answer it is right at right write L.C.M
Sam is keeping track of the distances at a long jump competition. The jumper from East side High jumped 7.85 meters. The jumper
2 answers:
You might be interested in
We need to work out the z-value of each question to work out the probability
Question a)
We have
X = 5 minutes
The mean, μ = 6.7 minutes
Standard deviation, σ = 2.2 minutes
z-score = (X - μ) / σ = (5 - 6.7) / 2.2 = -0.77
We want to find the probability of z < -0.77 so we read the z-table for the value of z on the left of -0.77 we have the probability P(z< -0.77) = 0.2206
The table is attached in picture 1 below
The probability of assembly time less than 5 minutes is 0.2206 = 22.06%
Question b)
z-score = (10-5) / 2.2 = 2.27
Reading the z-table for P(z<2.27) = 0.9884
The table reading is shown in the second picture below
The probability the assembly time will be less than 10 minutes is 0.9884
We can use this information to find the probability of the assembly time will be more than 10 minutes = 1 - 0.9884 = 0.0116 = 1.16%
Question c)
The value of z between 5 minutes and 10 minutes is
P(-0.77<z<2.27) = P(z<2.27) - P(z< -0.77)
P(-0.77<z<2.27) = 0.9884 - 0.2206 = 0.7678 = 76.78%
Question a)
We have
X = 5 minutes
The mean, μ = 6.7 minutes
Standard deviation, σ = 2.2 minutes
z-score = (X - μ) / σ = (5 - 6.7) / 2.2 = -0.77
We want to find the probability of z < -0.77 so we read the z-table for the value of z on the left of -0.77 we have the probability P(z< -0.77) = 0.2206
The table is attached in picture 1 below
The probability of assembly time less than 5 minutes is 0.2206 = 22.06%
Question b)
z-score = (10-5) / 2.2 = 2.27
Reading the z-table for P(z<2.27) = 0.9884
The table reading is shown in the second picture below
The probability the assembly time will be less than 10 minutes is 0.9884
We can use this information to find the probability of the assembly time will be more than 10 minutes = 1 - 0.9884 = 0.0116 = 1.16%
Question c)
The value of z between 5 minutes and 10 minutes is
P(-0.77<z<2.27) = P(z<2.27) - P(z< -0.77)
P(-0.77<z<2.27) = 0.9884 - 0.2206 = 0.7678 = 76.78%
Answer:
0.25
Step-by-step explanation:
1. Add up all of the students.
20+20+25+7+13+5+3+2+5=100
2. Make a ratio of that number and the number of all the students in Music/Drama
MUSIC/DRAMA: 7+13+5=25
3. Simplify
= 0.25
Given :
Nathan smiled at 50 people one day and recorded that 36 people smiled back at him.
To Find :
How many people should Nathan expect to return a smile if he smiles at 650 people over a period of time.
Solution :
Ratio of people smiled by total number of people is :
Now, it is given that we have to use the given conditions.
Therefore, ratio will be same :
Therefore, number of smiles Nathan expect to return is 468.