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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
Answer:
a) factors of polynomial are: (x+1)(x-2)(x-5)
b) the required polynomial in standard form is:
Step-by-step explanation:
The polynomial has zeros at -1,2,5
Part A) Write the three factors of the polynomial
we have x=-1, x=2 and x=5 as zeros of polynomial
so factors will be:
(x+1)=0, (x-2)=0, (x-5)=0
So, factors of polynomial are: (x+1)(x-2)(x-5)
Part B) Write the polynomial in standard form
For finding polynomial, we will multiply all the factors i.e
(x+1)(x-2)(x-5)
So, the required polynomial in standard form is: