Answer: the probability that a randomly selected value is greater than 161.9 is 0.977
Step-by-step explanation:
Since the distribution of values is normal, we would apply the formula for normal distribution which is expressed as
z = (x - µ)/σ
Where
x = randomly selected values.
µ = mean value
σ = standard deviation
From the information given,
µ = 165.6
σ = 18.7
We want to find the probability that a randomly selected value is greater than 161.9.. It is expressed as
P(x > 161.9) = 1 - P(x ≤ 161.9)
For x = 161.9
z = (161.9 - 165.6)/18.7 = - 0.2
Looking at the normal distribution table, the probability corresponding to the z score is 0.023
P(x > 161.9) = 1 - 0.023 = 0.977
You can first take out an x^2 from each number , 3x^6 - 39x^4 + 108x^2 = x^2 ( 3x^4 - 39x^2 + 108) Then just factor and solve for the zero's of the equation
The answer to this radical expression is -108h^2radical sign 5
30/4 = 7 2/4
= 7 1/2
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Step-by-step explanation:
(x-3)(x^2+2x+1)
=x^3-3x^2+2x^2-6x+x-3
=x^3+2x^2-3x^2-6x+x-3
=x^3-x^2-5x-3
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