It’s a little complicated but here’s how it works:
Imagine a table with the intervals
0:4 , 4:6 , 6:7 , 7:10 , 10:13 (10 year intervals)
Then we have different rows
Class width: 4 , 2 , 1 , 3 , 3
Freq density: 0.2 , 0.5 , 1.2 , 0.7 , 0.3
So now calculate frequency where freq = class width * density
Freq: 0.8 , 1 , 3.6 , 2.1 , 0.9
So to find median find cumulative frequency
(Add all freq)
Cfreq = 8.4 now divide by 2 = 4.2
So find the interval where 4.2 lies.
0.8 + 1 = 1.8 + 3.6 = 5.6
So 4.2 (median) will lie in that interval 60-70 years.
Answer: T = 4
Step-by-step explanation:
1. Write all the variables down
P = 8 V = 2X N = 2 R = 2X X = 3
2. Since you know that X = 3 substitute it in to find V and R
V = 2X = 2(3) = 6
R = 2X = 2(3) = 6
3. Find PV
PV = P x V
= 8 x 6
= 48
4. Find NRT
NRT = N x R x T
= 2 x 6 x T
= 12 x T
= 12T
5. Find T
PV = NRT
48 = 12T
12T = 48
divide both sides by 12
T = 48 ÷ 12
T = 4
Answer:
D: New York City is the correct answer
y/x-2=3/11 would be easier to work with if we were to put it into a more standard form, e. g., y = mx + b.
First, add 2 to both sides, to isolate y/x:
y/x = 3/11 + 22/11 = 25/11.
Next, mult. both sides by x, to get y by itself: y = (25/11)x.
This is your function.
Now make a table. You can choose any x values you want, and then calculate y for each one.
x y
0 0
1 25/11
3 (25/11)*3 = 75/11
Then we have three points on this line: (0,0), (1, 25/11), 3, 75/11). You could obtain more by choosing additional x values.
C=2 times j
c=2j
o=j-6
c+j+o=58
subsitte j-6 for o and 2j for c
2j+j+j-6=58
4j-6=58
add 6 to both sides
4j=64
divide both sides by 4
j=16
sub back
c=2j
c=2(16)
c=32
o=j-6
o=16-6
o=10
curtis=32
olivia=10
jonathan=16
equations are
c+j+o=58
c=2j
o=j-6
