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KatRina [158]
3 years ago
8

The points at x equals=​_______ and xequals=​_______ are the inflection points on the normal curve

Mathematics
1 answer:
uysha [10]3 years ago
8 0
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.






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Read 2 more answers
AB and BC form a right angle at point B. If A = (-3,-1) and B = (4,4), what is the equation of BC
m_a_m_a [10]

Answer:

y=-1.4x+9.6

Step-by-step explanation:

If A = (-3,-1) and B = (4,4),  the slope of AB is

\text{Slope}_{AB}=\dfrac{-1-4}{-3-4}=\dfrac{-5}{-7}=\dfrac{5}{7}

Two perpendicular lines have slopes that have product of -1:

\text{Slopee}_{AB}\cdot \text{Slope}_{BC}=-1\\ \\\dfrac{5}{7}\cdot \text{Slope}_{BC}=-1\\ \\\text{Slope}_{BC}=-\dfrac{7}{5}=-1.4

The equation of the line BC with slope -1.4 and passing through the point B(4,4) is

y-4=-1.4(x-4)\\ \\y-4=-1.4x+5.6\\ \\y=-1.4x+9.6

7 0
3 years ago
Given: ∠AOC, ∠BOC - complementary angles
34kurt

Answer:

  • 30
  • COB . . . . or . . . . BOC

Step-by-step explanation:

The reason given on the line of interest is "substitution," so the problem boils down to determining what was substituted for what. The previous statement says ...

  AOC + COB = 90

and the first part of the statement we're to complete has BOC + ___.

We see from the given statements that m∠AOC = m∠BOC + 30°, so it appears that is the substitution that has been made: AOC has been replaced by BOC + 30.

This means the first blank is filled with 30.

__

The second part of the previous statement is ...

  AOC + COB = 90

so we believe this (COB) should go in the second blank.

__

Then the line of interest would read ...

  BOC + 30 + COB = 90

_____

<em>Comment on the problem</em>

There is a curious mix of notations here. Usually, (as in the beginning of this problem) we refer to the measure of an angle using "m∠" in front of the angle designator, and we use a degree symbol to indicate the units of that measure. Part-way through the problem statement written here, those notations were dropped, and we're to assume they are intended. IMO, this is a poor way to demonstrate careful problem solving.

The substitution given for AOC is BOC+30, but the line into which that is substituted has AOC +COB = 90. This means the equation after substitution is ...

  BOC +30 +COB = 90

Since BOC and COB are the same angle, we can sort of fudge the "algebra" to get to  BOC=30, but if the problem were more carefully written, the angle would be referred to by consistent nomenclature:

  m∠AOC + m∠BOC = 90° . . . . . . . . . preferred angle designations

  (m∠BOC + 30°) + m∠BOC = 90° . . . . substitution for m∠AOC

  2(m∠BOC) = 60° . . . . . . algebra (subtract 30°, collect terms)

  m∠BOC = 30° . . . . . . . . algebra (divide by 2)

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