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.

7 0

2 years ago

Lucy has 4 inches of ribbon that she wants to cut into 6 equal sized pieces. What is true about the length of each piece of ribb

PolarNik [594]

Answer:

Option C

Step-by-step explanation:

Given

Total length of ribbon lucy has=4 inches

She wants to divide it into 6 equal sized pieces so

Length of each piece=4/6

This can also be written as:

Length of each piece=4*(1/6 )

Now we will look at the options one by one:

For A, Each piece has 6/4 inches of ribbon.. This statement is not true as we have to divide 4 inches into 6 pieces so 4 is supposed to be in the numerator.

For B, breaking down 4/6 gives us 4*(1/6), so the statement is not true.

For C , as we can see that the expression is 4*(1/6), each piece will have 1/6 of inches of ribbon.

So option C is the correct answer..

6 0

2 years ago