Answer: 5
Step-by-step explanation:
This question is incomplete, the complete question is;
The distribution of scores on a recent test closely followed a Normal Distribution with a mean of 22 points and a standard deviation of 2 points. For this question, DO NOT apply the standard deviation rule.
What proportion of the students scored at least 23 points on this test, rounded to five decimal places?
Answer:
proportion of the students that scored at least 23 points on this test is 0.30850
Step-by-step explanation:
Given the data in the question;
mean μ = 22
standard deviation σ = 2
since test closely followed a Normal Distribution
let
Z = x-μ / σ { standard normal random variable ]
Now, proportion of the students that scored at least 23 points on this test.
P( x ≥ 23 ) = P( (x-μ / σ) ≥ ( 23-22 / 2 )
= P( Z ≥ 1/2 )
= P( Z ≥ 0.5 )
= 1 - P( Z < 0.5 )
Now, from z table
{ we have P( Z < 0.5 ) = 0.6915 }
= 1 - P( Z < 0.5 ) = 1 - 0.6915 = 0.30850
P( x ≥ 23 ) = 0.30850
Therefore, proportion of the students that scored at least 23 points on this test is 0.30850
Using the distance formula, you get sqrt(36+25) which is sqrt(61) which is around 7.8
Answer:
- <u>Each digit has a value ten times the value of the digit to its right.</u>
Explanation:
The given number is 9,999,999.
Each digit has a value according to its place.
The right most 9 is in the place of the ones and its value is 9 × 10⁰ = 9 × 1 = 9.
The second 9 from the right is in the place of the tens and its value is 9 × 10 = 90.
The third 9 from the right end is in the place of the hundreds and its value is 9 × 100 = 900.
The next 9 is in the place of the thousands, so its value is 9,000.
So, each 9 has a value ten times the 9 to its right.