From given picture we can see that X is mid point of line PQ
So PQ=2PX
From given picture we can see that Y is mid point of line PR
So PR=2PY
Now consider triangles PRQ and PYX
Sides PQ and PX has same ratio as of sides PR and PY (as calculated above)
angle XPY= angle QPR (common angle)
Hence triangles PRQ and PYX are similar.
We know that ratio of the sides of similar triangles is always equal so we can write:

Plug the given values QR=8 and PQ=2PX





Hence final answer is XY = 4 units.
Answer:
The slope is 3 and the y-intercept is -5
If you need to y-intercept in coordinate form, then it's (0, -5)
Step-by-step explanation:
y = mx + b, the m is the slope and the b is the y-intercept
Answer:
C
Step-by-step explanation:
It crosses two points and its on the positive side so it'll be a positive 4 and 0
Answer:
a1 = 2
d = 3
an = 2 + (n - 1) * 3
a7 = 20
a59 = 176
Steps:
a1 is the initial value (when n equals 1), and since there are 2 crosses, it is 2.
d is the added value to each amount of crosses. And since the second amount is 5 and the third amount is 8, we can determine that each n is adding 3 crosses, therefore making d = 3.
The equation is simply plugging in the values for a1 and d.
A7 is simply plugging in 7 for n in the equation and solving for it. So;
a7 = 2 + (7 - 1) * 3
a7 = 2 + 6 * 3
a7 = 2 + 18
a7 = 20
And same thing as the last for 59 except substitute 59 in for where you put 7;
a59 = 2 + (59 - 1) * 3
a59 = 2 + 58 * 3
a59 = 2 + 174
a59 = 176
Answer: the student's score closest to 91 percentile.
Step-by-step explanation:
Since the scores on the standardized test are approximately normally distributed, we would apply the formula for normal distribution which is expressed as
z = (x - µ)/σ
Where
x = test scores.
µ = mean score
σ = standard deviation
From the information given,
µ = 480
σ = 90
If a student has a score of 600, then x = 600
For x = 600,
z = (600 - 480)/90 = 1.33
Looking at the normal distribution table, the probability corresponding to the z score is 0.91
the student's score closest to 91 percentile.