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(9x10^-4) (6 x 10^-5)

=   54 x 10^(-4+-5)

54 x 10 ^-9  = .000000054

Well.......5 or less down to the deuce is only 4 x 4(suits) = 16 cards out of 52 that will give me a winner.

16/52 = 30.7 % chance of winning

obviously YOU have a 69.3% chance of winning.......

SO let's set up a ratio to see if the 'house' has the advantage....

30.7/69.3   =?   12/37

.443   = ?   .324            it looks like that is a good bet to me !

Another way to look at it...if I play 100 times I will win 30.7 times and COLLECT 30.7 x 37 = 1135.90

       I will LOSE 69.3 times and have to PAY   69.3 x 12 = 831.60

             So if I play 100 times...I collect 1135.9   and PAY 831.6 for a net POSITIVE CASH FLOW of 304.30   WINNER!!!!

Incidentally  the Powerball Jackpot  for this Saturday's (1/9/15) drawing is over 700 MILLION for those of you old enough to play !   Yikes!!!!

Hi SHADES,

I as well as a guest, answered that same question somewhere  else.

Please do not ask the same question in 2 different places!  

It often means that you get 2 identical answers and someone else gets no answer to their question.!

Please follow the guidelines if you want to repost a question!   

Use a calculator 64 year old.

As guest points out, not necessarily......Example :

360 / 60 = 6

360 / 10  = 36

Well ....if this year's tuiton is 4% more than last year......that is 104% of last year's tuition

42500 x 104% =  42500 x 1.04 = 44,200.

No...it just depends on how close the two (divisor and dividend) are in magnitude....

examples

100/10   = 10

100/99 = 1.0101         See?     The two questions have the same number of digits but DIFFERENT number of digits in the answers/...

maybe a better example:

200/50 = 4

200/10 = 20

!5 = 44

\(\begin{array}{rcll} !n = n!\sum \limits_{k=0}^{n} \frac{(-1)^k}{k!} = n!\cdot \left( ~ 1 - \frac{1}{1!} +\frac{1}{2!} - \frac{1}{3!} + \cdots + (-1)^n \frac{1}{n!} ~ \right) \end{array}\)

Example:

\(\begin{array}{rcll} !6 = 6!\cdot \left( ~ 1 - \frac{1}{1!} +\frac{1}{2!} - \frac{1}{3!} +\frac{1}{4!} - \frac{1}{5!} +\frac{1}{6!}~ \right) = 265 \end{array}\)

\(\begin{array}{rcll} !5 &=& 5!\cdot \left( ~ 1 - \frac{1}{1!} +\frac{1}{2!} - \frac{1}{3!} +\frac{1}{4!} - \frac{1}{5!} ~ \right) \\ &=& 120\cdot \left( ~ 1 - 1 +\frac{1}{2} - \frac{1}{6} +\frac{1}{24} - \frac{1}{120} ~ \right) \\ &=& 120\cdot \left( ~ \frac{1}{2} - \frac{1}{6} +\frac{1}{24} - \frac{1}{120} ~ \right) \\ &=& 120\cdot \left( ~ \frac{3\cdot 4\cdot 5 - 4\cdot 5 + 5 - 1 }{120} ~ \right) \\ &=& 3\cdot 4\cdot 5 - 4\cdot 5 + 5 - 1 \\ &=& 60 - 20 + 5 - 1 \\ \mathbf{!5} & \mathbf{=} & \mathbf{44 } \end{array}\)

Oh I got it!! So trick question, the young boy is too lazy and uneducated to even do his homework so he just rolls over instead of walking to his bed since he's too fat to walk since he eats more than studying and working on his homework and his Mom gets mad at him for 5 seconds or so because she doesn't punish his son and makes his son become a lazy no life dummy. I think I got that one right, did I?

1 with a remainder of 11.     

50 x 10% =  50 x .1 = 5

With a machete.

