Ask question

Ask question

All categories

3 years ago

12

Suppose a sample of 256 was taken from a population with a standard deviation of 64 centimeters. What is the margin of error for

a 68% confidence interval?
A. 6
B. 8
C. 10
D. 4

2 answers:

Ganezh [65]3 years ago

8 0

I think the answer is b

Serhud [2]3 years ago

6 0

I feel like the answer is d

You might be interested in

Please help meeeee math

Sergeeva-Olga [200]

Q2. By the chain rule,

$\dfrac{dy}{dx} = \dfrac{dy}{dt} \cdot \dfrac{dt}{dx} = \dfrac{\frac{dy}{dt}}{\frac{dx}{dt}}$

We have

$x=2t \implies \dfrac{dx}{dt}=2$

$y=t^4+1 \implies \dfrac{dy}{dt}=4t^3$

The slope of the tangent line to the curve at $t=1$ is then

$\dfrac{dy}{dx} \bigg|_{t=1} = \dfrac{4t^3}{2} \bigg|_{t=1} = 2t^3\bigg|_{t=1} = 2$

so the slope of the normal line is $-\frac12$ . When $t=1$ , we have

$x\bigg|_{t=1} = 2t\bigg|_{t=1} = 2$

$y\bigg|_{t=1} = (t^4+1)\bigg|_{t=1} = 2$

so the curve passes through (2, 2). Using the point-slope formula for a line, the equation of the normal line is

$y - 2 = -\dfrac12 (x - 2) \implies y = -\dfrac12 x + 3$

Q3. Differentiating with the product, power, and chain rules, we have

$y = x(x+1)^{1/2} \implies \dfrac{dy}{dx} = \dfrac{3x+2}{2\sqrt{x+1}} \implies \dfrac{dy}{dx}\bigg|_{x=3} = \dfrac{11}4$

The derivative vanishes when

$\dfrac{3x+2}{2\sqrt{x+1}} = 0 \implies 3x+2=0 \implies x = -\dfrac23$

Q4. Differentiating with the product and chain rules, we have

$y = (2x+1)e^{-2x} \implies \dfrac{dy}{dx} = -4xe^{-2x}$

The stationary points occur where the derivative is zero.

$-4xe^{-2x} = 0 \implies x = 0$

at which point we have

$y = (2x+1)e^{-2x} \bigg|_{x=0} = 1$

so the stationary point has coordinates (0, 1). By its "nature", I assume the question is asking what kind of local extremum this point. Compute the second derivative and evaluate it at $x=0$ .

$\dfrac{d^2y}{dx^2}\bigg|_{x=0} = (8x-4)e^{-2x}\bigg|_{x=0} = -4 < 0$

The negative sign tells us this stationary point is a local maximum.

Q5. Differentiating the volume equation implicitly with respect to $t$ , we have

$V = \dfrac{4\pi}3 r^3 \implies \dfrac{dV}{dt} = 4\pi r^2 \dfrac{dr}{dt}$

When $r=5\,\rm cm$ , and given it changes at a rate $\frac{dr}{dt}=-1.5\frac{\rm cm}{\rm s}$ , we have

$\dfrac{dV}{dt} = 4\pi (5\,\mathrm{cm})^2 \left(-1.5\dfrac{\rm cm}{\rm s}\right) = -150\pi \dfrac{\rm cm^3}{\rm s}$

Q6. Given that $V=400\pi\,\rm cm^3$ is fixed, we have

$V = \pi r^2h \implies h = \dfrac{400\pi}{\pi r^2} = \dfrac{400}{r^2}$

Substitute this into the area equation to make it dependent only on $r$ .

$A = \pi r^2 + 2\pi r \left(\dfrac{400}{r^2}\right) = \pi r^2 + \dfrac{800\pi}r$

Find the critical points of $A$ .

$\dfrac{dA}{dr} = 2\pi r - \dfrac{800\pi}{r^2} = 0 \implies r = \dfrac{400}{r^2} \implies r^3 = 400 \implies r = 2\sqrt[3]{50}$

Check the sign of the second derivative at this radius to confirm it's a local minimum (sign should be positive).

$\dfrac{d^2A}{dr^2}\bigg|_{r=2\sqrt[3]{50}} = \left(2\pi + \dfrac{1600\pi}{r^3}\right)\bigg|_{r=2\sqrt[3]{50}} = 6\pi > 0$

Hence the minimum surface area is

$A\bigg_{r=2\sqrt[3]{50}\,\rm cm} = \left(\pi r^2 + \dfrac{800\pi}r\right)\bigg|_{r=2\sqrt[3]{50}\,\rm cm} = 60\pi\sqrt[3]{20}\,\rm cm^2$

Q7. The volume of the box is

$V = 8x^2$

(note that the coefficient 8 is measured in cm) while its surface area is

$A = 2x^2 + 12x$

(there are two $x$ -by- $x$ faces and four 8-by- $x$ faces; again, the coefficient 12 has units of cm).

When $A = 210\,\rm cm^2$ , we have

$210 = 2x^2 + 12x \implies x^2 + 6x - 105 = 0 \implies x = -3 \pm\sqrt{114}$

This has to be a positive length, so we have $x=\sqrt{114}-3\,\rm cm$ .

Given that $\frac{dx}{dt}=0.05\frac{\rm cm}{\rm s}$ , differentiate the volume and surface area equations with respect to $t$ .

$\dfrac{dV}{dt} = (16\,\mathrm{cm})x \dfrac{dx}{dt} = (16\,\mathrm{cm})(\sqrt{114}-3\,\mathrm{cm})\left(0.05\dfrac{\rm cm}{\rm s}\right) = \dfrac{4(\sqrt{114}-3)}5 \dfrac{\rm cm^3}{\rm s}$

$\dfrac{dA}{dt} = 4x\dfrac{dx}{dt} + (12\,\mathrm{cm})\dfrac{dx}{dt} = \left(4(\sqrt{114}-3\,\mathrm {cm}) + 12\,\mathrm{cm}\right)\left(0.05\dfrac{\rm cm}{\rm s}\right) = \dfrac{\sqrt{114}}5 \dfrac{\rm cm^2}{\rm s}$

5 0

1 year ago

17.34 , 17.04 , 17.7<br> from smallest to largest

zmey [24]

Answer:

17.04, 17.34, 17.7

7 0

3 years ago

Hello can you please help me posted picture of question

mr_godi [17]

Probability is calculated as:"Total number of Favorable outcomes divided by Total number of outcomes in the sample space".

The favorable outcomes in this case are the numbers divisible by 3. So the numerator lists the numbers that are divisible by 3, which are the favorable outcomes.

So, the answer to this question is option C

7 0

3 years ago

Two hundred one million,sixteen thousand eleven in number

BaLLatris [955]

201,016,011
How this helped!

8 0

2 years ago

Read 2 more answers

What is the standard form of this number?<br> 5,000,000 + 800,000 + 50,000 + 300

alexandr1967 [171]

Answer:

58503

Step-by-step explanation:

tell me if it's wrong please but I'm sure it's correct

7 0

2 years ago

Read 2 more answers