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%
B. 4.5=2m+g
f. 9.25=3m+2g
1.25=m
sub
b. 4.5=2(1.25)+g
f. 9.25=3(1.25)+2g
b. 4.5=2.5+g
f. 9.25=3.75+2g
minus 2.5 from both sides on the b. equaion and minus 3.75 from both sides on the f. equation
b. 2=g
f. 5.5=2g
divide both sides of f. equation by 2
b. 2=g
f. 2.75=g
2.75-2=0.75
finn paied $0.75 more for a bag of grapes
Answer:
0.27 repeating.
Step-by-step explanation:
So to solve this we just have to use division.
1 clearly doesnt go into 3 at least one time, so we can add a decimal point and add a 0 to make it 30.
11 goes into 30 2 times, so we have:
0.2
and 30-22=8
So we can add another 0 and make it 80.
Then 11 goes into 80 7 times. So we have:
0.27
and 80-77=3
So again, add the 0, we have 30.
11 goes into 30 2 times, so:
0.272
and 30-22=8
Add another 0 we get 80.
11 goes into 80 7 times.
So finally, we have:
0.2727.
This is a repeating decimal.
This can be shown as:
0.<u>27</u>
So this is your answer!
I was a bit confused with which one it was on your answer key, but knowing that it is 0.<u>27</u> I am guessing you can chose!
Hope this helps!
answer
10
step-by-step explanation
the equation given is sin(x) = cos(y) with x = 2k + 3 and y = 6k + 7
substitute in 2k+ 3 for x in sin(x) and substitute in 6k + 7 for y in cos(y)
sin(x) = cos(y)
sin(2k + 3) = cos(6k + 7)
we know that sin(x) = cos(90 -x)
sin(2k + 3)
= cos(90 - (2k + 3) )
= cos(90 - 2k - 3)
= cos(87 - 2k)
substitute this into the equation sin(2k + 3) = cos(6k + 7)
sin(2k + 3) = cos(6k + 7)
cos(87 - 2k) = cos(6k + 7)
87 - 2k = 6k +7
80 = 8k
k = 10
