All Answers 358468 Answers

https://www.wolframalpha.com/calculators/equation-solver/

7:00 PM.  Duh.

I need it in standord form 

Here is a pic:

5x+8y= -9     re-arrange in to   y= mx + b form

8y = -5x - 9

y = -5/8 x - 9/8       shows m₁  (the slope) = =5/8

Remember how to find pepindicular slope?

- 1 / m₁   will be perpindicular      -1 / (-5/8)  = 8/5

so the line we are looking for is   y = 8/5 x + b      Now sub in the point given (10,10) to find the value of b

                                                   10 = 8/5 (10) + b      b = -6

your line is       y = 8/5 x - 6        m+ b = 8/5-6 = - 4 2/5

Just by looking at it , I would say  y > x- 7

Welcome as a member Tiggsy.

I should have said this immediately but I did not realize straight away that you had joined!

I know you have been here forever but I finally talked you into becoming a member.  I am very pleased! 

Now I will be able to look back and find your fabulous answers (from this point forward)

Finally ... I have exhausted this question.  There was a lot for me to work through here. 

I do not remember seeing those formulas before that you used Tiggsy. Hopefully now they, and their proofs, are a little embedded into my brain.

Anyway, i finally fully understand both solutions so I am happy.      

Well, we can eliminate a few answers right away. 103% is not possible, so it's not that. And 95 and 91 are clearly too low if you flip it 200 times (but who would have the patience to do that?). However, it's also pretty clear that it would round to 100% instead of 99%. The reason is because they might have truncated the percent instead of rounding it.

I actually proved 4 equations in 8th grade, and #4 helps out a lot with this.

#4: \(1-(1-c)^t\) the probability of an event with \(c\) chance occurring at least once after \(t\) times.

For your problem, \(t=200\) and \(c=\frac12\). \(1-(1-\frac12)^{200}=1-(\frac12)^{200}=1-\frac{1}{BIG}=\frac{BIG-1}{BIG}\). I substituted 'BIG' for the result of the power because the number itself is 61 decimal digits. If we transfer the fraction generated into a percent and truncate at 60 decimal digits, then this is the percent: \(99.999999999999999999999999999999999999999999999999999999999937\)%. If I round, it will without a doubt round to \(100\)%. But if I truncate the entire decimal, I get \(99\)% (if you are wondering, truncating a fraction is basically cutting off its decimal after a certain point. In this case, I cut off all of its decimal, and probably so did the textbook.).

The 3rd answer is the right one.

You can see that 2 should occur about twice as often as each of the other 2.

Could I please have some help on this?

\(\begin{array}{|lrcll|} \hline n=0&\binom00^2 &=& 1 =\binom00 \\ n=1&\binom10^2 + \binom11^2 &=& 1^2+1^2 = 2 =\binom21 \\ n=2&\binom20^2 + \binom21^2 + \binom22^2 &=& 1^2+2^2+1^2 = 6 =\binom42 \\ n=3&\binom30^2 + \binom31^2 + \binom32^2 + \binom33^2 &=& 1^2+3^2+3^2+ 1^2 = 20 =\binom63 \\ n=4&\binom40^2 + \binom41^2 + \binom42^2 + \binom43^2 + \binom44^2 &=& 1^2+4^2+6^2+4^2+ 1^2 = 70 =\binom84 \\ &\ldots \\ &\dbinom{n}{0}^2 + \dbinom{n}{1}^2 + \cdots + \dbinom{n}{n}^2 &=& \dbinom{2n}{n} \\ \hline \end{array} \)