All categories

2 years ago

Use a z-Table to determine the percent of data specified.

Mathematics

2 answers:

Soloha48 [4]2 years ago

5 0

I think between b and d

kobusy [5.1K]2 years ago

4 0

I think c I might be wrong tho :)

You might be interested in

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

2 years ago

Read 2 more answers

What is the solution to the equation x - 5 = 15?

Nonamiya [84]

Answer:

x=20

Step-by-step explanation:

x - 5 = 15

x +5 = +5

x = 20

4 0

1 year ago

Read 2 more answers

Mai says that the equation 2x + 2 = x + 1 has no solution because the left-hand side is double the right hand side.

zhannawk [14.2K]

Answer:

h im billy

Step-by-step explanation:

7 0

1 year ago

Read 2 more answers

1 1/4 cups of flour divided by 1/8 cups of butter

Dimas [21]

5/32 is the answer
Hope this helped!

8 0

2 years ago

Read 2 more answers

if content settles for 1 1/2 hours before the liquid is remove .You fill the tank at 1:45 a.m when do you remove the liquid

artcher [175]

U fill the tank at 1:45 a.m......then u wait 1 1/2 hrs....

1:45 + 1 hr = 2:45.....2:45 + 30 minutes (because 1/2 hr = 30 minutes) =
3:15.....so u remove the liquid at 3:15 a.m.

3 0

2 years ago