All categories

2 years ago

okay so a circle has half a line of 11 in. , what's the circumference and area?? i'm not good at math and so confused

Mathematics

1 answer:

omeli [17]2 years ago

5 0

Answer:

C = 69.12 in.

A = 380.13 in.²

Step-by-step explanation:

Circumference of a circle is the length around the circle.

Circumference, C = 2πr

where,

Radius (r) = 11 in. (Given)

To find the circumference, substitute r = 11 into the circumference formula

C = 2*π*11

Input this into your calculator, you would get:

C = 69.1150384

C = 69.12 in. (Approximated to nearest hundredth)

Area = πr²

Plug in the value of r

A = π*11²

A = 380.132711

A = 380.13 in.² (to nearest hundredth)

You might be interested in

Please Help!!<br> Write the function I'm vertex form<br> y = -x^2-2x+6<br> y=

balu736 [363]

Rewrite in standard form and use the form to find the vertex.

(1, 5)

(x - 1)^2 + 5

7 0

3 years ago

Let f(x) = 2x – 1 and g(x) = 3x<br> Find g(f(x))

Gemiola [76]

6x-3

You have to put f(x) into g(x) to find the answer, so you rewrite it like this: 3(2x-1)

7 0

3 years ago

Classify each triangle by its sides.

Free_Kalibri [48]

Answer:

Step-by-step explanation:

None of the sides are equal. Isosceles means that 2 sides are equal. Equilateral means that all sides are equal.

3 0

2 years ago

The u.s. senate is composed of 2 senators from each state

Anit [1.1K]

The US Senate is composed of 2 Senators from each of the 50 states to give a total of 100 people in the Senate.

4 0

3 years ago

Suppose the horses in a large stable have a mean weight of 1467lbs, and a standard deviation of 93lbs. What is the probability t

krok68 [10]

Answer:

0.5034 = 50.34% probability that the mean weight of the sample of horses would differ from the population mean by less than 9lbs if 49 horses are sampled at random from the stable

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal probability distribution

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

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

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean $\mu$ and standard deviation $\sigma$ , the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$ .

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

In this problem, we have that:

$\mu = 1467, \sigma = 93, n = 49, s = \frac{93}{\sqrt{49}} = 13.2857$

What is the probability that the mean weight of the sample of horses would differ from the population mean by less than 9lbs if 49 horses are sampled at random from the stable?

This is the pvalue of Z when X = 1467 + 9 = 1476 subtracted by the pvalue of Z when X = 1467 - 9 = 1458.

X = 1476

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

By the Central Limit Theorem

$Z = \frac{X - \mu}{s}$

$Z = \frac{1476 - 1467}{13.2857}$

$Z = 0.68$

$Z = 0.68$ has a pvalue of 0.7517

X = 1458

$Z = \frac{X - \mu}{s}$

$Z = \frac{1458 - 1467}{13.2857}$

$Z = -0.68$

$Z = -0.68$ has a pvalue of 0.2483

0.7517 - 0.2483 = 0.5034

0.5034 = 50.34% probability that the mean weight of the sample of horses would differ from the population mean by less than 9lbs if 49 horses are sampled at random from the stable

5 0

2 years ago