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Twenty-five percent of all automobiles undergoing an emissions inspection at a certain inspection station fail the inspection. 

(d) What is the probability that among 25 randomly selected cars, the number that pass is within 1 standard deviation of the mean value? 

I think this is a trick question.....

The number that are within 1 standard deviation of the mean in a normal distribution, which I think is assumed here, is always  68%

earth?

Meth?

Wrong.

wealth?

Someone asked me to answer this, but i dont understand what you're trying to say.

Theres nothing "special" about this number. That was the first guest's question, and i answered it.

1. The matrix is not a square matrix, its length is 10¹² and its width is 12.

2. In my solution i HAD to count those zeroes (in the second one. In the first one i just used them to show we can look at it as a set of combinations)

3. You keep claimimg that the zeroes i added are exactly one half of the digits. Your calculation is wrong, because you are assuming A TON of things you think are right. I calculated it, and the number of added zeroes is less than 1/100 of all of the digits (after the zeroes were added)

Integrate the following with steps please:

integrate [sin^2 (x) cos^2 (x)] dx from x=0 to pi.

Thank you for any help.

Formula:

\(\begin{array}{|rcll|} \hline \cos^2(x) = \frac12\cdot \Big(1+\cos(2x)\Big) \\ \sin^2(x) = \frac12\cdot \Big(1-\cos(2x)\Big) \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline && \int \limits_{0}^{\pi} {\sin^2(x) \cos^2(x)\ dx} \\ &=& \int \limits_{0}^{\pi} {\frac12\cdot \Big(1-\cos(2x)\Big)\cdot \frac12\cdot \Big(1+\cos(2x)\Big)\ dx} \\ &=& \frac14\cdot \int \limits_{0}^{\pi} { \Big(1-\cos(2x)\Big)\cdot \Big(1+\cos(2x)\Big)\ dx} \\ &=& \frac14\cdot \int \limits_{0}^{\pi} { \Big(1-\cos^2(2x)\Big) \ dx} \quad & | \quad \cos^2(2x) = \frac12\cdot \Big(1+\cos(4x)\Big) \\ &=& \frac14\cdot \int \limits_{0}^{\pi} { \Big(1-\frac12\cdot \Big(1+\cos(4x)\Big) \ dx} \\ &=& \frac14\cdot \int \limits_{0}^{\pi} { \Big(1-\frac12 - \cos(4x)\Big) \ dx} \\ &=& \frac14\cdot \int \limits_{0}^{\pi} { \Big( \frac12 - \cos(4x)\Big) \ dx} \\ &=& \frac14\cdot [\frac{x}{2}-\frac{\sin(4x)}{4}]_{0}^{\pi} \\ &=& \frac18\cdot [ x - \frac{\sin(4x)}{2} ]_{0}^{\pi} \\ &=& \frac18\cdot [ \pi - \sin(4\pi)-(0-\frac{\sin(4\cdot 0)}{2} ) ] \\ &=& \frac18\cdot [ \pi - 0-(0-0) ] \\ &=& \frac18\cdot [ \pi ] \\ &=& \frac{\pi}{8}\\ \hline \end{array}\)

*edited

How many 3s..........

How many digits of 3 are there between 1 and 10^12, and what is the percentage of these 3's out of the total of 10^12 numbers, and why? Please explain.

Thank you for any help.

I agree with Alan the total count of 3s is in general 

between 1 and 10ⁿ =  \(n\cdot 10^{n-1}\)

between 1 and 10¹²  = \(12\cdot 10^{11}\)

My empirical solution:

\(\small{ \begin{array}{|lcr|c|rrrrrrrrrrrrr|r|} \hline &&& n & &12th&11th&10th&9th&8th&7th&6th&5th&4th&third&second&first & \text{Sum of 3's} \\ &&& & \text{3's in Postion} \\ \hline 1 &\ldots& 10 & 1 & & & & & & & & & & & & &1\times 3 & 1\\ 1 &\ldots& 100 & 2 & & & & & & & & & & & &10\times 3 &10\times 3 & 20 \\ 1 &\ldots& 1000 & 3 & & & & & & & & & & &100\times 3&100\times 3&100\times 3 & 300 \\ 1 &\ldots& 10000 & 4 & & & & & & & & & &1000\times 3&1000\times 3&1000\times 3&1000\times 3 & 4000 \\ \cdots \\ 1 &\ldots& 1000000000000 & 12 & &100000000000\times 3&100000000000\times 3&100000000000\times 3&100000000000\times 3&100000000000\times 3&100000000000\times 3&100000000000\times 3&100000000000\times 3&100000000000\times 3&100000000000\times 3&100000000000\times 3&100000000000\times 3 & 1200000000000 \\ \hline \end{array} } \)

Thanks a bunch man!

2. Well....if 1mm of rain falling on 1m^2 provides 1 liter of water......then 20 mm falling on the same area must provide 20 times as much water = 20 liters

So

10000L / (20L/1m^2 )  =   500m^2   is the required surface area 

I'm sorry, I forgot to say that the block is a pyramid. :)

current weight - old weight = gained weight

\(15\frac23-7\frac18=g \\~\\ = 15\frac{2*8}{3*8}-7\frac{1*3}{8*3} \\~\\ = 15\frac{16}{24}-7\frac{3}{24} \\~\\ = 8\frac{13}{24}\text{ pounds}\)

Here is the first one:

josh = 10 + alice

alice = 8 + tom     →  alice - 8 = tom

josh + alice + tom = 62

Substitute:

(10 + alice) + alice + (alice - 8) = 62

10 + alice + alice + alice - 8 = 62

2 + 3(alice) = 62

alice = 60/3 = 20

alice - 8 = tom

20 - 8 = tom = 12 cakes

You're not dumb!   

But this question is possible.

If the length to width is 5 to 4, then:

length = 5x cm

width = 4x cm

Where x is the unknown scale factor.

volume of a pyramid = (1/3) * length * width * height

1800 = (1/3) * 5x * 4x * height

(3)(1800) / (20x²) = height

height = 270/x² cm

I don't think you can get a more specific answer than that.

*edited to make it for a pyramid*

There isn't a single answer to this problem......for instance...

Length  = 10 cm

Width = 8  cm

Height  22.5 cm

Or

Length  = 5

Width = 4

Height = 90

Etc........

(2^2)^-2

\((2^2)^{-2}\\ =4^{-2}\\ =\frac{\color{blue}{1} *\color{black}4^{-2}}{1}\\ =\frac{\color{blue}{1} }{1*\color{black}4^{+2}}\\ =\frac{\color{blue}{1} }{\color{black}4^{2}}\\ =\frac{\color{blue}1 }{\color{black}16}\\ \)

it = 0.0625

It travels three times as far in three hours....so...

3 *  6.71*10^(8)  =

2.013 x 10⁹   miles

6x + 6  = 0   subtract 6 from both sides

6x  =   - 6      divide both sides by 6

x  =   -1

You are a genius. Thank you 

In one rotation, the linear distance covered  = 2 * pi * 13   = 26pi  inches

Converting 1200 ft to inches  = 1200 * 12 = 14400 in

So.....he number of rotations is :

14400 / [26 pi ]  ≈ 176  rotations