Answer:
1) 70 for both of them
2) Plot A's range is 100-50 = 50, Plot B's range is 100 - 40 = 60 so Plot A has the lesser range
3) Plot B's IQR is 85 - 60 = 25, Plot A's IQR is 80 - 65 = 15 so Plot B has the greater IQR
4) A: 80, B: 85
5) A: 65, B: 60
6) 50
7) 100
8) Plot B, see #2 for work
9) 50% - 50% of data is always between Q1 and Q3
10) 25%
11) 25%
The open circle means the inequality will be greater than or equal to (≥) or less than or equal to (≤).
A closed circle means the inequality will be greater than (>) or less than (<)
An arrow pointing right to increasingly positive values means the inequality is getting greater (> or ≥)
A narrowing pointing left to increasingly negative values means the inequality is getting lesser (< or ≤)
So for this graph with an open circle and rightward pointing arrow, “x” will be some number on the number line greater than the first point of -38:
x > -38
This is a geometric sequence because each term is twice the value of the previous term. So this is what would be called the common ratio, which in this case is 2. Any geometric sequence can be expressed as:
a(n)=ar^(n-1), a(n)=nth value, a=initial value, r=common ratio, n=term number
In this case we have r=2 and a=1 so
a(n)=2^(n-1) so on the sixth week he will run:
a(6)=2^5=32
He will run 32 blocks by the end of the sixth week.
Now if you wanted to know the total amount he runs in the six weeks, you need the sum of the terms and the sum of a geometric sequence is:
s(n)=a(1-r^n)/(1-r) where the variables have the same values so
s(n)=(1-2^n)/(1-2)
s(n)=2^n-1 so
s(6)=2^6-1
s(6)=64-1
s(6)=63 blocks
So he would run a total of 63 blocks in the six weeks.
Use this version of the Law of Cosines to find side b:
b^2 = a^2 + c^2 − 2ac cos(B)
We want side b.
b^2 = (41)^2 + (20)^2 - 2(41)(20)cos(36°)
After finding b, you can use the Law of Sines to find angles A and C or use other forms of the Law of Cosines to find angles A and C.
Try it....
