]3 years ago

5 0

Answer: 3/8

Step-by-step explanation: If you put it into the slope formula, you would get 18 - 9 / 19 - (-5) which simplifies to 9/24 and then to 3/8.

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Aline has a rise of 6 and a slope of<br> 1/20 What is the run

Softa [21]

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The equation is:

slope = rise/run

Rearrange for run:

run = rise/slope

Substitute values in:

run = 6/(1/20)

Solve:

run = 120

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3 0

3 years ago

On a linear X temperature scale, water freezes at −115.0°X and boils at 325.0°X. On a linear Y temperature scale, water freezes

belka [17]

Answer:

The current temperature on the X scale is 1150 °X.

Step-by-step explanation:

Let is determine first the ratio of change in X linear temperature scale to change in Y linear temperature scale:

$r = \frac{\Delta T_{X}}{\Delta T_{Y}}$

$r = \frac{325\,^{\circ}X-(-115\,^{\circ}X)}{-25\,^{\circ}Y - (-65.00\,^{\circ}Y)}$

$r = 11\,\frac{^{\circ}X}{^{\circ}Y}$

The difference between current temperature in Y linear scale with respect to freezing point is:

$\Delta T_{Y} = 50\,^{\circ}Y - (-65\,^{\circ}Y)$

$\Delta T_{Y} = 115\,^{\circ}Y$

The change in X linear scale is:

$\Delta T_{X} = r\cdot \Delta T_{Y}$

$\Delta T_{X} = \left(11\,\frac{^{\circ}X}{^{\circ}Y} \right)\cdot (115\,^{\circ}Y)$

$\Delta T_{X} = 1265\,^{\circ}X$

Lastly, the current temperature on the X scale is:

$T_{X} = -115\,^{\circ}X + 1265\,^{\circ}X$

$T_{X} = 1150\,^{\circ}X$

The current temperature on the X scale is 1150 °X.

5 0

3 years ago

Something divided by - 1 1/4 = 2.8

sergey [27]

-3 1/2 because if you do 2.8 times 1 and 1/4 you get 3.5 and 3.5 as a fraction is 3 1/2

5 0

3 years ago

Read 2 more answers

WHOEVER ANSWERS CORRECTLY I'LL GIVE BRAINLIEST!!!!!!!!!

Gennadij [26K]

Answer:

$5^1^5$

Step-by-step explanation:

$\frac{(5^3)^9}{5^1^2}$

$=\frac{5^2^7}{5^1^2}$

$=5^1^5$

8 0

2 years ago

McAllister et al. (2012) compared varsity football and hockey players with varsity athletes from noncontact sports to determine

jonny [76]

Answer:

$t _{critical} = 1.760$

t = 2.2450

d. 0.264

Step-by-step explanation:

The null hypothesis is:

$H_o: \mu_1 - \mu_2 = 0$

Alternative hypothesis;

$H_a : \mu_1 - \mu_2 > 0\\$

The pooled variance t-Test would have been determined if the population variance are the same.

$S_p^2 = \dfrac{(n_1-1)S_1^2+(n_2-1)S^2_2}{(n_1-1)+(n_2-1)}$

$S_p^2 = \dfrac{(8-1)2.507^2+(8-1)2.8282^2}{(8-1)+(8-1)}$

$S_p^2 = 7.14$

The t-test statistics can be computed as:

$t= \dfrac{(x_1-x_2)-(\mu_1 - \mu_2)}{\sqrt{Sp^2 ( \dfrac{1}{n} +\dfrac{1}{n_2})}}$

$t= \dfrac{(9-6)-0}{\sqrt{7.14 ( \dfrac{1}{8} +\dfrac{1}{8})}}$

$t= \dfrac{3}{1.336}$

t = 2.2450

Degree of freedom $df = (n_1 -1) + ( n_2 +1 )$

df = (8-1)+(8-1)

df = 7 + 7

df = 14

At df = 14 and ∝ = 0.05;

$t _{critical} = 1.760$

Decision Rule: To reject the null hypothesis if the t-test is greater than the critical value.

Conclusion: We reject $H_o$ and there is sufficient evidence to conclude that the test scores for contact address s less than Noncontact athletes.

To calculate r²

The percentage of the variance is;

$r^2 = \dfrac{t^2}{t^2 + df}$

$r^2 = \dfrac{2.2450^2}{2.2450^2 + 14}$

$r^2 = \dfrac{5.040025}{5.040025+ 14}$

$r^2 = 0.2647$

7 0

3 years ago