XxmathguyxX

ok.

\(3\times14=42\)

so....

there you go!

instead of writing all the 54's(which is in fact very painful!) since 54 is even and 100 is even, 4 must be the 100th digit

The probability is 4*C(13,3)/C(52,3) = 22/425.

guest is correct, maybe it will chage later?

1. 64000

2.78000

3. 55 hours

so according to the Pythaogorean Thereom,

\(a^2+b^2=c^2\)

in this case,

\(7^2+{PM}^2=18^2\)

so \(49+PM^2=324\)

\(PM^2=275\)

\(PM=5\sqrt11\)

and geno was literally 8 seconds before me!

its basically 2 iscoceles triangles if think about it, so if bac=76, then abd=180=152=28 

bdc=180-76=104 

dbc=180-104=76 76/2=38

28+38=66

WAITWAITWAIT

In this situation it is much easier to calculate the probability of the event we are looking for (he hits at least one target) by calculating the probability of its complement (he misses every target), and subtracting from 1.

In other words, we can use this strategy:

P(at least one hit)=1-P(miss all 10)

Calculations:

P(at least one hit)

=1-P(miss all 10)

=1-(0.9)^10

≈1-0.349

≈0.65

Answer:

P(at least one hit)≈0.65

oops....

geno did it 2 minutes before me lel :)