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3 years ago

Write an equation for the perpendicular line to the line whose equation is 4y-5x=20 that contains the point (-2,-3)

Mathematics

1 answer:

yuradex [85]3 years ago

7 0

Answer:

4x + 5y = - 23

Step-by-step explanation:

the equation of a line in slope-intercept form is

y = mx + c ( m is the slope and c the y-intercept )

rearrange 4y - 5x = 20 into this form

add 5x to both sides

4y = 5x + 20 ( divide all terms by 4 )

y = $\frac{5}{4}$ x + 5 ← in slope-intercept form

with slope m = $\frac{5}{4}$

given a line with slope m then the slope of a line perpendicular to it is

$m_{perpendicular}$ = - $\frac{1}{m}$ , hence

$m_{perpendicular}$ = - 1 / $\frac{5}{4}$ = - $\frac{4}{5}$

y = - $\frac{4}{5}$ x + c ← is the partial equation

to find c substitute (- 2, - 3) into the partial equation

- 3 = $\frac{8}{5}$ + c ⇒ c = - $\frac{23}{5}$

y = - $\frac{4}{5}$ x - $\frac{23}{5}$ ← in slope-intercept form

multiply all terms by 5

5y = - 4x - 23 ( add 4x to both sides )

4x + 5y = - 23 ← in standard form

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4 years ago

A research study uses 800 men under the age of 55. Suppose that 30% carry a marker on the male chromosome that indicates an incr

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Answer:

a) There is a 12.11% probability that exactly 1 man has the marker.

b) There is a 85.07% probability that more than 1 has the marker.

Step-by-step explanation:

There are only two possible outcomes: Either the men has the chromosome, or he hasn't. So we use the binomial probability distribution.

Binomial probability

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.\pi^{x}.(1-\pi)^{n-x}$

In which $C_{n,x}$ is the number of different combinatios of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And $\pi$ is the probability of X happening.

In this problem, we have that:

30% carry a marker on the male chromosome that indicates an increased risk for high blood pressure, so $\pi = 0.30$

(a) If 10 men are selected randomly and tested for the marker, what is the probability that exactly 1 man has the marker?

10 men, so $n = 10$

We want to find P(X = 1). So:

$P(X = x) = C_{n,x}.\pi^{x}.(1-\pi)^{n-x}$

$P(X = 1) = C_{10,1}.(0.30)^{1}.(0.7)^{9} = 0.1211$

There is a 12.11% probability that exactly 1 man has the marker.

(b) If 10 men are selected randomly and tested for the marker, what is the probability that more than 1 has the marker?

That is $P(X > 1)$

We have that:

$P(X \leq 1) + P(X > 1) = 1$

$P(X > 1) = 1 - P(X \leq 1)$

We also have that:

$P(X \leq 1) = P(X = 0) + P(X = 1)$

$P(X = 0) = C_{10,0}.(0.30)^{0}.(0.7)^{10} = 0.0282$

$P(X \leq 1) = P(X = 0) + P(X = 1) = 0.0282 + 0.1211 = 0.1493$

Finally

$P(X > 1) = 1 - P(X \leq 1) = 1 - 0.1493 = 0.8507$

There is a 85.07% probability that more than 1 has the marker.

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