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Did you make this one up? ANY value of x is a solution!!

Future Population =FP

Present Population=PP

FP =PP x [1 + R]^N

FP =43,433 x [1 + 0.069]^49

FP =43,433 x     1.069^49

FP =43,433 x      26.2970816751.....

FP =1,142,161 - the future population of the town after 49 years.

$1,833 x (100% - 11%) =

$1,833 x 89% =

$1,833 x 0.89 =$1,631.37 - what he paid after 11% sale.

x * 1.052 =$115,705

x =$115,705 / 1.052 =$109,985.74 - what Lola was earning before 5.2% raise.

See the answer here: https://web2.0calc.com/questions/help-one-more-problem

Consider two positive even integers less than 15 (not necessarily distinct).

When the sum of these two numbers is added to their product, how many different possible values may result?

\(\text{Let $n_1\lt 15$ is an positive even integer.} \\ \text{Let $n_2\lt 15$ is an positive even integer.} \\ \text{The sum of these two numbers is added to their product $n_1+n_2+n_1\cdot n_2 $.} \)

\(\text{Because $n_1$ is even, we can set: $n_1 = 2i \Rightarrow i_{\text{min}} = 1 $ and $i_{\text{max}} = 7 $ } \\ \text{Because $n_2$ is even, we can set: $n_2 = 2j \Rightarrow j_{\text{min}} = 1 $ and $j_{\text{max}} = 7 $ } \)

\(\begin{array}{|rcll|} \hline && n_1+n_2+n_1\cdot n_2 \\ &=& 2i + 2j + 2i2j \\ &=& 2(i+j+2ij) \quad & 1 \leq i \leq 7 \qquad 1 \leq j \leq 7 \\ \hline \end{array} \)

\(2(i+j+2ij) \quad 1 \leq i \leq 7 \qquad 1 \leq j \leq 7 \\ \begin{array}{|c|r r r r r r r|} \hline (i,j) & 1 & 2 & 3 & 4 & 5 & 6 & 7 \\ \hline 1 & 8 & 14 & 20 & 26 & 32 & 38 & 44 \\ 2 & & 24 & 34 & \color{red}44 & 54 & 64 & 74 \\ 3 & & & 48 & 62 & 76 & 90 & 104 \\ 4 & & & & 80 & 98 & 116 & 134 \\ 5 & & & & & 120 & 142 & 164 \\ 6 & & & & & & 168 & 194 \\ 7 & & & & & & & 224 \\ \hline \end{array}\)

Only 44 is double.

There may result 27 different possible values

There are 4! (or 24) "words" that can be formed using each of the letters a, b, c and d once.

If these "words" are alphabetized, which one is 17th?

sorted:
\(\begin{array}{|r|r|} \hline 1. & abcd \\ 2. & abdc \\ 3. & acbd \\ 4. & acdb \\ 5. & adbc \\ 6. & adcb \\ 7. & bacd \\ 8. & badc \\ 9. & bcad \\ 10.& bcda \\ 11.& bdac \\ 12.& bdca \\ 13.& cabd \\ 14.& cadb \\ 15.& cbad \\ 16.& cbda \\ \color{red}{17.} & \color{red}{cdab} \\ 18.& cdba \\ 19.& dabc \\ 20.& dacb \\ 21.& dbac \\ 22.& dbca \\ 23.& dcab \\ 24.& dcba \\ \hline \end{array}\)

If $a * b = a^b + b^a$, for all positive integer values of $a$ and $b$,
then what is the value of $2 * 6$?

\(\begin{array}{|rcll|} \hline a * b& =& a^b + b^a \quad & | \quad a = 2 \text{ and } b= 6 \\ a * b & =& 2^6 + 6^2 \\ a * b & =& 64+36 \\ \mathbf{a * b} & \mathbf{=}& \mathbf{100} \\ \hline \end{array}\)

LMFAO!!! What name are you going to give him? Do you have a short list?

I usually just call them “dumbf**ks” A little generic and unrefined, but fitting as a class name.

The operation $*$ is defined for non-zero integers as follows: $a * b = \frac{1}{a} + \frac{1}{b}$.

If $a+b = 9$ and $ a \times b = 20$,

what is the value of $a*b$?

Express your answer as a common fraction.

Compute:

\(1 - 2 + 3 - 4 + \dots + 2005 - 2006 + 2007\)

\(\small{ \begin{array}{|r|rcrcrcrcccrcrcrl|} \hline \text{sum }= & 1 &-& 2 &+& 3 &-& 4 &+& \dots &+& 2005 &-& 2006 &+& 2007 \\ \text{sum }= & 2007 &-& 2006 &+& 2005 &-& 2004 &+& \dots &+& 3 &-& 2 &+& 1 \\ \hline 2\cdot \text{sum } = & 2008 &-& 2008 &+& 2008 &-& 2008 &+& \dots &+& 2008 &-& 2008 &+& 2008 \\ 2\cdot \text{sum } = & & & 0 &+& && 0 &+& \dots &+& && 0 &+& 2008 \\ 2\cdot \text{sum } = & & & & & && & & & & && & & 2008 & \quad | \quad : 2 \\ \text{sum } = & & & & & && & & & & && & & \mathbf{1004} \\ \hline \end{array} } \)

(1) If we let x be how much the drill was when Lily bought it and y the price regularly, we can make the equation x+17=y. We also know y*0.08=17. So the second equation turns into 1700/8=y so y is 212.50. Then we know x is 195.50, so lily paid $195.50 for the dill.

