What is the measure in degrees of angle R?

How do you do this I am so confused please help me

jok3333 [9.3K]

Rsc . rrr . rrc rrs . rsr rcr crs src scr csr ccc sss
these are all the possible outcomes
12 total outcomes

5 0

3 years ago

Find the volume of the cylinder with a diameter of 12 inches and a height of 10

kati45 [8]

Answer:

360 pi in ^3

Step-by-step explanation:

The volume of a cylinder is given by

V = pi r^2 h

We know the diameter is 12 so the radius is 1/2 the diameter

r = d/2 = 12/2 = 6

V = pi (6)^2 * 10

V = pi (36)*10

V = 360 pi in ^3

We can approximate pi by 3.14

V =1130.4 in ^3

Or we can approximate pi by using the pi button

V =1130.973355 in ^3

6 0

2 years ago

Given: PQ ⊥ QR , PR=20, SR=11, QS=5 Find: The value of PS.

dlinn [17]

Answer:

The value of the side PS is 26 approx.

Step-by-step explanation:

In this question we have two right triangles. Triangle PQR and Triangle PQS.

Where S is some point on the line segment QR.

Given:

PR = 20

SR = 11

QS = 5

We know that QR = QS + SR

QR = 11 + 5

QR = 16

Now triangle PQR has one unknown side PQ which in its base.

Finding PQ:

Using Pythagoras theorem for the right angled triangle PQR.

PR² = PQ² + QR²

PQ = √(PR² - QR²)

PQ = √(20²+16²)

PQ = √656

PQ = 4√41

Now for right angled triangle PQS, PS is unknown which is actually the hypotenuse of the right angled triangle.

Finding PS:

Using Pythagoras theorem, we have:

PS² = PQ² + QS²

PS² = 656 + 25

PS² = 681

PS = 26.09

PS = 26

7 0

3 years ago

Persons having Raynaud's syndrome are apt to suffer a sudden impairment of blood circulation in fingers and toes. In an experime

Citrus2011 [14]

Answer:

There is enough evidence to say that the true average heat output of persons with the syndrmoe differs from the true average heat output of non-sufferers.

Step-by-step explanation:

We have to perform a hypothesis test on the difference between means.

The null and alternative hypothesis are:

$H_0: \mu_1-\mu_2=0\\\\H_a: \mu_1-\mu_2$

μ1: mean heat output for subjects with the syndrome.

μ2: mean heat output for non-sufferers.

We will use a significance level of 0.05.

The difference between sample means is:

$M_d=\bar x_1-\bar x_2=0.63-2.09=-1.46$

The standard error is

$s_{M_d}=\sqrt{\sigma_1^2/n_1+\sigma_2^2/n_2}=\sqrt{0.3^2/9+0.5^2/9}=\sqrt{ 0.038 } \\\\ s_{M_d}=0.194$

The t-statistic is

$t=\dfrac{M_d}{s_{M_d}}=\dfrac{-1.46}{0.194}=-7.52$

The degrees of freedom are

$df=n_1+n_2-2=9+9-2=16$

The critical value for a left tailed test at a significance level of 0.05 and 16 degrees of freedom is t=-1.746.

The t-statistic is below the critical value, so it lies in the rejection region.

The null hypothesis is rejected.

There is enough evidence to say that the true average heat output of persons with the syndrmoe differs from the true average heat output of non-sufferers.

5 0

2 years ago

Mr. Nichols put $1,200 in a savings account at a local bank. The simple annual interest rate was 5%. How much interest did his a

satela [25.4K]

$\bf ~~~~~~ \textit{Simple Interest Earned} \\\\ I = Prt\qquad \begin{cases} I=\textit{interest earned}\\ P=\textit{original amount deposited}\to& \$1200\\ r=rate\to 5\%\to \frac{5}{100}\to &0.05\\ t=years\to &3 \end{cases} \\\\\\ I=(1200)(0.05)(3)\implies I=180$

8 0

3 years ago