Find the equation of the line perpendicular to x−5y=15 that passes through the point (−2,5).

Mathematics

1 answer:

defon3 years ago

7 0

Answer:

Step-by-step explanation:

-5y = -x + 15

5y = x - 15

y = 1/5x - 3

perp. -5

y - 5 = -5(x + 2)

y - 5 = -5x - 10

y = -5x - 5

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For i≥1 , let Xi∼G1/2 be distributed Geometrically with parameter 1/2 . Define Yn=1n−−√∑i=1n(Xi−2) Approximate P(−1≤Yn≤2) with l

murzikaleks [220]

Answer:

The answer is "0.68".

Step-by-step explanation:

Given value:

$X_i \sim \frac{G_1}{2}$

$E(X_i)=2 \\$

$Var (X_i)= \frac{1- \frac{1}{2}}{(\frac{1}{2})^2}\\$

$= \frac{ \frac{2-1}{2}}{\frac{1}{4}}\\\\= \frac{ \frac{1}{2}}{\frac{1}{4}}\\\\= \frac{1}{2} \times \frac{4}{1}\\\\= \frac{4}{2}\\\\=2$

Now we calculate the $\bar X \sim N(2, \sqrt{\frac{2}{n}})\\$

$\to \frac{\bar X - 2}{\sqrt{\frac{2}{n}}} \sim N(0, 1)\\$

$\to \sum^n_{i=1} \frac{X_i - 2}{n} \times\sqrt{\frac{n}{2}}} \sim N(0, 1)\\\\\to \sum^n_{i=1} \frac{X_i - 2}{\sqrt{2n}} \sim N(0, 1)\\$

$\to Z_n = \frac{1}{\sqrt{n}} \sum^n_{i=1} (X_i -2) \sim N(0, 2)\\$

$\to P(-1 \leq X_n \leq 2) = P(Z_n \leq Z) -P(Z_n \leq -1) \\\\$

$= 0.92 -0.24\\\\= 0.68$

6 0

3 years ago

Find an equation of the line that passes through the points (-5, -3) and (3, 1)

Mkey [24]

$\bf (\stackrel{x_1}{-5}~,~\stackrel{y_1}{-3})\quad (\stackrel{x_2}{3}~,~\stackrel{y_2}{1}) ~\hfill \stackrel{slope}{m}\implies \cfrac{\stackrel{rise} {\stackrel{y_2}{1}-\stackrel{y1}{(-3)}}}{\underset{run} {\underset{x_2}{3}-\underset{x_1}{(-5)}}}\implies \cfrac{1+3}{3+5}\implies \cfrac{4}{8}\implies \cfrac{1}{2}$

$\bf \begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-\stackrel{y_1}{(-3)}=\stackrel{m}{\cfrac{1}{2}}[x-\stackrel{x_1}{(-5)}]\implies y+3=\cfrac{1}{2}(x+5) \\\\\\ y+3=\cfrac{1}{2}x+\cfrac{5}{2}\implies y=\cfrac{1}{2}x+\cfrac{5}{2}-3\implies y = \cfrac{1}{2}x-\cfrac{1}{2}$