Find the area of triangle ABC with vertices A(2,1), B (12,2), C (12,8). Hence, or otherwise find the perpendicular distance from
1 answer:
<h2>The area of a triangle is =54 square units</h2><h2>The perpendicular distance from B to AC is = 
</h2>
Step-by-step explanation:
Given a triangle ABC with vertices A(2,1),B(12,2) and C(12,8)
The area of a triangle is=
=
= = 54 square units
The length of AC =
=
= units
Let the perpendicular distance from B to AC be = x
According To Problem
⇔ units
Therefore the perpendicular distance from B to AC is =
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Answer:
(2,4) is a solution to this system of equations
Step-by-step explanation:
Given system of equation are
To find the solution of the given system of equations
To Check that (2,4) is a solution to this system or not
Solving equations (1) and (2)
From equation (1) and y=2x
Now substitute y=2x is equation (2)
10-5x=0
-5x=-10
Substitute x=2 in equation (1)
y=2x
y=2(2)
Therefore y=4
Therefore the solution is (2,4)
Therefore (2,4) is a solution to the system of equations.
Answer:
A. P(x<3.85 or x>4.15)= P(x<3.85)+P(x>4.15) = 0.1336
Step-by-step explanation:
Working with an ordinary Normal Distribution of probability and trying to find the probabilities asked in it could be difficult, because there´s no easy method to find probabilities in a generic Normal Distribution (with mean μ=4 and STD σ=0.1). The recommended approach to this question is to use a process called "Normalize", this process let us translate the problem of any Normal Distribution to a Standard Normal Distribution (μ=0 and σ=1) where there´s easier ways to find probabilities in there. The "Normalization" goes as follows:
Suppose you want to know P(x<a) of the Normal Distribution you are working with:
P(x<a)=P( (x-μ)/σ < (a-μ)/σ )=P(z<b) ( b=(a-μ)/σ )
Where μ is the mean and σ is the STD of your Normal Distribution. Notice P(z<b) now it´s a probability in a Standard Normal Distribution, now we can find it using the available method to do so. My favorite is a chart (It´s attached to this answer) that contains a lot of probabilities in a Standard Normal Distribution. Let´s solve this as an example
A. We want to find the probability of the cork being defective (P(x<3.85) + P(x>4.15)). Now we find those separated and, then, add them for our answer.
Let´s begin with P(x<3.85), we start by normalizing that probability:
P(x<3.85)= P( (x-μ)/σ < (3.85-4)/0.1 )= P(z<-1.5)
And now it´s time to use the chart, it works like this: If you want P(z<c) and the decimal expansion of c=a.bd... , then:
P(z<c)=(a.b , d)
Where (a.b , d) are the coordinates of the probability in the chart. Keep in mind that will only work with "<" (It won´t work directly with P(z>c)) and we will do some extra work in those cases.
P(z<-1.5) is in the coordinates (-1.5 , 0)
P(z<-1.5)= 0.0668
P(x<3.85)= 0.0668
Now we are looking for P(x>4.15), let´s Normalize it too:
P(x>4.15)=P( (x-μ)/σ < (4.15-4)/0.1 )=P(z>1.5)
But remember the chart only work with "<", so we need to use a property of probability:
P(z>1.5)= 1 - P(z<1.5)
Using the chart:
P(z<1.5)=0.9332 (1.5 , 0)
P(z>1.5)= 1 - 0.9332
P(z>1.5)= 0.0668
P(x>4.15)= 0.0668
And our final answer will be:
P(x<3.85 or x>4.15)= P(x<3.85)+P(x>4.15) = 0.1336
