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3 years ago

Find the slope of the line through each pair of points (6, -10) , (-15 , 15)

Mathematics

1 answer:

dem82 [27]3 years ago

5 0

<h3>Answer:</h3>

$\boxed{\boxed{\pink{\bf \leadsto The \ slope \ of \ the \ line \ is \ \dfrac{-25}{21}.}}}$

<h3>Step-by-step explanation:</h3>

Two points are given to us and we need to find the slope of the line . The slope of the line passing through points $\bf (x_1,y_1 ) \ \& \ (x_2,y_2)$ is given by ,

$\qquad\boxed{\red{\bf Slope = tan\theta=\dfrac{y_2-y_1}{x_2-x_1}}}$

Here , the points are ,

( 6 , -10 )
( -15 , 15 )

$\bf\implies Slope = \dfrac{y_2-y_1}{x_2-x_1} \\\\\bf\implies Slope =\dfrac{15-(-10)}{-15-6} \\\\\bf\implies Slope = \dfrac{15+10}{-21}\\\\\bf\implies Slope =\dfrac{-1(25)}{-1(-21)}\\\\ \bf\implies\boxed{\red{\bf Slope =\dfrac{-25}{21}}}$

<h3>★ Hence the slope of the line joining the two points is -25/21 .</h3>

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d. Compute the expected number of withdrawals.

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Step-by-step explanation:

This is a binomial problem and the formula for binomial is:

$P(X = x) = nCx p^{x} q^{n - x}$

a) Compute the probability that 2 or fewer will withdraw

First we need to determine, given 2 students from the 20. Which is the probability of those 2 to withdraw and all others to complete the course. This is given by:

$P(X = x) = nCx p^{x} q^{n - x}\\P(X = 2) = 20C2(0.3)^2(0.7)^{18}\\P(X = 2) =190 * 0.09 * 0.001628413597\\P(X = 2) = 0.027845872524$

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$P(X = x) = nCx p^{x} q^{n - x}\\P(X = 4) = 20C4(0.3)^4(0.7)^{16}\\P(X = 4) = 4845 * 0.0081 * 0.003323293056\\P(X = 4) = 0.130420974373\\P(X = 4) = 0.1304$

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First we will compute the probability that exactly 3 students withdraw, which is given by

$P(X = x) = nCx p^{x} q^{n - x}\\P(X = 3) = 20C3(0.3)^3(0.7)^{17}\\P(X = 3) = 1140 * 0.027 * 0.002326305139\\P(X = 3) = 0.071603672205\\P(X = 3) = 0.0716$

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d) Compute the expected number of withdrawals.

E(X) = 3/10 * 20 = 6

Expected number of withdrawals is the 30% of 20 which is 6.

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