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Elis [28]
2 years ago
10

According to Consumer Digest (July/August 1996), the probable location of personal computers (PC) in the home is as follows: Adu

lt bedroom: 0.03 Child bedroom: 0.15 Other bedroom: 0.14 Office or den: 0.40 Other rooms: 0.28 (a) What is the probability that a PC is in a bedroom? (b) What is the probability that it is not in a bedroom? (c) Suppose a household is selected at random from households with a PC; in what room would you expect to find a PC?
Mathematics
1 answer:
Andreas93 [3]2 years ago
6 0

Answer:

a) 0.32

b) 0.68

c) Office or den

Step-by-step explanation:

a) The probability that a PC is in a bedroom is the sum of the probabilities of a PC being is an adult bedroom, child bedroom or other bedroom:

P(B) =P(adult)+P(child)+P(other)\\P(B) = 0.03+0.15+0.14\\P(B) =0.32

b) The probability that a PC is not in a bedroom is 100% minus the probability of it being in a bedroom:

P(Not\ B) = 1- P(B)\\P(Not\ B) =1-0.32\\P(Not\ B) =0.68

c) The expected room to find a PC from a randomly selected household is the room with highest likelihood of having a PC according to Consumer Digest. The Office or den, is the most probable room with a 0.40 chance.

You would expect to find a PC in the Office or den.

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Use lagrange multipliers to find the points on the given surface that are closest to the origin. y2 = 64 + xz
Inessa [10]
The distance between an arbitrary point on the surface and the origin is

d(x,y,z)=\sqrt{x^2+y^2+z^2}

Recall that for differentiable functions g(x) and h(x), the composition g(h(x)) attains extrema at the same points that h(x) does, so we can consider an augmented distance function

D(x,y,z)=x^2+y^2+z^2

The Lagrangian would then be

L(x,y,z,\lambda)=x^2+y^2+z^2+\lambda(y^2-64-xz)

We have partial derivatives

\begin{cases}L_x=2x-\lambda z\\L_y=2y+2y\lambda\\L_z=2z-\lambda x\\L_\lambda=y^2-64-xz\end{cases}

Set each partial derivative to 0 and solve the system to find the critical points.

From the second equation it follows that either y=0 or \lambda=-1. In the first case we arrive at a contradiction (I'll leave establishing that to you). If \lambda=-1, then we have

\begin{cases}2x+z=0\\2z+x=0\end{cases}\implies x=0,z=0

This means y^2=64\implies y=\pm8

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2 years ago
Find a single expression that represents the area of the outer ring of the circle if the area of the whole circle is represented
sp2606 [1]

Answer:  The answer is A_o=2\pi\sqrt{\dfrac{7}{22}(25x^2-12x-9)}.


Step-by-step explanation: Given that the area of the whole circle is represented by the expression

A_c=25x^2-12x-9.

We are to find the area of the outer ring of the circle, i.e., to find the circumference of the circle.

Now, if 'r' represents the radius of the circle, then we have

A_c=\pi r^2\\\\\Rightarrow \dfrac{22}{7}r^2=25x^2-12x-9\\\\ \Rightarrow r^2=\dfrac{7}{22}(25x^2-12x-9)\\\\\Rightarrow r=\sqrt{\dfrac{7}{22}(25x^2-12x-9)}.

Thus, the area of the outer ring is

A_o=2\pi r=2\pi\sqrt{\dfrac{7}{22}(25x^2-12x-9)}.


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Step-by-step explanation:

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A random sample from a normal population is obtained, and the data are given below. Find a 90% confidence interval for . 114 157
melamori03 [73]

Answer:

103.160 \leq \sigma \leq 187.476

And the upper bound rounded to the nearest integer would be 187.

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population mean or variance lies between an upper and lower interval".  

The margin of error is the range of values below and above the sample statistic in a confidence interval.

The Chi Square distribution is the distribution of the sum of squared standard normal deviates .

\bar X represent the sample mean for the sample  

\mu population mean (variable of interest)

s represent the sample standard deviation

n represent the sample size  

Solution to the problem

Data given: 114 157 203 257 284 299 305 344 378 410 421 450 478 480 512 533 545

The confidence interval for the population variance \sigma^2 is given by the following formula:

\frac{(n-1)s^2}{\chi^2_{\alpha/2}} \leq \sigma \leq \frac{(n-1)s^2}{\chi^2_{1-\alpha/2}}

On this case we need to find the sample standard deviation with the following formula:

s=sqrt{\frac{\sum_{i=1}^{17} (x_i -\bar x)^2}{n-1}}

And in order to find the sample mean we just need to use this formula:

\bar x =\frac{\sum_{i=1}^n x_i}{n}

The sample mean obtained on this case is \bar x= 362.941 and the deviation s=132.250

The next step would be calculate the critical values. First we need to calculate the degrees of freedom given by:

df=n-1=17-1=16

Since the Confidence is 0.90 or 90%, the value of \alpha=0.1 and \alpha/2 =0.05, and we can use excel, a calculator or a tabel to find the critical values.  

The excel commands would be: "=CHISQ.INV(0.05,16)" "=CHISQ.INV(0.95,16)". so for this case the critical values are:

\chi^2_{\alpha/2}=26.296

\chi^2_{1- \alpha/2}=7.962

And replacing into the formula for the interval we got:

\frac{(16)(132.250)^2}{26.296} \leq \sigma \leq \frac{(16)(132.250)^2}{7.962}

10641.959 \leq \sigma^2 \leq 35147.074

Now we just take square root on both sides of the interval and we got:

103.160 \leq \sigma \leq 187.476

And the upper bound rounded to the nearest integer would be 187.

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Answer:

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