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aniked [119]
2 years ago
5

Is a measure of 26 inches​ "far away" from a mean of 16 ​inches? As someone with knowledge of​ statistics, you answer​ "it depen

ds" and request the standard deviation of the underlying data. ​(a) Suppose the data come from a sample whose standard deviation is 2 inches. How many standard deviations is 26 inches from 16 ​inches? ​(b) Is 26 inches far away from a mean of 16 ​inches? ​(c) Suppose the standard deviation of the underlying data is 7 inches. Is 26 inches far away from a mean of 16 ​inches?
Mathematics
1 answer:
alexandr402 [8]2 years ago
3 0

Answer: Hello!

In a normal distribution, between the mean and the mean plus the standar deviation, there is a 34.1% of the data set, between the mean plus the standar deviation, and the mean between two times the standard deviation, there is a 16.2% of the data set, and so on.

If our mean is 16 inches, and the measure is 26 inches, then the difference is 10 inches between them.

a) if the standar deviation is 2 inches, then you are 10/2 = 5 standar deviations from the mean.

b) yes, is really far away from the mean, in a normal distribution a displacement of 5 standar deviations has a very small probability.

c) Now the standar deviation is 7, so now 26 is in the range between 1 standar deviation and 2 standar deviations away from the mean.

Then this you have a 16% of the data, then in this case, 26 inches is not far away from the mean.

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A cylinder shaped can needs to be constructed to hold 500 cubic centimeters of soup. The material for the sides of the can costs
iogann1982 [59]

Answer:

r=3.628cm

h=12.093cm

Step-by-step explanation:

For this problem we are going to use principles, concepts and calculations from multivariable calculus; mainly we are going to use the Lagrange multipliers method. This method is thought to help us to find a extreme value of a multivariable function 'F' given a restriction 'G'. F represents the function that we want to optimize and G is just a relation between the variables of which F depends. The Lagrange method for just one restriction is:

\nabla F=\lambda \nabla G

First, let's build the function that we want to optimize, that is the cost. The cost is a function that must sum the cost of the sides material and the cost of the top and bottom material. The cost of the sides material is the unitary cost (0.03) multiplied by the sides area, which is A_s=2\pi rh for a cylinder; while the cost of the top and bottom material is the unitary cost (0.05) multiplied by the area of this faces, which is A_{TyB}=2\pi r^2 for a cylinder.

So, the cost function 'C' is:

C=2\pi rh*0.03+2\pi r^2*0.05\\C=0.06\pi rh+0.1\pi r^2

The restriction is the volume, which has to be of 500 cubic centimeters:

V=500=\pi r^2h\\500=\pi hr^2

So, let's apply the Lagrange multiplier method:

\nabla C=\lambda \nabla V\\\frac{\partial C}{\partial r}=0.06\pi h+0.2\pi r\\\frac{\partial C}{\partial h}=0.06\pi r\\\frac{\partial V}{\partial r}=2\pi rh\\\frac{\partial V}{\partial h}=\pi r^2\\(0.06\pi h+0.2\pi r,0.06\pi r)=\lambda (2\pi rh,\pi r^2)

At this point we have a three variable (h,r, λ)-three equation system, which solution will be the optimum point for the cost (the minimum). Let's write the system:

0.06\pi h+0.2\pi r=2\lambda \pi rh\\0.06\pi r=\lambda \pi r^2\\500=\pi hr^2

(In this kind of problems always the additional equation is the restricion, in this case, V=500).

Let's divide the first and second equations by π:

0.06h+0.2r=2\lambda rh\\0.06r=\lambda r^2\\500=\pi hr^2

Isolate λ from the second equation:

\lambda =\frac{0.06}{r}

Isolate h from the third equation:

h=\frac{500}{\pi r^2}

And then, replace λ and h in the first equation:

0.06*\frac{500}{\pi r^2} +0.2r=2*(\frac{0.06}{r})r\frac{500}{\pi r^2} \\\frac{30}{\pi r^2}+0.2r= \frac{60}{\pi r^2}

Multiply all the resultant equation by \pi r^{2}:

30+0.2\pi r^3=60\\0.2\pi r^3=30\\r^3=\frac{30}{0.2\pi } =\frac{150}{\pi}\\r=\sqrt[3]{\frac{150}{\pi}}\approx 3.628cm

Then, find h by the equation h=\frac{500}{\pi r^2} founded above:

h=\frac{500}{\pi r^2}\\h=\frac{500}{\pi (3.628)^2}=12.093cm

4 0
3 years ago
In a certain community, 36 percent of the families own a dog and 22 percent of the families that own a dog also own a cat. In ad
zysi [14]

Answer:  a) 0.0792   b) 0.264

Step-by-step explanation:

Let Event D = Families own a dog .

Event C = families own a cat .

Given : Probability that families own a dog : P(D)=0.36

Probability that families own a dog also own a cat : P(C|D)=0.22

Probability that families own a cat : P(C)= 0.30

a) Formula to find conditional probability :

P(B|A)=\dfrac{P(A\cap B)}{P(A)}\\\\\Rightarrow P(A\cap B)=P(B|A)\times P(A)   (1)

Similarly ,

P(C\cap D)=P(C|D)\times P(D)\\\\=0.22\times0.36=0.0792

Hence, the probability that a randomly selected family owns both a dog and a cat : 0.0792

b) Again, using (2)

P(D|C)=\dfrac{P(C\cap D)}{P(C)}\\\\=\dfrac{0.0792}{0.30}=0.264

Hence, the conditional probability that a randomly selected family owns a dog given that it owns a cat = 0.264

5 0
2 years ago
What is five percent of 150
3241004551 [841]
7.5 .. multiply .05 times 150
6 0
2 years ago
Read 2 more answers
Expand the following.... please help
Evgesh-ka [11]
= (3/8) (16x + - 24)
=(3/8) (16x) +(3/8)(-24)
=6x - 9
4 0
3 years ago
ASAP help I don’t have time! It detects if it’s right or not:(((
Leno4ka [110]

Answer:

I'm pretty sure it's B! :)

7 0
2 years ago
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