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There are \(\binom{8}{4}\) ways to choose 4 different members.

Yes it should be \(-1+12i\).

∑[ 1/(3^(2*n-1)) +2/(3^(2*n)), n, 1, ∞]==converges to 5/8

oops I made a typo on this when calculating, ignore. 

Floor{[100/5]+[100/25]+[200/5]+[200/25]+[200/125]+[300/5]+[300/25]+[300/125]+[400/5]+[400/25]+[400/125]}==246 zeros:

144000000 0000000000 0000000000 0000000000 0000000000 0000000000 0000000000 0000000000 0000000000 0000000000 0000000000 0000000000 0000000000 0000000000 0000000000 0000000000 0000000000 0000000000 0000000000 0000000000 0000000000 0000000000 0000000000 0000000000 0000000000==246 zeros.

Well, 48+5=53.
53+48=101
101-97=4
So 4 students take both.

Test of correctness: Spanish=44                                                44+49+4=97
                                Chinesse=49
                                 Both:4
 

can you explain to me how you got that answer? Though I do appreciate the attempt!

Thank you!

That'll be 2 ;)

THX !!!

Nice answer  and Welcome  !!!

B

                sqrt 14              

C       3             D

BC =  sqrt [ (sqrt 14)^2 - 3^2 ]   = sqrt  5

tan B  =   CD  /   BC  =     3 /  sqrt 5  =   3sqrt (5)  /  3

[(x - 1)(x - 2) (x - 3)]  / [  x ( x  -1) ( x -2) ]

Any x that makes the denominator =   0    is not in the domain

So      x = 0    x =  1  and  x  = 2   are not in the domain

 (We will have  "holes" at x = 1   and x = 2   and a vertical asymptote at x =  0  )

Note that if  we have the parabola

y= (1/2)(x -1)(x + 2)

The roots  are 1  and -2  and the y intercept = -1

So.....the area of this  triangle  =  (1 - - 2) /  (  2 * 1 )  =  1.5

If we let k  = 2

Then   (k + 1) ( 1*k)  =   (2 + 1)  (1(2))  / 2 =   twice the area of this triangle 

Also  if we let k =  -2   then  ( -2  - 1) ( 1*-2))/ 2  =  twice the area 

y = (1/2)(x - 3.5)(x + k)

This implies  that

(k + 3.5) (3.5k) / 2  =  3

k ≈ .436

k≈ -3.93559  

Also

( k - 3.5) (3.5k)/2 = 3

k ≈  - .435

k ≈ 3.93559

(100/5 + 100/25 + 200/5 + 200/25 + 200/125 +300/5 + 300/25 + 300/125 + 400/5 + 400/5 + 400/5)= 387 zeroes.

50/81 as a fraction. 

I hope this is correct. (This is my first question or answer on this site! I'm in 8th grade and got 17 on the AMC 8 and got 78th place in Virginia in state Mathcounts.)
There are 5!=120 possible combinations of those 5 digits. The smallest is 13458 and the largest is 85431. Simply add the two numbers and multiply the sum by 60. 60(13458+85431) = 5933340.

A/[XYZ] = 9/16.

1. A/[XYZ] = 45/96

2. Angle ETF = 96 degrees.

it says it's wrong... thanks for your time though

On average, 1 out of 10 times you will draw the Superball (and this wins all by itself so we don't care what the other balls are) for a winning probability of 0.10.

Now, for the 9 out of ten times that you do not draw the Superball:

We will have to get a losing blue ball, I'll call that B (and the probability of getting that losing ball is 0.90).

We can win by getting two or more of the white balls. I'll classify a winning white ball (whose probability is 1/10 = 010)

as W and a losing white ball as L (whose probability is 9/10 = 0.90).

So, we can still win if we have this:

B x W x W x W   =   (0.90) x (0.10) x (0.10) x (0.10)   =  0.0009

B x W x W x L    =   (0.90) x (0.10) x (0.90) x (0.10)   =  0.0081

B x W x L x W    =   ...

B x L x W x W    =   ...

Since these are independent, add the above 5 results together:  0.10 + 0.009 + 0.0081 + ... + ...  =  313/2500.

it is 12i-1

Re-write as \(a+b+8\)

By Vieta's, we know that \(a+b= -{b \over a} \). Plugging the values in, we find that \(a+b=-{1 \over 7}\)

We now have to add 8 to this total, and we find the final answer is \(\color{brown}\boxed{7 {6 \over 7}}\)

Hi Melody,

the questioner should have asked for "alphanumeric characters".

Rudi    !

There are \(50^2 = 2500\) cases

To solve this problem, we will count the ways of getting at least 1 multiple of 8. 

1 multiple of 8: 6 ways for the multiple of 8, 44 ways for the non-multiple, and 2 ways to order multiple and non-multiple, so \(44 \times 6 \times 2 = 528\)

2 multiples of 8: 6 ways for the first number, 6 ways for the second (we don't order because they are the same case), so \(6 \times 6 = 36\)

Thus, the total probability is \({{528+36} \over 2500 }= \color{brown}\boxed{0.2256}\)