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For the following normal distribution, give the x-values of the inflection points of the curve (the points where the curve's concavity changes). x ~ N(0, 52)

Answers

Answer 1

Answer:

Answer:

The x-values of the inflection points of the curve are x = -52 and x = 52.

Step-by-step explanation:

Suppose we have a normal curve with mean $\mu$ and standard deviation $\sigma$

The x-values of the inflection points of the curve are $x = \mu - \sigma$ and $x = \mu + \sigma$

x ~ N(0, 52)

This means that $\mu = 0, \sigma = 52$

$x = 0 - 52 = -52$

$x = 0 + 52 = 52$

The x-values of the inflection points of the curve are x = -52 and x = 52.

Answer 2

Answer:

Final answer:

The inflection points in a normal distribution occur at one standard deviation above and below the mean. For the distribution x ~ N(0, 52), the inflection points are at -√52 and √52.

Explanation:

In a normal distribution, the inflection points occur at one standard deviation above and below the mean. In the given normal distribution, x ~ N(0, 52), the mean (μ) is 0 and the standard deviation (σ) is the square root of the variance, i.e., √52.

Therefore, the x-values for the inflection points would be μ - σ and μ + σ, which are -√52 and +√52, respectively.

Learn more about Inflection Points here:

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What is the absolute value of -6.8

Answers

Answer: 6.8

Step-by-step explanation:

absolute value turns the number into the positive value.

The correct answer is 6.8

What is the solution to x+5>-7

Answers

According to the given data we have the following inequality:

$x+5>-7$

To find the solution of the inequality above we would make the following:

$\begin{gathered} \mathrm{Subtract\: }5\mathrm{\: from\: both\: sides} \n x+5-5>-7-5 \end{gathered}$ $\begin{gathered} \text{Finally, simplify},\text{ and so:} \n x>-12 \end{gathered}$

Therefore, the solution to x+5>-7 would be x>-12

Evaluate the expression if b = 5 and c = -4.
3c - 2b

Answers

Answer:

3(-4) - 2(5) = -12 - 10 = -22

Determine whether the improper integral converges or diverges, and find the value of each that converges.∫^0_-[infinity] 5e^60x dx

Answers

Answer:

The improper integral converges.

$\displaystyle \int\limits^0_(- \infty) {5e^(60x)} \, dx = (1)/(12)$

General Formulas and Concepts:
Calculus

Limit

Limit Rule [Variable Direct Substitution]: $\displaystyle \lim_(x \to c) x = c$

Differentiation

Derivatives
Derivative Notation

Derivative Property [Multiplied Constant]: $\displaystyle (d)/(dx) [cf(x)] = c \cdot f'(x)$

Derivative Rule [Basic Power Rule]:

f(x) = cxⁿ
f’(x) = c·nxⁿ⁻¹

Integration

Integrals

Integration Rule [Reverse Power Rule]: $\displaystyle \int {x^n} \, dx = (x^(n + 1))/(n + 1) + C$

Integration Rule [Fundamental Theorem of Calculus 1]: $\displaystyle \int\limits^b_a {f(x)} \, dx = F(b) - F(a)$

Integration Property [Multiplied Constant]: $\displaystyle \int {cf(x)} \, dx = c \int {f(x)} \, dx$

Integration Method: U-Substitution

Improper Integral: $\displaystyle \int\limits^(\infty)_a {f(x)} \, dx = \lim_(b \to \infty) \int\limits^b_a {f(x)} \, dx$

Step-by-step explanation:

Step 1: Define

Identify.

$\displaystyle \int\limits^0_(- \infty) {5e^(60x)} \, dx$

Step 2: Integrate Pt. 1

[Integral] Rewrite [Integration Property - Multiplied Constant]: $\displaystyle \int\limits^0_(- \infty) {5e^(60x)} \, dx = 5 \int\limits^0_(- \infty) {e^(60x)} \, dx$
[Integral] Rewrite [Improper Integral]: $\displaystyle \int\limits^0_(- \infty) {5e^(60x)} \, dx = \lim_(a \to - \infty) 5 \int\limits^0_(a) {e^(60x)} \, dx$

Step 3: Integrate Pt. 2

Identify variables for u-substitution.

Set u: $\displaystyle u = 60x$
[u] Differentiate [Derivative Properties and Rules]: $\displaystyle du = 60 \ dx$
[Bounds] Swap: $\displaystyle \left \{ {{x = 0 \rightarrow u = 0} \atop {x = a \rightarrow u = 60a}} \right.$

Step 4: Integrate Pt. 3

[Integral] Rewrite [Integration Property - Multiplied Constant]: $\displaystyle \int\limits^0_(- \infty) {5e^(60x)} \, dx = \lim_(a \to - \infty) (1)/(12) \int\limits^0_(a) {60e^(60x)} \, dx$
[Integral] Apply Integration Method [U-Substitution]: $\displaystyle \int\limits^0_(- \infty) {5e^(60x)} \, dx = \lim_(a \to - \infty) (1)/(12) \int\limits^0_(60a) {e^(u)} \, du$
[Integral] Apply Exponential Integration: $\displaystyle \int\limits^0_(- \infty) {5e^(60x)} \, dx = \lim_(a \to - \infty) (1)/(12) e^u \bigg| \limits^0_(60a)$
Evaluate [Integration Rule - Fundamental Theorem of Calculus 1]: $\displaystyle \int\limits^0_(- \infty) {5e^(60x)} \, dx = \lim_(a \to - \infty) (1 - e^(60a))/(12)$
[Limit] Evaluate [Limit Rule - Variable Direct Substitution]: $\displaystyle \int\limits^0_(- \infty) {5e^(60x)} \, dx = (1 - e^(60(-\infty)))/(12)$
Rewrite: $\displaystyle \int\limits^0_(- \infty) {5e^(60x)} \, dx = (1)/(12) - (1)/(12e^(60(\infty)))$
Simplify: $\displaystyle \int\limits^0_(- \infty) {5e^(60x)} \, dx = (1)/(12)$

∴ the improper integral equals $\displaystyle \bold{(1)/(12)}$ and is convergent.

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Learn more about improper integrals: brainly.com/question/14413972

Learn more about calculus: brainly.com/question/23558817

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Topic: AP Calculus BC (Calculus I + II)

Unit: Integration

Answer:

$\int_(-\infty)^0 5 e^(60x) dx = (1)/(12)[e^0 -0]= (1)/(12)$

Step-by-step explanation:

Assuming this integral:

$\int_(-\infty)^0 5 e^(60x) dx$

We can do this as the first step:

$5 \int_(-\infty)^0 e^(60x) dx$

Now we can solve the integral and we got:

$5 (e^(60x))/(60) \Big|_(-\infty)^0$

$\int_(-\infty)^0 5 e^(60x) dx = (e^(60x))/(12)\Big|_(-\infty)^0 = (1)/(12) [e^(60*0) -e^(-\infty)]$

$\int_(-\infty)^0 5 e^(60x) dx = (1)/(12)[e^0 -0]= (1)/(12)$

So then we see that the integral on this case converges amd the values is 1/12 on this case.

Find the volume V of the solid obtained by rotating the region bounded by the given curves about the specified line.y = 6 sin x, y = 6 cos x, 0 ≤ x ≤ π/4; about y = −1