Hope this helps!! :)

If 5 is x and y is 0 and we plug them in, we have 0 < -5+5, which is 0 < 0, so (5,0) is NOT a solution.

Sum =6 x [1 - 3/2] / [1 - 3/2]

Sum =6 x [1 - 59,049 /1024] / [ - 1/2]

Sum =6 x [-58,025 /1,024] / [-1/2]

Sum =6 x [-58,015/1,024 x -2/1]

Sum =6 x [116,050/1,024]

Sum = 696,300 / 1,024       divide both sides by 4

Sum = 174,075 / 256

You could also use arithmetic sequence formula to sum them up as two separate sequences:

1 + 3 + 5 + 7 +...............+ 2007 =1,008,016.00

-2 - 4 - 6 - 8 - ................- 2006 =-1,007,012.00

1,008,016.00 + (-1,007,012.00) =1,004

Must the point have integer coordinates?

If so, list all the possible points, count how many are not above the x-axis and count the total number of points, and divide the answers.

If not, there are an infinite number of points possible. The number of points on the x-axis can be can be added to those below the x-axis without increasing the (infinite) number. Then, there will be as many points below the x-axis as above the x-axis, so the answer will be 1/2. 

This isn't a geometric series ...

(1 - 2) + (3 - 4) + (5 - 6) + ... + (2005 - 2006) + 2007

   -1     +     -1   +     -1   + ... +          -1           + 2007

There are 1003 groups that result in - 1, so:   (1003)(-1) + 2007  =  -1003 + 2007  =  1004

^{If someone asked me to turn my back while they rolled two dice onto a table top, and then asked me how many 1's I would expect to see when I turned round to look at them, I would say that I expected to see none.}

Really?  That’s blòódy amazing! So then, if you watched the dice roll then you would expect to see ones every time.  You must have telekinetic powers.

Perhaps, when you are not watching, you are unable to control your telekinetic powers to produce a one (1), and in fact cause the dice to roll to every number except a one.  That is still impressive!

Of course, you might just have very low expectations.  Are you pessimistic by nature?

If you are not using telekinetic powers and this still happens then maybe you’ve entered the quantum realm, where the very act of measuring something changes the result.  This relates to the Heisenberg Uncertainty Principle. (I’d explain this, but I’m not sure how to dumb-it-down to your education and intelligence level.)

However, if you resist using your telekinetic powers, and leave the dice to roll naturally –without influence, then the mathematical expectation for rolling two dice and having at least one (1) “single pip” show on one die is (1/3). (Note: in case you didn’t know, a “single pip” means a “one” on a die.)

Anyway, this question is about mathematical expectation, not your personal expectations.

What you posted is a Binomial CDF. This is the Cumulative Distribution Function for a binomial probability, not mathematical expectation. The mathematical expectation does not change –no matter how many times you do this experiment. If it did change, then the Binomial CDF would also change.

It’s great to have (another) “erudite” idiot to troll! We genetically enhanced chimps truly enjoy trolling “erudite” idiots. If you stay around long enough, I’ll give you a name.

The expectation is 92%, that you will continue posting for at least six weeks. What’s the Binomial CDF for this?

My dog, Mr. Peabody, and cat, DC Copper, have reviewed my post, now they want to shoot craps. DC says he wants to test his telekinetic powers for rolling snake eyes. This should be entertaining!

GA

Hi Melody,

I calculated “the expected number of 1's obtained” on a single roll of two dice.

If this were a roll of one die, then the expectation would be (1/6). However there are two dice, so the expectation is weighted by a factor of two (2), giving 2*(1/6)=(1/3). 

Though mathematical expectation and probability are closely related, they are not the same thing.  One notable difference is a probability is always a value between zero (0) and one (1), and an expectation may be equal to any (real) number.  This value depends on the quantity of interest and on the measured probability.  Some probabilities do not have a mathematical expectation, but all mathematical expectations have at least one associated probability.

  .

My troll post, below, contrasts the mathematical expectation to that of a Binomial CDF. 

GA

It is a derangement question, so using the formula:

!5=[5!/e]=44

so the probability is 44/120,  which cancels to 11/30.

It is a Pythagorean triple. The length of YZ is 6, as 10²-8²=6²

As we are concerned with the angle at X, the hypotenuse is the side XZ and the opposite is the side YZ.

Using SOHCAHTOA, we know that sinx=O/H, meaning that sinx=6/10

Therefore our answer is 6/10 or 0.6 :-)

Remember that square rooting is equivalent to raising to the power of a half.

1. The operations would cancel leaving x-1

2. This would be (y⁶)^1/2 which, using the power law of indices, would leave y³

3. We would need to square each side, subtract 1 from each side, square each side, subtract 2 from each side and then square each side, in that order. That would leave n=49

bumo.............

Sorry, fixed it!

Ginger, you have counted the probability of getting 2 ones twice, that is why your answer is too high :)