Answer:
Class interval 10-19 20-29 30-39 40-49 50-59
cumulative frequency 10 24 41 48 50
cumulative relative frequency 0.2 0.48 0.82 0.96 1
Step-by-step explanation:
1.
We are given the frequency of each class interval and we have to find the respective cumulative frequency and cumulative relative frequency.
Cumulative frequency
10
10+14=24
14+17=41
41+7=48
48+2=50
sum of frequencies is 50 so the relative frequency is f/50.
Relative frequency
10/50=0.2
14/50=0.28
17/50=0.34
7/50=0.14
2/50=0.04
Cumulative relative frequency
0.2
0.2+0.28=0.48
0.48+0.34=0.82
0.82+0.14=0.96
0.96+0.04=1
The cumulative relative frequency is calculated using relative frequency.
Relative frequency is calculated by dividing the respective frequency to the sum of frequency.
The cumulative frequency is calculated by adding the frequency of respective class to the sum of frequencies of previous classes.
The cumulative relative frequency is calculated by adding the relative frequency of respective class to the sum of relative frequencies of previous classes.
First step is to plot all the given coordinates in the xy plane.
Notice that after plotting the points A, B, C and D, we have got the figure ABCD is a trapezium with bases AD = 4 units, BC = 6 units and height: DE= 5 units.
Formula to find the area of trapezium is:
Area = 
= 
=
= 5 * 5
= 25 square units.
So, third choice is correct.
Hope this helps you!
Answer:
P(x) = (x - 1)(x + 2)(x - 3)(3x + 1)
Step-by-step explanation:
Since P(1) = 0 and P(- 2) = 0, then
(x - 1) and (x + 2) are factors of P(x)
(x - 1)(x + 2) = x² + x - 2 ← is also a factor of P(x)
dividing 3
- 5x³ - 17x² + 13x + 6 by x² + x - 2 gives
P(x) = (x - 1)(x + 2)(3x² - 8x - 3) = (x - 1)(x + 2)(x - 3)(3x + 1)
Answer:
9x^2 - 3x - 2, R -7.
Step-by-step explanation:
x - 5) 9x^3 - 48x^2 + 13x + 3(9x^2 - 3x - 2 <---- Quotient.
9x^3 - 45x^2
- 3x^2 + 13x
-3x^2 + 15x
-2x + 3
-2x + 10
-7 <----- Remainder
