All categories

4 years ago

In circle P, diameter QS measures 20 centimeters.

Mathematics

2 answers:

Drupady [299]4 years ago

8 0

Answer:

19.9

Step-by-step explanation:

kari74 [83]4 years ago

6 0

check the picture below.

the line QS is just a flat-line and therefore it has 180° on either side, so the angle QPR is 57°, as you see there, and since we know the diameter is 20, that simply means the radius of the circle is 10.

$\bf \textit{arc's length}\\\\ s=\cfrac{\theta \pi r}{180}~~ \begin{cases} r=radius\\ \theta =angle~in\\ \qquad degrees\\ ------\\ r=10\\ \theta =57 \end{cases}\implies s=\cfrac{(57)(\pi )(10)}{180}\implies s=\cfrac{19\pi }{6}$

You might be interested in

The inverse of the function f(x) = { x + 10 is shown.

velikii [3]

Answer:

Step-by-step explanation:

f(x) = x + 10

Replace f(x) with y.

y = x + 10

Now, remember that in an inverse function you flip the x and y, so

x = y + 10

Solve for y

y = x - 10

because it's 2x... we must multiply by 2.

2y = 2x - 20

Original equation:

h(x) = 2x - 0

h(x) is 2y..

4 0

3 years ago

A rancher uses 1400 acres of his property for cattle grazing. If the area of the property is 1600 acres, what percent is used fo

zepelin [54]

Answer: 87.5%

Step-by-step explanation:

From the question, we are informed that a rancher uses 1400 acres of his property for cattle grazing amd that the area of the property is 1600 acres. To get the percentage used for cattle grazing, we divide 1400 by 1600 and then multiply the result by 100%. Mathematically, this will be expressed as:

= 1400/1600 × 100

= 0.875 × 100%

= 87.5%

The percent that is used for cattle grazing is 87.5%.

4 0

3 years ago

Suppose that the probability that any particle emitted by a radioactive material will penetrate a certain shield is 0.01. If 10

AnnyKZ [126]

Answer:

a)P=0.42

b) $n\geq 297$

Step-by-step explanation:

We have a binomial distribution, since the result of each experiment admits only two categories (success and failure) and the value of both possibilities is constant in all experiments. The probability of getting k successes in n trials is given by:

$P=\begin{pmatrix}n\\ k\end{pmatrix} p^k(1-p)^{n-k}=\frac{n!}{k!(n!-k!)}p^k(1-p)^{n-k}$

a) we have k=2, n=10 and p=0.01:

$P=\frac{10!}{2!(10!-2!)}0.01^2(1-0.01)^{10-2}\\P=\frac{10!}{2!*8!}0.01^2(0.99)^{8}\\P=45*0.01^2(0.99)^8=0.42$

b) We have, $1-(1-p)^n=P$ , Here P is the probability that at least one particle will penetrate the shield, this probabity has to be equal or greater than 0.95. Therefore, this will be equal to subtract from the total probability, the probability that the particles do not penetrate raised to the total number of particles.

$1-0.99^n\geq 0.95\\0.99^n\leq 1-0.95\\0.99^n\leq 0.05\\n\geq 297$

4 0

3 years ago

Why is 130 oz bigger than 8 lbs

nordsb [41]

130 oz is bigger than 8 pounds because 1 pound is equal to 16 oz

3 0

4 years ago

Use the given degree of confidence and sample data to construct a confidence interval for the population proportion p. n = 195,

Semenov [28]

Given:
n = 195, sample size.
x = 162, successes in the sample

The proportion is
p = x/n = 162/195 = 0.8308

n* p = 195*0.8308 = 162
n*(1-p) = 195*(1 - 0.8308) = 33
If n*p >= 10, and n*(1-p) >= 10, then the sample proportions will have a normal distribution. This condition is satisfied.

The proportion mean is
μ = 0.8308
The proportion standard deviation is
$\sigma = \sqrt{ \frac{p(1-p)}{n} } = \sqrt{ \frac{0.8308(1-0.8308)}{195} } =0.0269$

σ/√n = 0.0269/√195 = 0.00192

At the 95% confidence level, the interval for the population proportion is
(μ - 1.96(σ/√n), μ + 1.96(σ/√n))
= (0.8308 - 1.96*0.00192, 0.8308 + 1.96*0.00192)
= (0.827, 0.8345)

Answer: The 95% confidence interval is (0.827, 0.835)

5 0

4 years ago