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Katarina [22]
3 years ago
6

What’s the line contains the point (-5,7) and is perpendicular to a line with a slope of 5/3

Mathematics
2 answers:
inysia [295]3 years ago
7 0
Slope. perp. -3/5

y - 7 = -3/5(x + 5)

y - 7 = -3/5x - 3

y = -3/5x + 4
11Alexandr11 [23.1K]3 years ago
5 0

Answer:

\large\boxed{y=-\dfrac{3}{5}+4}

Step-by-step explanation:

\text{Let}\ k:y=m_1x+b_1,\ \text{and}\ l:y=m_2x+b_2,\ \text{then}\\\\l\ \perp\ k\iff m_1m_2=-1\to m_2=-\dfrac{1}{m_1}\\===========================\\\\\text{We have}\ m_1=\dfrac{5}{3},\ \text{therefore}\ m_2=-\dfrac{1}{\frac{5}{3}}=-\dfrac{3}{5}.\\\\l:y=-\dfrac{3}{5}x+b\\\\\text{Put the coordinates of the point (-5, 7) to the equation of a line }\ l:\\\\7=-\dfrac{3}{5}(-5)+b\\\\7=3+b\qquad\text{subtract 3 from both sides}\\\\4=b\to b=4\\\\\text{Fimally we have the equation:}\\\\l:y=-\dfrac{3}{5}+4

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Since we're given the standard deviation of the true distribution we don't need a t distribution or anything like that. n=45 is big enough (more than 30 or so) that we can substitute the normal distribution for the t distribution anyway.


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