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3 years ago

Given John's data collection method and findings, what conclusion can be made?

Mathematics

2 answers:

Vikentia [17]3 years ago

6 0

Answer:

Step-by-step explanation:

edg2020

Snezhnost [94]3 years ago

4 0

OB Neither of the claims are correct

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Suppose that Y has density function

zvonat [6]

I'm assuming

$f(y)=\begin{cases}ky(1-y)&\text{for }0\le y\le1\\0&\text{otherwise}\end{cases}$

(a) f(x) is a valid probability density function if its integral over the support is 1:

$\displaystyle\int_{-\infty}^\infty f(x)\,\mathrm dx=k\int_0^1 y(1-y)\,\mathrm dy=k\int_0^1(y-y^2)\,\mathrm dy=1$

Compute the integral:

$\displaystyle\int_0^1(y-y^2)\,\mathrm dy=\left(\frac{y^2}2-\frac{y^3}3\right)\bigg|_0^1=\frac12-\frac13=\frac16$

So we have

k / 6 = 1 → k = 6

(b) By definition of conditional probability,

P(Y ≤ 0.4 | Y ≤ 0.8) = P(Y ≤ 0.4 and Y ≤ 0.8) / P(Y ≤ 0.8)

P(Y ≤ 0.4 | Y ≤ 0.8) = P(Y ≤ 0.4) / P(Y ≤ 0.8)

It makes sense to derive the cumulative distribution function (CDF) for the rest of the problem, since F(y) = P(Y ≤ y).

We have

$\displaystyle F(y)=\int_{-\infty}^y f(t)\,\mathrm dt=\int_0^y6t(1-t)\,\mathrm dt=\begin{cases}0&\text{for }y$

Then

P(Y ≤ 0.4) = F (0.4) = 0.352

P(Y ≤ 0.8) = F (0.8) = 0.896

and so

P(Y ≤ 0.4 | Y ≤ 0.8) = 0.352 / 0.896 ≈ 0.393

(c) The 0.95 quantile is the value φ such that

P(Y ≤ φ) = 0.95

In terms of the integral definition of the CDF, we have solve for φ such that

$\displaystyle\int_{-\infty}^\varphi f(y)\,\mathrm dy=0.95$

We have

$\displaystyle\int_{-\infty}^\varphi f(y)\,\mathrm dy=\int_0^\varphi 6y(1-y)\,\mathrm dy=(3y^2-2y^3)\bigg|_0^\varphi = 0.95$

which reduces to the cubic

3φ² - 2φ³ = 0.95

Use a calculator to solve this and find that φ ≈ 0.865.

8 0

3 years ago

The radius of a right circular cylinder is increasing at the rate of 7 in./sec, while the height is decreasing at the rate of 6

Arlecino [84]

Answer:

$\approx \bold{6544\ in^3/sec}$

Step-by-step explanation:

Given:

Rate of change of radius of cylinder:

$\dfrac{dr}{dt} = +7\ in/sec$

(This is increasing rate so positive)

Rate of change of height of cylinder:

$\dfrac{dh}{dt} = -6\ in/sec$

(This is decreasing rate so negative)

To find:

Rate of change of volume when r = 20 inches and h = 16 inches.

Solution:

First of all, let us have a look at the formula for Volume:

$V = \pi r^2h$

Differentiating it w.r.to 't':

$\dfrac{dV}{dt} = \dfrac{d}{dt}(\pi r^2h)$

Let us have a look at the formula:

$1.\ \dfrac{d}{dx} (C.f(x)) = C\dfrac{d(f(x))}{dx} \ \ \ (\text{C is a constant})\\2.\ \dfrac{d}{dx} (f(x).g(x)) = f(x)\dfrac{d}{dx} (g(x))+g(x)\dfrac{d}{dx} (f(x))$

$3.\ \dfrac{dx^n}{dx} = nx^{n-1}$

Applying the two formula for the above differentiation:

$\Rightarrow \dfrac{dV}{dt} = \pi\dfrac{d}{dt}( r^2h)\\\Rightarrow \dfrac{dV}{dt} = \pi h\dfrac{d }{dt}( r^2)+\pi r^2\dfrac{dh }{dt}\\\Rightarrow \dfrac{dV}{dt} = \pi h\times 2r \dfrac{dr }{dt}+\pi r^2\dfrac{dh }{dt}$

Now, putting the values:

$\Rightarrow \dfrac{dV}{dt} = \pi \times 16\times 2\times 20 \times 7+\pi\times 20^2\times (-6)\\\Rightarrow \dfrac{dV}{dt} = 22 \times 16\times 2\times 20 +3.14\times 400\times (-6)\\\Rightarrow \dfrac{dV}{dt} = 14080 -7536\\\Rightarrow \dfrac{dV}{dt} \approx \bold{6544\ in^3/sec}$