Answer:
-48
Step-by-step explanation:
We know that the probability density function of a variable that is normally distributed is f(x) = 1/(σ√2π) * exp[1/2 (x – µ). Its inflection point is the point where f"(x) = 0.
Taking the first derivative, we get f'(x) = –(x–µ)/(σ³/√2π) exp[–(x–µ)²/(2σ²)] = –(x–µ) f(x)/σ².
The second derivative would be f"(x) = [ –(x–µ) f(x)/σ]' = –f(x)/σ² – (x–µ) f'(x)/σ² = –f(x)/σ² + (x-µ)² f(x)/σ⁴.
Setting this expression equal to zero, we get
–f(x)/σ² + (x-µ)² f(x)/σ⁴ = 0
Multiply both sides by σ⁴/f(x):
–σ² + (x-µ)² = 0
(x-µ)² = σ²
x-µ= + – σ
x = µ +– σ
So the answers are x = µ – σ and x = µ + σ.
There are 6×6×6 = 216 possible outcomes. Of those, 5×5×5 = 125 of them will be outcomes that do not have any occurrences of 6. The number that will have at least one occurrence of 6 is 216 - 125 = 91.
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5×5×5 counts the number of ways that the numbers 1–5 might be rolled on 3 rolls.
Should be D, if you type all of it in the calculator .
