All Answers 358180 Answers

This question is stated correctly. 94,000 is a future value.

Let's convert the years into months. (12*14=168)

Now let's set up our notation.

A=F(A/F, i , n)

A=94,000(A/F, 5.7%, 168)

Now time for the formulas.

A= 94000((.057)/(1.057^168-1))

A = $0.48355 (This is the monthly deposit to achieve $94,000 in 14 years [168 months])

On the 10th day, Mr. Santos should have sold =2^(10-1)=2^9=512 tickets.

The sum of the first five terms of a Geometric Sequence is 155 and the fifth term is 80. What is the 6th term?

a₁- first term. a_n=q^n-1*a₁. Sum of first 5 terms- a₁*(q⁵-1)/(q-1)=155

fifth term=80=a₅=a₁*q⁴----->a₁=80/q⁴

n = 1

\(\sum_{r=2}^{2014}(\frac{1}{\sum_{k=1}^rk})=\frac{2013}{2015}\\ \text{ add } \frac{2}{2015} \text{ to get } 1\)

5, 10, 20, 40, 80, 160, 320, 640......and so on.

1st term=5, common ratio=2

what is 12+22+32+......+n2 ? you need to express it using a simple expression

=[n(n+1)(2n +1)] / 6

lets call the ratio q, ok?

the sum of the first term+the second term+.....+the nth term=a₁*(qⁿ-1)/(q-1). a₁=1, so the sum=(qⁿ-1)/(q-1). we know the nth term is 128, so q^n-1=128, so qⁿ=128*q

128*q-1/q-1=256

128*q-1=256*q-256

-1=128*q-256

128*q=255

q=255/128

the series is impossible

Whole numbers are all the nonnegative integers

\(0, 1, 43, 234678, 356545789765\) are all whole numbers

\(-5, \frac{1}{2}, \pi, 3.9, 6\frac{2}{3}\) are not whole numbers. 

-5 is a negative integer

½ is not an integer

\(\pi\) is not an integer

3.9 is not an integer

\(6\frac{2}{3}\) is not an integer. 

All this time, I have not answered the question at hand: Is 0 a whole number?

Yes, 0 is a whole number. 

What? I was on an hour ago.  Didn’t you see my brilliant post?

Logging on is a pain in my chimp rump --I mean my delicate heinie. The security certificates corrupt and I have to reinstall them. They are probably corrupted from all the allergy-causing blarney that passes through them.

You did not give much explanation here Heureka :)

That is not like you ://

I mean, it seems that you're all enjoying his riddles, and some of his riddles are very interesting, so you can check that out!

Ginger ale! welcome back! long time no see :(.

But why arent you showing your face? we all know you have a user, you can use it too, see:

user.

staggering!

JB, if you are going to imitate me in your troll posts at least target posts with wrong answers.  Mr BB gave the right answer. Least wise it exactly matches the answer by Presh Talwalkar here:

OH NO!! NOT again!!

You have got to be kidding!!  It’s very obvious the answer is less than 1000/4! (< 41)

I’m developing an extreme allergy to . . . . to . . . .BLARNEY!!

Tiggsy, your presentation is amazing. This clearly indicates all the mathematical relations, including the 10% “adjustment” for double counts.

Here’s a condensed presentation

\(\text{Prb = }\left(\dfrac{\binom{3}{2}\binom{17}{1} + \binom{3}{3}}{\binom{20}{3}} * \dfrac{9}{10}\right) + \dfrac{1}{10} \approx 0.1411 \text{ or } 14.11%\)

Melody’s Venn diagram also makes clear why the adjustment only affects the intersection of the two events and not the events separately.

At first blink, this question seems straightforward, and relatively simple to solve. However, it has subtleties that require the mathematician to have practiced skills in combinatorics and statistics to fish out the nuances that affect the final counts. Melody and Tiggsy obviously have these skills. 

Someone previously asked this question on here:

n=1/1+2 + 1/1+2+3 + ...... + 1/1+2+3+....+2014 + 2/2015, n=?

edited

continued fraction ?

\(\begin{array}{|rcll|} \hline n =\cfrac{1}{1+2+\cfrac{1}{1+2+3+\cfrac{1}{ 1+2+3+4+\ldots \cfrac{1}{1+2+3+4+\ldots + 2014 +\frac{2}{2015}}}}} \\ \hline \end{array} \)

n = 0.31606040

Thanks Alan, but could you please expand upon the meaning of this.  Or give an example maybe.

This is just an if/then that applies to a specific function f, isn't it?     

I really do not understand well at all :(

\(x\in D(f) \cdot \ni\cdot f(x)=6\)

Might be:

or:

if D stands for domain, this could mean something like:  "the set of x's belonging to the domain of the function f implies that there is an f(x) equal to 6".

What is xED(f)*backwards E* f(x)=6 mean?

\(xED(f)*\ni* f(x)=6\)

\(\ni \text{ means owns, or has member} \)

ref:   

In an ORDERED code lock, there are 10 possible choices for each of the four spots in the code, making for a total of 10^4=10,000 possible codes.
In this UNORDERED code lock, we have to avoid double counting since codes like 1000 and 0100 are equivalent unlock codes as they are the same digits in a different order.
So, one way to count unique unlock codes is to count by the number of unique digits in the unlock code.
Will make a table for various scenarios:

Unique digit(s)   ways to choose digit(s)    possible patterns       # diff. codes

1                             10C1=10                          1:aaaa                               10

2                             10C2=45                          3:aaab,abbb,aabb  45*3=135

3                             10C3=120                        3:abca,abcb,abcc 120*3=360

4                              10C4=210                       1:abcd                              210

Now, will just sum up the number of different codes                            ____

                                                                                                                 715    

And that is my take on it!!.                    

Whole Numbers are simply the numbers 0, 1, 2, 3, 4, 5, ... (and so on)

No Fractions!