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luda_lava [24]
3 years ago
14

What is the slope of the line through (1, 9) and (–3, 16)?

Mathematics
2 answers:
zhannawk [14.2K]3 years ago
6 0

Answer:

-7/4.

Step-by-step explanation:

This is  the difference in the y coordinates  / corresponding difference in the x coordinates.

here it is (16 - 9) / (-3-1)

= 7 / -4

= -7/4.

tatiyna3 years ago
5 0

For this case we have that by definition, the slope of a line is given by:

m = \frac {y_ {2} -y_ {1}} {x_ {2} -x_ {1}}

Where:

(x_ {1}, y_ {1}) and(x_ {2}, y_ {2})are two points through which the line passes.

We have as data that:

(x_ {1}, y_ {1}) :( 1,9)\\(x_ {2}, y_ {2}): (- 3,16)

Substituting we have:

m = \frac {16-9} {- 3-1} = \frac {7} {- 4} = - \frac {7} {4}

Thus, the slope of the line is- \frac {7} {4}

Answer:

m = - \frac {7} {4}

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1.3°F * 10 = 13°F

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Step-by-step explanation:

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5 0
3 years ago
Tennis elbow is thought to be aggravated by the impact experienced when hitting the ball. The article "Forces on the Hand in the
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Answer:

Step-by-step explanation:

Hello!

The objective is to study whether there is a greater force after impacting on one- handed backhand drive in advanced tennis players than in intermediate tennis players.

Sample 1: Advanced tennis players

X₁: Force (N) on the hand just after impact on a one- handed backhand drive for an advanced tennis player.

n₁= 6

X[bar]₁= 40.29 N

S₁= 11.29

Sample 2: Intermediate players

X₂: Force (N) on the hand just after impact on a one- handed backhand drive for an intermediate tennis player.

n₂= 8

X[bar]₂= 21.40

S₂= 8.30

Assuming that both variables have a normal distribution and both population variances are equal, to compare these two populations is best to do so trough their population means using a t-test for independent samples.

If the force is greater for the advanced players than for the intermediate players, then you'd expect the population mean for the advanced players to be greater than the population mean for the intermediate players:

H₀: μ₁ ≤ μ₂

H₁: μ₁ > μ₂

α: 0.05

t= \frac{(X_[bar]_1-X[bar]_2)-(Mu_1-Mu_2)}{Sa\sqrt{\frac{1}{n_1} +\frac{1}{n_2} } } ~~t_{n_1+n_2-2}

Sa= \sqrt{\frac{(n_1-1)S_1^2+(n_2-1)S_2^2}{n_1+n_2-2} } = \sqrt{\frac{5*127.51+7*68.92}{6+8-2} }= 9.66

t_{H_0}= \frac{(40.29-21.40)-0}{9.66\sqrt{\frac{1}{6} +\frac{1}{8} } } = 3.62

Using the p-value approach, the decision rule is

If p-value ≤ α, reject the null hypothesis

If p-value > α, do not reject the null hypothesis

The p-value for this test is 0.00024, it is less than the level of significance, so the decision is to reject the null hypothesis.

This means that at a 5% significance level you can conclude that the average force experienced on the hand after a one-handed backhand drive for advanced players is greater than the average force experienced on the hand after a one-handed backhand drive for intermediate players.

I hope this helps!

3 0
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