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3 years ago

8. The endpoints of ST are $(-3, 2) and T(5,8). Find the coordinates of

Mathematics

1 answer:

Nezavi [6.7K]3 years ago

6 0

Answer:

Option (A)

Step-by-step explanation:

If the extreme ends of a segment are $(x_1, y_1)$ and $(x_2,y_2)$ ,

Coordinates of the mid-point will be = $[\frac{(x_1+x_2)}{2},\frac{(y_1+y_2)}{2}]$

And the length of the segment is = $\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

Coordinates of the extreme ends of the ST are S(-3, 2) and T(5, 8).

Therefore, coordinates of the midpoint of ST will be = $(\frac{-3+5}{2},\frac{2+8}{2})$

= $(1,5)$

Length of ST = $\sqrt{(5+3)^2+(8-2)^2}$

= $\sqrt{64+36}$

= 10 units

Option (A) will be the answer.

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2 years ago

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A 20 foot ladder leaning against a wall is used to reach a window that is 17 feet above the ground. How far from the wall is the

Liono4ka [1.6K]

Answer:

The wall is 10.5 foot far from the bottom of the ladder.

Step-by-step explanation:

Given:

Height of the ladder leaning against wall (Hypotenuse) = 20 foot.

Height from where the window is above from the ground (Leg1) = 17 feet.

To find the distance of the bottom of the ladder far from the wall (Leg2) = ?

Now, by using the pythagorean theorem:

$Hypotenuse^{2} = (Leg1)^{2} + (Leg2)^{2}$

$20^{2} =17^{2}+(Leg2) ^{2}$

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by squaring both sides

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Therefore, the wall is 10.5 foot far from the bottom of the ladder.

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Step-by-step explanation:

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Answer:

$t=\frac{2.35-1.83}{\sqrt{\frac{0.88^2}{10}+\frac{0.54^2}{7}}}}=1.507$

$p_v =P(t_{(15)}>1.507)=0.076$

If we compare the p value and the significance level given $\alpha=0.01$ we see that $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and the group CTS, NOT have a significant higher mean compared to the Normal group at 1% of significance.

Step-by-step explanation:

1) Data given and notation

$\bar X_{CTS}=2.35$ represent the mean for the sample CTS

$\bar X_{N}=1.83$ represent the mean for the sample Normal

$s_{CTS}=0.88$ represent the sample standard deviation for the sample of CTS

$s_{N}=0.54$ represent the sample standard deviation for the sample of Normal

$n_{CTS}=10$ sample size selected for the CTS

$n_{N}=7$ sample size selected for the Normal

$\alpha=0.01$ represent the significance level for the hypothesis test.

t would represent the statistic (variable of interest)

$p_v$ represent the p value for the test (variable of interest)

State the null and alternative hypotheses.

We need to conduct a hypothesis in order to check if the mean for the group CTS is higher than the mean for the Normal, the system of hypothesis would be:

Null hypothesis: $\mu_{CTS} \leq \mu_{N}$

Alternative hypothesis: $\mu_{CTS} > \mu_{N}$

If we analyze the size for the samples both are less than 30 so for this case is better apply a t test to compare means, and the statistic is given by:

$t=\frac{\bar X_{CTS}-\bar X_{N}}{\sqrt{\frac{s^2_{CTS}}{n_{CTS}}+\frac{s^2_{N}}{n_{N}}}}$ (1)

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine whether the means of two groups are equal to each other".

Calculate the statistic

We can replace in formula (1) the info given like this:

$t=\frac{2.35-1.83}{\sqrt{\frac{0.88^2}{10}+\frac{0.54^2}{7}}}}=1.507$

P-value

The first step is calculate the degrees of freedom, on this case:

$df=n_{CTS}+n_{N}-2=10+7-2=15$

Since is a one side right tailed test the p value would be:

$p_v =P(t_{(15)}>1.507)=0.076$

We can use the following excel code to calculate the p value in Excel:"=1-T.DIST(1.507,15,TRUE)"

Conclusion

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3 years ago