Recent questions in Post Secondary

Samples
Answered

Nathanial Levine
2022-10-03

Critical Point
Answered

Lisantiom
2022-10-03

$f(x,y)={x}^{2}y{e}^{-x-y}.$

$x=0$ and $x=2$ both satisfy ${f}_{x}=0,{f}_{y}=0$, and when $x=2,y=1$. But when $x=0,y$,$x=0,y$ is arbritary. So how can I determine whether the critical point at $0$ is max/min/saddle?

de Broglie Equation
Answered

spatularificw2
2022-10-02

A detailed explanation behind larger atomic size with larger Planck’s constant, h.

Intervals of Increase and Decrease
Answered

Jensen Mclean
2022-10-02

I have the function $f(x)=6+\frac{6}{x}+\frac{6}{{x}^{2}}$ and I want to find the intervals where it increases or decreases. The problem is that when I find ${f}^{\prime}(x)=0$, which becomes $x=-2$. Once I put -2 on a number line, and find whether the numbers higher and lower than -2 produce positive or negative numbers when plugged into f′(x), I get both positives and negatives for different numbers when $x>-2$. For example, ${f}^{\prime}(-1)>0$ and ${f}^{\prime}(5)<0$.

I am not sure if this is because the function has a vertical asymptote at $x=0$ and a horizontal attribute at $y=6$. If so, I'd like to know what I need to do to handle them.

How I found the first derivative:

$\frac{d}{dx}\frac{6}{x}=\frac{[0]-[6\cdot 1]}{{x}^{2}}=\frac{-6}{{x}^{2}}$

$\frac{d}{dx}\frac{6}{{x}^{2}}=\frac{[0]-[6\cdot 2x]}{({x}^{2}{)}^{2}}=\frac{-12}{{x}^{3}}$

${f}^{\prime}(x)=\frac{-6}{{x}^{2}}-\frac{-12}{{x}^{3}}$

How I found the asymptotes:

When you combine the fractions of f(x), $f(x)=\frac{6{x}^{2}+6x+6}{{x}^{2}}$. When set equal to zero, ${x}^{2}$ has an x-value of zero. Therefore, f(x) has a vertical asymptote at $x=0$.

Once the fractions are combined, and when everything but the coefficients of the leading terms on the numerator and denominator are removed, $f(x)=\frac{6}{1}$. So f(x) has a horizontal asymptote at $y=6$.

How I found the value of ${f}^{\prime}(x)=0$:

$\frac{-6}{{x}^{2}}-\frac{12}{{x}^{3}}=0$

Add $\frac{12}{{x}^{3}}$ to both sides

$\frac{6}{{x}^{2}}=\frac{-12}{{x}^{3}}$

Multiply both sides by ${x}^{3}$.

$6x=-12$

Divide both sides by 6.

$x=\frac{-12}{6}=-2$

Samples
Answered

Aidyn Crosby
2022-10-02

de Broglie Equation
Answered

Aryan Lowery
2022-10-02

Frame of reference
Answered

Janessa Benson
2022-10-02

Euler's Method
Answered

mriteyl
2022-10-02

Integrals
Answered

Austin Rangel
2022-10-02

Confidence intervals
Answered

odcinaknr
2022-10-02

In one of my assignments I have to "test" if the confidence intervals for a set of parameters in a mixed effect model is accurate. I'm asked to simulate from fittet parameters and there after refit them using the same model many times, and lastly take 2.5% and 97.5% quantiles of them and compare with the original CIs. My question is, how does this procedure in anyway measure how accurate my original confidence intervals are?

Intervals of Increase and Decrease
Answered

solvarmedw
2022-10-02

$f(x)={x}^{3}-4{x}^{2}+2$, which of the following statements are true:

(1) Increasing in $(-\mathrm{\infty},0)$, decreasing in $(\frac{8}{3},+\mathrm{\infty})$.

(2) Increasing in both $(-\mathrm{\infty},0)$, decreasing in $(\frac{8}{3},+\mathrm{\infty})$.

(3) decreasing in both $(-\mathrm{\infty},0)$, and $(\frac{8}{3},+\mathrm{\infty})$.

(4) Decreasing in $(-\mathrm{\infty},0)$, Increasing in $(\frac{8}{3},+\mathrm{\infty})$.

(5) None of the above.

${f}^{\prime}(x)=0=3{x}^{2}-8x=0\Rightarrow x=\frac{8}{3},x=0$ are the singular point/point of inflection.Could anyone tell me what next?

Raw Data
Answered

garnirativ8
2022-10-02

Confidence intervals
Answered

samuelaplc
2022-10-02

If I got, let's say, a 95 % confidence interval for the mean and a 95 % confidence interval for the variance.

Would it then be wrong to conclude:

The 95 % confidence interval for the mean contains with at least 95 % probability the true mean?

and

The 95 % confidence interval for the variance contains with at least 95 % probability the true variance?

What would be a more correct/precise way to express what the confidence intervals stand for?

Random variables
Answered

fofopausiomiava
2022-10-02

Random variables
Answered

Haiden Meyer
2022-10-02

Laplace transform
Answered

miniliv4
2022-10-02

$$1+\frac{1}{9}+\frac{1}{25}+\frac{1}{49}+\cdots $$

Random variables
Answered

Bridger Holden
2022-10-02

Survey Questions
Answered

aurelegena
2022-10-02

In a survey on Popsicle flavor preferences of kids aged 3-5, it was found that:

- 22 like strawberry.

- 25 like blueberry.

- 39 like grape.

- 9 like blueberry and strawberry.

- 17 like strawberry and grape.

- 20 like blueberry and grape.

- 6 like all flavors.

- 4 like none.

Apparently the answer is 50, but I can't seem to figure out how to arrive at this

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