Answer:
congruent
Step-by-step explanation:
If two lines intersect to form a linear pair of congruent angles, then the lines are perpendicular. When two adjacent angles form a linear pair, their non-shared sides form a straight line (m).
i'm not sure tho
I think it's C.
Let me know!
I'll talk you through it so you can see why it's true, and then
you can set up the 2-column proof on your own:
Look at the two pointy triangles, hanging down like moth-wings
on each side of 'OC'.
-- Their long sides are equal, OA = OB, because both of those lines
are radii of the big circle.
-- Their short sides are equal, OC = OC, because they're both the same line.
-- The angle between their long side and short side ... the two angles up at 'O',
are equal, because OC is the bisector of the whole angle there.
-- So now you have what I think you call 'SAS' ... two sides and the included angle of one triangle equal to two sides and the included angle of another triangle.
(When I was in high school geometry, this was not called 'SAS' ... the alphabet
did not extend as far as 'S' yet, and we had to call this congruence theorem
"broken arrow".)
These triangles are not congruent the way they are now, because one is
the mirror image of the other one. But if you folded the paper along 'OC',
or if you cut one triangle out and turn it over, it would exactly lie on top of
the other one, and they would be congruent.
So their angles at 'A' and at 'B' are also equal ... those are the angles that
you need to prove equal.
Answer:
(a) The probability that <em>Y</em>₂₀ exceeds 1000 is 3.91 × 10⁻⁶.
(b) <em>n</em> = 28.09
Step-by-step explanation:
The random variable <em>Y</em>ₙ is defined as the total numbers of dollars paid in <em>n</em> years.
It is provided that <em>Y</em>ₙ can be approximated by a Gaussian distribution, also known as Normal distribution.
The mean and standard deviation of <em>Y</em>ₙ are:

(a)
For <em>n</em> = 20 the mean and standard deviation of <em>Y</em>₂₀ are:

Compute the probability that <em>Y</em>₂₀ exceeds 1000 as follows:

**Use a <em>z </em>table for probability.
Thus, the probability that <em>Y</em>₂₀ exceeds 1000 is 3.91 × 10⁻⁶.
(b)
It is provided that P (<em>Y</em>ₙ > 1000) > 0.99.

The value of <em>z</em> for which P (Z < z) = 0.01 is 2.33.
Compute the value of <em>n</em> as follows:

The last equation is a quadratic equation.
The roots of a quadratic equation are:

a = 16
b = -805.4289
c = 10000
On solving the last equation the value of <em>n</em> = 28.09.