They don't ALL do that.....but we had that thread before....you need to know the limitations/operating characteristics of your tools (calculator) to use them correctly and get correct answers......

For THIS calculator you have to use parentesis    (-5)^2    to get the answer you expect  : 25

if you put in -5^2   you will get  -  (5^2)      -25    

OTHER scientific calculators may or may not do this......MOST of them will give you  25    SOME (Including this one) will give you    -25     I think this is a problem.....but only because I am used to the calculators that give you  +25   I was an engineer in a former life.....all of my calculators gave +25    but some will give you -25

Melody said it before:  KNOW your tools.

~jc

What are the first 4 terms of f(n-1)+15

If f(1) = 7

\(\begin{array}{rcll} f(2) &=& f(2-1)+15 \\ f(2) &=& f(1)+15 \qquad | \qquad f(1) = 7\\ f(2) &=& 7+15\\ f(2) &=& 22 \\\\ f(3) &=& f(3-1)+15 \\ f(3) &=& f(2)+15 \qquad | \qquad f(2) = 22\\ f(3) &=& 22+15\\ f(3) &=& 37 \\\\ f(4) &=& f(4-1)+15 \\ f(4) &=& f(3)+15 \qquad | \qquad f(3) = 37\\ f(4) &=& 37+15\\ f(4) &=& 52 \\\\ \cdots \end{array}\)

arithmetic sequence: \(\boxed{ ~ a_n=a_1+(n-1)\cdot d ~}\)

\(\begin{array}{rcll} f(n) &=& f(n-1)+15 \\ f(n)- f(n-1) &=& 15 \\ d &=& 15\\\\ \end{array}\)

\(\begin{array}{rcll} a_1 = f(1) &=& 7 \\ d &=& 15 \\\\ a_n &=& 7 + (n-1)\cdot 15 \\ a_n &=& 7 + 15n -15 \\ \mathbf{ a_n} & \mathbf{ = } & \mathbf{ -8 + 15n } \end{array}\)

CPhill answered your question accurately some 3 hours ago. Just look at the answer at the bottom of this page, especially if are not the person who asked the same question.

I answered this one here :   https://web2.0calc.com/questions/statue-of-liberty

Exactly! You should always do it first on paper, then check it on calc!

Did you forget to put an problem?

Place numerator over denominator. Use ex. Guest gave you...

Just put one number over another like this: 3/4, 5/7, 13/4, 1/5........etc.

45x^5+6x^4+3x^3+8x+12 by 3x-12

This is certainly an unusual one!!!!

First, we want  to factor the denominator  as 3[x - 4]

Then, we want to divide all the coefficients by 3   [don't forget to include the "missing"  power of x^2....it's coefficient will be 0

[45x^5 + 6x^4 + 3x^3 + 0x^2 + 8x + 12]  / [3 (x - 4)] =

This results in :

[15x^5 + 2x^4 + x^3 + 0x^2 + (8/3)x + 4] / (x - 4)

Now, proceed normally

4  [   15        2        1          0          8/3              4          ]

                  60      248     996      3984          47840/3  

       --------------------------------------------------------------

         15     62      249    996       11960/3      47852/3

And the resultant polynomial  is :

15x^4 + 62x^3 + 249x^2 + 996x + 11960/3  +   remainder of 47852 / 3

21^2=13^2+29^2-2*13*29*cosA      and we're trying to find angle "A"

441  = 169 + 841  - 2(13)(29) cosA

441 = 1010 - 754 cosA     subtract 1010 from each side

-569 =  -754 cosA      divide both sides by -754

569/754  = cosA       we need to use the cosine inverse, here

cos-1 (569/754) = A = about  41°

Circumference/Pi=Diameter

10% of 50 = 1/10 of 50 = 5  Hope this helps   :)

To find the % discount :

( [Old Price - New Price] / Old Price x 100)  %

 ( [  (27 - 8) / 27 ]  x 100 ) %  =   about 70.37%  discount