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OlgaM077 [116]
3 years ago
11

−10x+5=−6x−35 ooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooo

Mathematics
2 answers:
Zarrin [17]3 years ago
8 0

Answer:

x=10

Step-by-step explanation:

Add 6x to both sides.

−4x+5=−35

Subtract 5 from both sides.

−4x=−40  

Divide both sides by -4.

\frac{-4x}{-4}=\frac{-40}{-4}

x=10

hope this helps

Scorpion4ik [409]3 years ago
5 0
-10x+5=-6x-35
Add 6 both sides
-10x+5=6x-35
Combine -10x and 6x to get -4x
4x+5=-35
Subtract 5 from both side
-4x = -35-5
Subtract 5 from -35 to get -40
-4x=-40
Divide both side by 4
-4/-40
Divide -4 and -40 will give you 10
x = 10
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Step-by-step explanation:

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

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