36*3 = 108
HOPE THIS HELPS!!!!
Answer:
first of all 44% of 600 people is equal to 264 .
the statistics is misleading because 264 people can't unless the taxi angecy included themselves
The fraction of the numbers in the distribution that are between 50 and 70 is; 95/100
<h3>How to use the empirical rule in statistics?</h3>
We are given;
Mean = 60
Standard deviation = 5
To find the numbers in the distribution that are between 50 and 70, it means that the numbers are going to be 2 standard deviations from the mean because 2σ = 2 * 5 = 10 which is the difference of both pairs from the mean.
According to empirical rule, 2 standard deviations from the mean is approximately 95/100 or 95%.
Complete question is;
In a certain distribution, the mean is 60 with a standard deviation of 2. At least what traction of the numbers are between the following pair of numbers? At least of the numbers in the distribution are between 50 and 70
Read more about Statistics Empirical Rule at; brainly.com/question/10093236
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1 and 1/3 time 2? i thinks that's what you want?
or is it 1/3x squared?
I;ll presume its 1 and 1/3 time 2
first way:
Multiple whole numbers first 1 x 2 = 2
Multiple fraction by the two 1/3 x 2 = 2/3
now add them together
2 and 2/3
second way
Make 1 and 1/3 improper = 4/3
make 2 into a a/b form = 2/1
now 4/3 x 2/1 = 8/3 now make a mixed number 2 and 2/3
Answer:
ü=2i+2j+0k
Step-by-step explanation:
The given plane 2x + 2y + 2 = 3 can also be written as:
2x+2y=3-2
2x+2y=1
The general equation for a plane is Ax+By+Cz=D and by definition the normal vector of that plane is n=Ai+Bj+Ck
Where i,j,k are the unit vectors
In order to demostrate that the vector n is normal to the plane, let R1=(a1,b1,c1) and R2=(a2,b2,c2) be two vectors that are in the plane.
If R1 ∈ Ax+By+Cz=D then Aa1+Bb1+Cc1=D
If R2 ∈ Ax+By+Cz=D then Aa2+Bb2+Cc2=D
Therefore, the vector R1R2=R2-R1=(a2-a1)i+(b2-b1)j+(c2-c1)k
You can apply the dot product. <em>If the dot product of the two vectors is zero then the vectors are normal.</em>
So, the vector which components are A,B,C is normal to the plane becase it is normal to any vector contained in the plane.
In this case:
A=2, B=2, C=0
ü=2i+2j+0k