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Probability of Type I error = 1 - Confidence Level. For example, assigning a confidence level of 95% means you're only giving yourself 1 - 0.95 =0.05, or 5% chance of making a type I error, that is, a 5% window of making the mistake of rejecting the null hypothesis (status quo) when it is actually true.

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We can use Slovin's formula , but what is the population N of MBA students?

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Hmm. Get back to you soon. Got a class to teach.
Back now.
14.
a.) . This corresponds to the open interval (0.39404, 0.45962).
b.) No, it is not reasonable to conclude from the sample that the president is doing a good job. In fact we could say that the true proportion may be less than 50%. (As opposed to the null hypothesis that the true proportion is equal to or greater than 50%.)
15.
a.) $63/hr.
b.) Using the t-distribution with 22 degrees of freedom, the C.I. is . This corresponds to the open interval (59.81, 66.19).
c.)We use the standard normal table again. We want . This should give , or . We can put n = 101.

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(3x-4)(x+1)

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(2x + 1)(x + 4)

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Adding corresponding sides, we get 3y = 3, or y = 1.
Substituting into the 1st equation, x - 2 = -1, or x = 1.

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a. The line is a vertical line (same x-coordinates). The equation is x = 3.
b. ,
,
,
.

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y > 9/10, after dividing both sides by 10. (Direction of inequality doesn't change.)

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100-6 >-5/3x>-6,
,
,
.

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Significance levels differ among industries because different industries hold different quality standards. The null hypothesis is more likely to be rejected at 0.10 significance level. As the significance level increases to α = 0.10 from α = 0.01, type I error is more likely to occur (that is, it is more likely that we reject a statement that may actually be true.) To reduce the likelihood of a type I error, continuous and periodic sampling within the industry is suggested (quality control) using as large (but manageble)a size as possible.

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Let t = 0. Then the system becomes
0 = A-2,
0 = B+C+3,
1 = -A+2D,
1 = -B-2C-C, or 1 = -B - 3C.
The 1st equation gives A = 2.
Substitution into the 3rd equation gives 1 = -2 + 2D, or D = 3/2.
eliminating B from the 2nd and 4th equations gives C = 1.
Finally, substitution into the 2nd equation gives B=-4.

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|2x+1| >= 5 is the same as
or .
or .
or .
In interval notation, (, -3]U[2, ).

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Of the 100 students, 100*0.40 = 40 have brown hair. Of these, 40*0.30 = 12 are boys. So 12 boys have brown hair.

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sin(105) = sin(60 + 45) = sin(60)*cos(45) + cos(60)*sin(45)
=
=+
=

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let x = # gallons pure water that must be added. Then
10*0.30 + x*0.00 = (10 + x)0.25,
3 = 2.5 + 0.25x,
0.5 = 0.25x
x = 2 gallons of water must be added.
***Note: x*0.00 denotes the weight contribution of the water to the salt content of the newly mixed solution. x is in gallons, and its salt concentration is 0.00 weight units per gallon, so that x*0.00 is in weight units.***

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How many scored in what game? =D

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The left-hand side is reducible to sinx. Hence we solve sinx = 1. Only one angle satisfies that, x = 90 degrees.

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QIV.

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False. Just substitute the coordinates of the given point into the equation.

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from the inverse relationship, where k is the constant of proportionality. When y = 7, x = 3, so . Hence k = 63, and therefore the variation equation is .

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,
,
,
,
,
. The first factor gives
, or , giving = 30 or 150 degrees.
The second factor does not give any solutions. (Why?)

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,
Hence, or . Using the quadratic formula:
. The values are, to 3 decimal places, 0.001 (from 0.000962432), or -7.0001 (from -7.000962432). We have to eliminate the 2nd value, because it won't satisfy the original equation. The answer is 0.001.

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For the first function must make sure that 7x >0, or x>0 (after dividing both sides of the inequality by 7).
For the second function, . This is just the set of all real numbers except x = 4.

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All of the 1x1x1 cubes hidden inside the 6x6x6 cube.

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Prove:. Now

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By the property of a normal distribution, the probability that a brook trout's length a.)exceeds the mean is 0.50,
b.) exceeds the mean by at least 1 standard deviation is about 0.16 (by adjusting the 68-95-99.7 rule),
c.)exceeds the mean by at least 2 standard deviations is about 0.025(by adjusting the 68-95-99.7 rule),
d.)is within 2 sd's is 0.95.

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Choose a particular topic and ask a question... =)

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Assuming the 3 events are mutually independent (i.e., performance in the next round is not affected by the previous round), the probability is . The answer is D.

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I got a feeling you meant the following:
p(X=0)=.25,
P(X=1)=.125
P(X=2)=.125 and
P(X=3)=.5, right?
The cumulative distribution function is
F(X = 0) = 0.25
F(X = 1) = 0.375
F(X = 2) = 0.5
F(X = 3) = 1.
It's easy to plot the probability mass function for this distribution.

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There are 2 o's, 2 n's, 1 l, and 1 d in the word "london". Then from a formula from combinatorics , there are ways.