Given that lines a and b are parallel, what angles formed on line b when cut by the transversal are congruent with ∠2?
1 answer:
Answer:
The angles formed on line b when cut by the transversal are congruent with ∠2 are
Step-by-step explanation:
Consider the provided information.
If transversal line crossed by two parallel lines, then, the corresponding angles and alternate angles are equal .
The angles on the same corners are called corresponding angle.
Alternate Angles: Angles that are in opposite positions relative to a transversal intersecting two lines.
∠2 and ∠6 are corresponding angles
Therefore, ∠2 = ∠6
∠2 and ∠7 are alternate exterior angles
Therefore, ∠2 = ∠7
Hence, the angles formed on line b when cut by the transversal are congruent with ∠2 are
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1) Mean: 6.5
2) Standard deviation: 3.2
Step-by-step explanation:
1)
A uniform random distribution is the distribution of a variable that can assume n different values, and each of these values is equally likely to occur.
This means that the probability of occurring of each of these values is the same:
For a uniform random distribution, the mean of the distribution is simply equal to the median value of the distribution, which is exactly at the center of the distribution itself. It can be calculated as
where
a is the lowest value of the distribution
b is the highest value of the distribution
Here the distribution consists of all values ranging from 1 to 12, so:
a = 1
b = 12
Therefore the mean value is
2)
For a uniform random distribution, it can be dimostrated that the variance of the distribution is
For the distribution in this problem, we have:
a = 1
b = 12
Therefore the variance is
The standard deviation is the square root of the variance, therefore for this distribution it is:
Learn more about mean and standard deviation:
#LearnwithBrainly
Answers:
- System 1: x = 3 and y = -5
- System 2: x = 1 and y = 2
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Explanation:
For now, let's focus on system 1.
We have identical copies of "2y" in each equation, which means that if we were to subtract the equations straight down, then the y terms cancel out. That allows us to solve for x.
- The x terms subtract to 7x-x = 6x
- The y terms subtract to 0 and go away
- The right hand sides subtract to 11 minus (-7) = 11-(-7) = 11+7 = 18
After doing those subtractions, we have the new equation 6x = 18 which solves to x = 3. Divide both sides by 6 to isolate x.
Once we know x, we can find out y.
7x+2y = 11
7(3)+2y = 11
21+2y = 11
2y = 11-21
2y = -10
y = -10/2
y = -5
or we can say
x+2y = -7
3+2y = -7
2y = -7-3
2y = -10
y = -10/2
y = -5
We should lead to the same y value either way. This helps confirm we have the correct x value.
Overall, the solution to system 1 is (x,y) = (3, -5)
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Now onto system 2.
Unfortunately, we don't have identical y terms this time. But what we can do is double everything in the second equation to go from -x+y = 1 to -2x+2y = 2.
The new equivalent system is
Now we have the same situation as before: Subtract straight down, solve for x.
- The x terms subtract to 7x-(-2x) = 7x+2x = 9x
- The y terms go away when subtracting
- The right hand sides subtract to 11-2 = 9
So we have 9x = 9 and that solves to x = 1.
We'll use this x to find the y value.
7x+2y = 11
7(1)+2y = 11
7+2y = 11
2y = 11-7
2y = 4
y = 4/2
y = 2
Or you could pick on the second equation
-x+y = 1
-1+y = 1
y = 1+1
y = 2