we know that
if two lines are parallel
then
their slopes are equal
Step 1
Find the slope of the line 2x + 5y = 4
2x + 5y = 4------> 5y=4-2x-----> y=(4/5)-(2/5)x------> slope m=(-2/5)
step 2
with m=-2/5 and the point (5, –4) find the equation of the line
y-y1=m*(x-x1)------> y+4=(-2/5)*(x-5)----> y=(-2/5)x+2-4
y=(-2/5)x-2
therefore
the answer is the option D
Answer:
We conclude that If Tawnee increases the length and width of the playground by a scale factor of 2, the perimeter of the new playground will be twice the perimeter of the original playground.
Step-by-step explanation:
We know that the perimeter of a rectangle = 2(l+w)
i.e.
P = 2(l+w)
Here
Given that the length and width of the playground by a scale factor of 2
A scale factor of 2 means we need to multiply both length and width by 2.
i.e
P = 2× 2(l+w)
P' = 2 (2(l+w))
= 2P ∵ P = 2(l+w)
Therefore, we conclude that If Tawnee increases the length and width of the playground by a scale factor of 2, the perimeter of the new playground will be twice the perimeter of the original playground.
Asimply measure its length. What else could you measure? After all, length is the only feature a segment has. You’ve got your short, your medium, and your long segments.
Step-by-step explanation:
length
<h3>
Answer: Choice B</h3>
=========================================================
Explanation:
The rule we use is
![\Large a^{m/n} = \sqrt[n]{a^m} = \left(\sqrt[n]{a}\right)^m](https://tex.z-dn.net/?f=%5CLarge%20a%5E%7Bm%2Fn%7D%20%3D%20%5Csqrt%5Bn%5D%7Ba%5Em%7D%20%3D%20%5Cleft%28%5Csqrt%5Bn%5D%7Ba%7D%5Cright%29%5Em)
where 'a' is the base, m stays in the role of the exponent, and n plays the role of the root index (eg: n = 3 is a cube root, n = 4 is a fourth root, and so on).
So for instance,
![\Large 2^{3/4} = \sqrt[4]{2^3} = \left(\sqrt[4]{2}\right)^3](https://tex.z-dn.net/?f=%5CLarge%202%5E%7B3%2F4%7D%20%3D%20%5Csqrt%5B4%5D%7B2%5E3%7D%20%3D%20%5Cleft%28%5Csqrt%5B4%5D%7B2%7D%5Cright%29%5E3)
or in this case,
![\Large t^{5/8} = \sqrt[8]{t^5} = \left(\sqrt[8]{t}\right)^5](https://tex.z-dn.net/?f=%5CLarge%20t%5E%7B5%2F8%7D%20%3D%20%5Csqrt%5B8%5D%7Bt%5E5%7D%20%3D%20%5Cleft%28%5Csqrt%5B8%5D%7Bt%7D%5Cright%29%5E5)
A = event the person got the class they wanted
B = event the person is on the honor roll
P(A) = (number who got the class they wanted)/(number total)
P(A) = 379/500
P(A) = 0.758
There's a 75.8% chance someone will get the class they want
Let's see if being on the honor roll changes the probability we just found
So we want to compute P(A | B). If it is equal to P(A), then being on the honor roll does not change P(A).
---------------
A and B = someone got the class they want and they're on the honor roll
P(A and B) = 64/500
P(A and B) = 0.128
P(B) = 144/500
P(B) = 0.288
P(A | B) = P(A and B)/P(B)
P(A | B) = 0.128/0.288
P(A | B) = 0.44 approximately
This is what you have shown in your steps. This means if we know the person is on the honor roll, then they have a 44% chance of getting the class they want.
Those on the honor roll are at a disadvantage to getting their requested class. Perhaps the thinking is that the honor roll students can handle harder or less popular teachers.
Regardless of motivations, being on the honor roll changes the probability of getting the class you want. So Alex is correct in thinking the honor roll students have a disadvantage. Everything would be fair if P(A | B) = P(A) showing that events A and B are independent. That is not the case here so the events are linked somehow.
