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Mathematics

1 answer:

likoan [24]3 years ago

7 0

Answer:

The slopes of the sides of the square ABCD are -1/3 and 3

Step-by-step explanation:

The coordinates of the vertices are;

A(-2, 5) and B(4, 3), therefore, we have;

The slope of segment AB, m, is given as follows;

$Slope, \, m =\dfrac{y_{2}-y_{1}}{x_{2}-x_{1}}$

Where, (x₁, y₁) = (-2, 5) and (x₂, y₂) = (4, 3), we have;

$Slope, m \ of \ line \ AB, \ m_{AB}=\dfrac{3-5}{4-(-2)} = \dfrac{-2}{6} = -\dfrac{1}{3}$

Therefore, given that the adjacent sides of a square are perpendicular, we have;

The slope of a line perpendicular to another line with slope, m is equal to the negative inverse of the slope of the line

Therefore, the given that the slope of a line = m, the slope of the perpendicular = -1/m

Therefore;

$The \ slope \ of \ the \ perpendicular \ to \ the \ line \ AB = -\dfrac{1}{The \ slope \ of \ the \ line \ AB }$

$The \ slope \ of \ the \ perpendicular \ to \ the \ line \ AB = -\dfrac{1}{-\dfrac{1}{3} } = 3$

Given that the one other side is parallel to the line AB, whereby the slope of parallel lines are equal, and that two sides are perpendicular to the line AB, we have, the slopes of the sides of the square ABCD are;

-1/3 and 3.

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Aneli [31]

Answer:

D. H0: p ≤0.36 H1: p>0.36

$$$(Right tail test) Z_{\alpha}=Z_{0.1}=1.28155$

$Z_p=-0.25516$

D. Reject H0. There is sufficient evidence to support the hypothesis that the proportion of admissions officers who visit an applying students' social networking page has increased in the past year

Step-by-step explanation:

To solve this problem, we run a hypothesis test about the population proportion.

$$$Sample proportion: $\bar P=0.47\\Sample size n=100$\\Significance level \alpha=0.10$\\(Left tail test) Z_{1-\alpha}=Z_{0.90}=-1.28155$\\(Right tail test) Z_{\alpha}=Z_{0.1}=1.28155$\\(Two-tailed test) $Z_{1-\alpha/2}=Z_{0.95}=-1.64485$ and $ Z_{\alpha/2}=Z_{0.05}=1.64485\\\\$

The appropriate hypothesis system for this situation is:

$H_0:\pi_0=0.35\\H_a:\pi_0 > 0.35\\\\$Proportion in the null hypothesis is:\\\pi_0=0.35\\\\$

$$$The test statistic is $Z=\frac{(\bar P-\pi_0)\sqrt{n}}{\sqrt{\pi_0(1-\pi_0)}}\\$The calculated statistic is Z_c=\frac{(0.47-0.35)\sqrt{100}}{\sqrt{0.35(1-0.35)}}=-0.25516\\p-value = P(Z \geq Z_c)=0.01620\\\\$

Since, the calculated statistic $Z_c$ is greater than critical $Z_{\alpha}$ , the null hypothesis should be rejected. There is enough statistical evidence to state that the proportion of admissions officers who visit an applying students' social networking page has increased in the past year.