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See https://web2.0calc.com/questions/geometry_63869

Nice try Mmm but you forgot some of the posibilities   

The Boomtown Bears are playing against the Tipton Toros in a baseball tournament. The winner of the tournament is the first team that wins three games. The Bears have a probability of 1/5 of winning each game. Find the probability that the Bears win the tournament.

I will assume that it is imposible to draw.

P(B wins) = 1/5

P(B wines 3 games out of 3) =    1/   5^3                                                      =  25 / 5^5

P(B wins 3 games out of 4) =  4C3 * (1/5)^3 * (4/5) = 4 * 4 / 5^4 = 16 / 5^4  =  80 / 5^5

P ( B wins 3 games out of 5)  =  5C3 * (1/5)^3 *( 4/5)^2 = 10 *  16 /  5^5      =   160 / 5^5

add them together    25+80+160 = 265

265 /  5^5  simplifies to   53/625  =  0.0848

So the prob that the bears win is  8.48%

Expand   (2y - 5 + y^2)^6:

y^12 + 12 y^11 + 30 y^10 - 140 y^9 - 585 y^8 + 792 y^7 + 4164 y^6 - 3960 y^5 - 14625 y^4 + 17500 y^3 + 18750 y^2 - 37500 y + 15625
 

LCM(30, 70) ==210 hours

210 hours / 30 hours ==7 poems - he can write

7  x  $100 ==$700 - he can earn from 7 poems

840 hours  -  210 hours ==630 hours left

630 hours /70 hours ==9 short stories he can write.

9  x  $250 ==$2,250 he can earn from 9 short dtores.

So then, he should write: 7 poems + 9 short stories ==16

His total earning from the 16 works: $700  +  $2,250 ==$2,950

(2, 3, 5, 6, 7, 10, 11, 12, 13, 14, 15, 17, 18, 19, 20, 21, 22, 23, 24, 26, 28, 29, 30, 31, 33, 34, 35, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 50, 51, 52, 53, 54, 55, 56, 57, 58, 59, 60, 61, 62, 63, 65, 66, 67, 68, 69, 70, 71, 72, 73, 74, 75, 76, 77, 78, 79, 80, 82, 83, 84, 85, 86, 87, 88, 89, 90, 91, 92, 93, 94, 95, 96, 97, 98, 99, 101, 102, 103, 104, 105, 106, 107, 108, 109, 110, 111, 112, 113)

113 is the 100th term on list

43=23+20

The answer is 15%.

There are \({8 \choose 4} = 70\) ways to choose dogs for the group of 5. 

There are \({4 \choose 3} = 4\) ways to choose the dogs for the group of 3. 

Then there is \({1 \choose 1} = 1\) ways to choose the dogs for the group of 1. 

So, there are \(70 \times 4 \times 1 = \color{brown}\boxed{280}\)

I entered \color{brown}\boxed{20\text{%}}

To box it, just do \boxed{}

How did you box your answer?

\({14 \over 70} = {2 \over 10} = \color{brown}\boxed{20\text{%}}\)

Percent is in the form of:

$$x/100$$
 

So we try to get:

$$14/70 = x/100$$

We simplify the right side fraction:

$$2/10 = x/100$$
 

We multiply the right side by 10 on the numerator and denominator:

$$20/100 = x / 100$$
 

So x = 20

So your answer is 20%

~Hmmmm

The answer is 4.

10 - 99 : 11
100 - 999 : 112
1000 - 9999 : 1124
10000 - 99999 : 11248
100000 - 999999 : 112496
1000000 - 9999999 : 1124992
10000000 - 99999999 : 11249985
100000000 - 999999999 : 112499976
1000000000 - 9999999999 : 1124999972

any one have a correct answer

The answer is 36*pi.

This is the case of 4 DISTINCT balls into 3 IDENTICAL boxes with no restrictions.

sumfor(i, 0, 3,(4 nCr i))==[1 + 4 + 6 + 4] ==15 ways