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3 years ago

10

Sarah and get 2 friends split a bag of candy evenly. They each ate 13 pieces of candy and there were 2 pieces left over. How man

y pieces of candy were originally in the bag

1 answer:

PSYCHO15rus [73]3 years ago

4 0

I cant tell if the problem wasn’t writing correctly when u said “sarah and get 2 friends?”
i think you might of meant sarah and ___ get two friends?? please correct me but if it was meant to have another name after sarah then this would be the answer:
There was 54 original pieces

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What is the are of a hexagon with a radius of 5 inches?

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The area of a hexagon with a radius of 5 inches is 86.603.

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3 years ago

Name the sampling method used in each of the following situations:

elixir [45]

Answer:

a) Convenient

b) Cluster

c) Stratified

d) Systematic

e) Random

Step-by-step explanation:

Samples may be classified as:

Convenient: Sample drawn from a conveniently available pool.

Random: Basically, put all the options into a hat and drawn some of them.

Systematic: Every kth element is taken. For example, you want to survey something on the street, you interview every 5th person, for example.

Cluster: Divides population into groups, called clusters, and each element in the cluster is surveyed.

Stratified: Also divides the population into groups. However, then only some elements of the group are surveyed.

Situation a:

Asks all from a single group(sitting passengers), that is, convenint sample.

Situation b:

First she separates the students by row, then from each row, she chooses all students. So this is Cluster sampling.

Situation c:

The manager separates the stores by location, and then in each location, random customers are chosen. This is an example of Stratified sampling.

Situation d:

Marks for every fourth patron, that is, every 4th element. So Systematic sampling.

Situation e:

People chosen at random, so Random sampling.

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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

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

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