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:
0.43715
Step-by-step explanation:
We solve using z score calculator
z = (x-μ)/σ, where
x is the raw score
μ is the population mean = $276,000
σ is the population standard deviation = 32,000
For x = $276,000
z = 276,000 - 276,000/32000
z = 0
Probability value from Z-Table:
P(x = 276000) = 0.5
For x = $325,000
z = 325,000 - 276,000/32000
z = 1.53125
Probability value from Z-Table:
P(x = 325000) = 0.93715
The probability that the next house in the community will sell for between $276,000 and $325,000 is
P(x = 325000) - P(x = 276000)
= 0.93715 - 0.5
= 0.43715
Answer:
B. f(x) = -x + 11
Step-by-step explanation:
First if you rearrange the equation you get f(x) = -x + 11, like this:
x + y = 11
Minus x from both sides
y = -x + 11
F(x) and y are interchangeable so we just put f(x) in
f(x) = -x + 11
Each of the graphs are attached and we see that they are the same!
<em>Hope this helps!!</em>
<em>- Kay</em>
