There are 10 letters in Peppermill but we have 3p,2l,2e. We must take this into consideration.
10!/(3!2!2!)=151200
Hope you have a calculator with !
Answer:
y= -1/4 x+4
Step-by-step explanation:
1. Solve for y
2. Use the same equation you just got without the 7 (b value) and plug in the new point (8,2) into the x and y
3. Solve for b
4. Final equation includes the slope found in step 1 and the new b value
7b + 3 - 4b = 3 - 3(b + 4)
3b + 3 = 3 - 3b - 12
3b + 3b = 3 - 12 - 3 = -12
6b = -12
b = -12/6 = -2
b = -2
Answer:
0.1715
Step-by-step explanation:
From previous experience 340/1982 is the past probability that a person has high blood pressure.
The division of that leads to 0.1715 that the next person will have high blood pressure. That number is roughly (very roughly) 1 person in 6. So if 6 people come in 1 will have high blood presure.
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.