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Thank you, you are correct Guest.  I made an arithmetical mistake!  I've now edited my reply to correct the error.

That is ok Lightning. The oether was someone else in your class I expect.

It would be nice if you added a 'thank you' for Alan and me for answering you though.   

Do not repeat your questions !!

At least not in close succession. 

When you do repeat them then put the address of the original in and ask people to answer you there!

Go online to this site and use the formulas listed in the table at the bottom:

 Permutations and Combinations > Distributing Balls into Boxes

If s is the smallest of the five integers, the mean is \((5s+10)/5=s+2\) (not (s+5) as previously suggested). We therefore require \(s+4<2(s+2)\), which reduces to \(s>0\). Thus the smallest integer must be 1 or greater.

As a reality check note that if s=1 we have the set 1,2,3,4,5 with average 3 and \(2\times3\) is indeed greater than 5, but with s=0 we have 0,1,2,3,4 with mean 2 and since \(2\times2=4\) the inequality just fails.

Thanks Alan. Just take the reciprocal of the decimal as: 1/0.0002020202020....=4,950. Or, the reciprocal of 4,950 as: 1/4,950 =0.00020202020..........etc.

Here is my attempt at distributing the 19 balls. Please be warned that this may be wrong! Take it at your own peril !!!!:

1-    8 + 7 + 3 + 1 = 19
2-   8 + 7 + 2 + 2 = 19
3-   8 + 6 + 4 + 1 = 19
4-   8 + 6 + 3 + 2 = 19
5-    8 + 5 + 5 + 1 = 19
6-   8 + 5 + 4 + 2 = 19
7-   8 + 5 + 3 + 3 = 19
8-   8 + 4 + 4 + 3 = 19
9-    7 + 7 + 4 + 1 = 19
10-  7 + 7 + 3 + 2 = 19
11-  7 + 6 + 5 + 1 = 19
12-  7 + 6 + 4 + 2 = 19
13-  7 + 6 + 3 + 3 = 19
14-  7 + 5 + 5 + 2 = 19
15-  7 + 5 + 4 + 3 = 19
16-  6 + 6 + 5 + 2 = 19
17-  6 + 6 + 4 + 3 = 19
18-  6 + 5 + 5 + 3 = 19

The capacity of the boxes is counted from left to right as: 8, 7, 5, 3

Let s be the smallest integer. Then the 5 numbers are: s, s+1, s+2, s+3, s+4

Their average is (5s + 10)/5 or s + 2

So: s + 4 < 2(s + 2)

s + 4 < 2s + 4

s >0

So the smallest integer greater than 0 is 1.

Edited to correct mistake.  Thanks to Guest below for noting this.

Looks like this is meant to be: 0.0002020202....

So let        x = 0.0002020202...

 Then  100x = 0.0202020202...

Subtract the first from the second: 99x = 0.02, so x = 0.02/99

Multiply numerator and denominator by 100:   x = 2/9900

Simplify:  x = 1/4950

I didn't double post this one. I would have realized the answer was already answered. Plus, I have a work from 9pm to 3am. So it coudn't be me.

I didn't double post this one. I would have realized the answer was already answered. Plus, I have a work from 9pm to 3am. So it coudn't be me.

Sorry, can't read your question as written. Write it in "plaintext". Thanks.

This problem appears in a Mathematical Site on the Internet. It is classified under "Probability problem #113." Both the problem and the solution are posted there. Here is the main site:

http://www.cut-the-knot.org/content.shtml

Note: The small error in CPhill's counting of EVEN sixes neglected to take "0" into account as follows: C(10,0)*(5/6)^10*(1/6)^0 =0.1615056......., which would give a total of: 0.50867....etc., so that both odd sixes and even sixes must ADD up to 1.

1- What is the probability of an odd number of sixes turning up in a random toss of 10 fair dice?

We  can either have 1 - 3 - 5 - 7  or  9 6s

We have a (1/6) probability of a six on any roll  and a  (5/6) probability of  "not 6" on any roll...so....

P( one 6)  = C(10,1)(5/6)^9 (1/6) = 9765625/30233088  ≈ 32.3%

P(three 6s)  =  C(10,3) (5/6)^7 ( 1/6)^3   =  390625/2519424 ≈ 15.5%

P ( five 6s)  = C(10,5)(5/6)^5 (1/6)^5   = 21875/1679616 ≈ 1.3%

P ( seven 6s)  = C(10,7) (5/6)^3 (1/6)^7  = 625/2519424 ≈ .02%

P ( nine 6s)  = C(10,9)(5/6) (1/6)^9   = 25/30233088 ≈  0%

Adding all the fractions, we get   58025/118098  ≈ 49.1%

2-What is the probability of an even number of sixes turning up in a random toss of 10 fair dice?

We can have 2 - 4 - 6 - 8  or 10  6s

P(two 6s) =   C(10,2) (5/6)^8 (1/6)^2  = 1953125/6718464 ≈ 29.1%

P(four 6s)  =  C(10,4) (5/6)^6(1/6)^4  = 546875/10077696 ≈ 5.42%

P( six 6s)  = C(10,6) (5/6)^4 (1/6)^6  = 21875/10077696 ≈ .21%

P(eight 6s) = C(10,8) (5/6)^2(1/6)^8  = 125/6718464 ≈ 0%

P (ten 6s)  =  (1/6)^10  = 1/60466176 ≈ 0%

Adding all these fractions, we get   20991751/60466176  ≈ 34.7%

They are not equal....notice that P( one 6 ) > P (two 6s)  and  P ( three 6s) > P (four 6s)  and P( five 6s) > P (six 6s)...etc.

