Ask question

Ask question

All categories

3 years ago

10

The time required for a citizen to complete the 2010 U.S. Census ""long"" form is normally distributed with a mean of 40 minutes

and a standard deviation of 10 minutes. What is the third quartile (in minutes) for the time required to complete the form?

1 answer:

Nimfa-mama [501]3 years ago

3 0

Answer:

46.75 minutes

Step-by-step explanation:

Mean time (μ) = 40 minutes

Standard deviation (σ) = 10 minutes

The third quartile is at the 75th percentile of a normal distribution. According to a z-score table, the corresponding z-score for the 75th percentile is z=0.675.

The z-score for any given time, 'X', is:

$z=\frac{X- \mu}{\sigma}$

The third quartile for the time required to complete the form is:

$0.675=\frac{X- 40}{10}\\X= 40+6.75=46.75\ minutes$

You might be interested in

Determine the slope of the line through the given points. If the slope is undefined, so state. (3,2) and (6,11)

agasfer [191]

Answer:

The slope is 3

Step-by-step explanation:

We can find the slope by using

m = (y2-y1)/(x2-x1)

= (11 -2)/(6-3)

= 9/3

= 3

6 0

3 years ago

Read 2 more answers

WILL GIVE BRAINLIEST!!!

Mnenie [13.5K]

The answer to your question is 34.13%

8 0

3 years ago

I'm having trouble with #2. I've got it down to the part where it would be the integral of 5cos^3(pheta)/sin(pheta). I'm not sur

Butoxors [25]

$\displaystyle\int\frac{\sqrt{25-x^2}}x\,\mathrm dx$

Setting $x=5\sin\theta$ , you have $\mathrm dx=5\cos\theta\,\mathrm d\theta$ . Then the integral becomes

$\displaystyle\int\frac{\sqrt{25-(5\sin\theta)^2}}{5\sin\theta}5\cos\theta\,\mathrm d\theta$
$\displaystyle\int\sqrt{25-25\sin^2\theta}\dfrac{\cos\theta}{\sin\theta}\,\mathrm d\theta$
$\displaystyle5\int\sqrt{1-\sin^2\theta}\dfrac{\cos\theta}{\sin\theta}\,\mathrm d\theta$
$\displaystyle5\int\sqrt{\cos^2\theta}\dfrac{\cos\theta}{\sin\theta}\,\mathrm d\theta$

Now, $\sqrt{x^2}=|x|$ in general. But since we want our substitution $x=5\sin\theta$ to be invertible, we are tacitly assuming that we're working over a restricted domain. In particular, this means $\theta=\sin^{-1}\dfrac x5$ , which implies that $\left|\dfrac x5\right|\le1$ , or equivalently that $|\theta|\le\dfrac\pi2$ . Over this domain, $\cos\theta\ge0$ , so $\sqrt{\cos^2\theta}=|\cos\theta|=\cos\theta$ .

Long story short, this allows us to go from

$\displaystyle5\int\sqrt{\cos^2\theta}\dfrac{\cos\theta}{\sin\theta}\,\mathrm d\theta$

to

$\displaystyle5\int\cos\theta\dfrac{\cos\theta}{\sin\theta}\,\mathrm d\theta$
$\displaystyle5\int\dfrac{\cos^2\theta}{\sin\theta}\,\mathrm d\theta$

Computing the remaining integral isn't difficult. Expand the numerator with the Pythagorean identity to get

$\dfrac{\cos^2\theta}{\sin\theta}=\dfrac{1-\sin^2\theta}{\sin\theta}=\csc\theta-\sin\theta$

Then integrate term-by-term to get

$\displaystyle5\left(\int\csc\theta\,\mathrm d\theta-\int\sin\theta\,\mathrm d\theta\right)$
$=-5\ln|\csc\theta+\cot\theta|+\cos\theta+C$

Now undo the substitution to get the antiderivative back in terms of $x$ .

$=-5\ln\left|\csc\left(\sin^{-1}\dfrac x5\right)+\cot\left(\sin^{-1}\dfrac x5\right)\right|+\cos\left(\sin^{-1}\dfrac x5\right)+C$

and using basic trigonometric properties (e.g. Pythagorean theorem) this reduces to

$=-5\ln\left|\dfrac{5+\sqrt{25-x^2}}x\right|+\sqrt{25-x^2}+C$

4 0

3 years ago

Read 2 more answers

shutvik [7]

Answer:

6.5 x 10^6 To answer this question, you need to divide the mass of the sun by the mass of mercury. So 2.13525 x 10^30 / 3.285 x 10^23 = ? To do the division, divide the mantissas in the normal fashion 2.13525 / 3.285 = 0.65 And subtract the exponents. 30 - 23 = 7 So you get 0.65 x 10^7 Unless the mantissa is zero, the mantissa must be greater than or equal to 0 and less than 10. So multiply the mantissa by 10 and then subtract 1 from the exponent, giving 6.5 x 10^6 So the sun is 6.5 x 10^6 times as massive as mercury.

:}

8 0

3 years ago

Read 2 more answers

8x-3=3x+17<br> Solve for x

Olin [163]

Answer:

x = 4

Step-by-step explanation:

3 0

3 years ago

Read 2 more answers