Factorize the numerator and denominator, then simplify:
10/45 = (2×5)/(9×5) = 2/9
Now,
1/9 = 1/10 + 1/90
1/90 = 1/10 × 1/9 = 1/10 × (1/10 + 1/90) = 1/100 + 1/900
1/900 = 1/100 × 1/9 = 1/100 × (1/10 + 1/90) = 1/1000 + 1/9000
and so on, which is to say
1/9 = 1/10 + 1/100 + 1/1000 + …
or
1/9 = 0.111…
so that multiplying both sides by 2 gives
2/9 = 0.222…
The data given as a whole would be called ungrouped data. Now to get the variance, you will need the formula:
s^2= <u>Σ(x-mean)^2</u>
n
x = raw data
mean = average of all data
n = no. of observations
s^2 = variance
Now we do not have the mean yet, so you have to solve for it. All you need to do is add up all the data and divide it by the number of observations.
Data: <span>90, 75, 72, 88, 85 n= 5
</span>Mean=<u>Σx</u>
n
Mean = <u>90+75+72+88+85 </u> = <u>410</u> = 82
5 5
The mean is 82. Now we can make a table using this.
The firs column will be your raw data or x, the second column will be your mean and the third will be the difference between the raw data and the mean and the fourth column will be the difference raised to two.
90-82 = 8
8^2 =64
75-82 = -7
-7^2 =49
72-82 = -10
-10^2=100
88-82=6
6^2 = 12
85-82=3
3^2=9
Now you have your results, you can now tabulate the data:
x mean x-mean (x-mean)^2
90 82 8 64
75 82 -7 49
72 82 -10 100
88 82 6 36
85 82 3 9
Now that you have a table, you will need the sum of (x-mean)^2 because the sigma sign Σ in statistics, means "the sum of."
64+49+100+36+9 = 258
This will be the answer to your question. The value of the numerator of the calculation will be 258.
<u>
</u>
A because the origin is at 0,0 and point A is at 0,0
Answer:
Step-by-step explanation:
First digit 2,3,4,5 = 4 different digits.
Second digit 2,3,4,5 = 4 different digits.
Third digit = 1,2,3,6,7,9 = 6 different digits.
Total number of possibilities = 4*4*6 = 96
Answer:
Step-by-step explanation:
The pediatrician said that the z-score for the boy's height, relative to other 2-year-olds in the country, was 1.59
He explains to the parents of the boy, that the extreme 5% of cases often require special attention.
(A) Does this child fall into that group?
Yes, this child falls into that group. He is among the extreme 5% of cases at the tail of the distribution.
(B) What do you need to assume about the heights of 2-year-olds, in order to find your answer to part (A)?
Assume that the distribution of heights is skewed, the heights are independent of one another, and the number of heights is very large.