Thus it appears that  P(2n - 1  6s)  > P(2n  6s)  for  n = 1 - 5  

Let  the center of the circle with a radius of 5 be centered at  (0,0)  and  let  A  = (-5,5)

The midpoint of  EF  is  (5 cos (45°), -5sin(45°) )  = ( 5/√2, -5/√2)

Let this midpoint be  = G

So   the height of triangle AEF    =  GA  =  √ [( 5/√2 + 5) ^2+ ( 5 +5/√2)^2 ]    = 5 + 5√2 units =

5 ( 1 + √2)   units   =  5 (√2 + 1) units

Let H  be the point of tangency where BD touches the circle

And  FG  = FH since these are tangents drawn to a circle from the same exterior point, F

And angle HOF  = 22.5°  and angle OFH  = 67.5°...so...using the Law of  Sines

FH / sin (22.5°)  = OF / sin (67.5°)

FH  = 5 sin (22.5°) / sin (67.5°)

FH = 5 sin (22.5°) / cos (22.5°)

FH  = 5 * tan (22.5°)  = 5  * [  1  - cos (45°) ] /[ sin (45°) ]  =  5 *  [ 2 - √2] / [ √2]  units  =

5 (√2 -1)    = FG

So...the area of  AEF  =   (GA) ( FG) =  5 (√2 + 1) * 5 ( √2 -1)  units^2  =

25 (2 - 1)  =  

25 units^2

The area of  square ABCD   =10*10  =  100 units^2

So  area   ΔAEF  =   25 / 100   =  25%    of    square ABCD

Here's a pic :

Your answer is better than mine Alan. it is better explained.

I rarely touch questions that are presented so poorly. Maybe that is why Lightning reposted. 

Or maybe this question was posted by someone else in his class.

Ref for others :  https://web2.0calc.com/questions/help_51906

That is not a problem Alan   

Maybe 2 answers are better than one and it certainly is not your fault if Lightning double posted.

Brenda is going from (-4, 5) to (5, -4), but she needs to stop by the origin on the way. How far does she have to travel?

Hi Logic!

a) (-4,5) to (0,0)

b) (0,0) to (5,-4)

\(d=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}\)

\(d_a=\sqrt{(-4-0)^2+(5-0)^2}=\sqrt{16+25}\\ d_a=\sqrt{41}\)

\(d_b=\sqrt{(0-5)^2+(0-(-4))^2}=\sqrt{25+16}\\ d_b=\sqrt{41}\)

\(d_a+d_b=2\cdot \sqrt{41}=12.806..\)

Brenda has to travel 12,806 times the local unit of length.

Is the origin on the way the origin of the coordinate system?
Or does Brenda travel back to the starting point?

Then:

twice (-4,5) to (5,-4)

\(d=2\cdot \sqrt{(-4-5)^2+(5-(-4))^2}=2\cdot\sqrt{81+81}\\ d=2\cdot \sqrt{2\cdot 3^4}=2\cdot 3^2\cdot \sqrt{2}\\ \color{blue}d=18\cdot\sqrt{2}=25.4558..\)

Brenda has to travel 25.456 times the local unit of length.

Greatings

  !

Sorry Melody, I didn't realise you were working on this.

This question seems to have appeared twice!  I answered the other one at https://web2.0calc.com/questions/help-please_18822#r1.

A partial fractions problem:

.

\frac{x + 7}{x^2 - x - 2} = \frac{A}{x - 2} + \frac{B}{x + 1}

\(\frac{x + 7}{x^2 - x - 2} = \frac{A}{x - 2} + \frac{B}{x + 1}\\ A(x+1)+B(x-2)=x+7\\ Ax+Bx=x \qquad A-2B=7\\ A+B=1 \qquad A=7+2b\\ 7+2B+B=1\\ 7+3B=1\\ 3B=-6\\ B=-2\\ A=3\\~\\ check\\ 3(x+1)-2(x-2)=x+7 \quad great \)

answer   (3,-2)

I think it is  531441

Let the number of people at the beginning be x

The number after question 1 is answered is x/3

the number after question 2 is answered is x/(3^2)

...

the number left after question 12 answered is  x/(3^12)

\(\frac{x}{3^{12}}=1\\ x=3^{12}\\ x=531441\)

"Find all values of  t such that \(\lfloor t\rfloor = 2t + 3\) . If you find more than one value, then list the values you find in increasing order, separated by commas."

This graph should help.  HOWEVER, be careful to check all the values of t that seem to be suggested by the graph!!

.

What is the remainder of 5^{2010} when it is divided by 7?

\(5^{2010}\:mod\;7\\ \equiv (-2)^{2010}\:mod\;7\\ \equiv2^{2010}\:mod\;7\\ \equiv 2^{3*670}\:mod\;7\\ \equiv 8^{670}\:mod\;7\\ \equiv 1^{670}\:mod\;7\\ \equiv 1\:mod\;7\\\)

so the remainder is 1

First take $17.60 and divide it by the number of slices to get what each slice costs.

$17.60/16 = $1.10

Multiply the cost of one slice by the number of slices eaten

$1.10 x 6 = $6.60

Brad ate $6.60 of the pizza.