All Answers 352718 Answers

don't bump please, someone will get to it and bumping doesn't help.

How can a probability be 28 2/5 ?

0 <= Pr[anything] <= 1

\(P[\text{both same color}] = \\ P[\text{2 red}]+P[\text{2 blue}] = \\ \dfrac{5}{9}\dfrac{4}{8} + \dfrac{4}{9}\dfrac{3}{8} = \\ \dfrac{20 + 12}{72} = \\ \dfrac{4}{9}\)

#6) more diophantine equations

\(\text{Let the number of candies in the box = }N\\ N-3 \pmod{7}=0\\ N-6 \pmod{8} = 0\\ N+3 \pmod{5} = 0\)

\(N \pmod{7}=3,~N\pmod{8}=6\\ 7m+3 = 8n+6\\ m = \dfrac{8n+3}{7}\in \mathbb{Z}\\ 8n+3 \pmod{7}=0\\ n \pmod{7}=4\)

\(N=8(7k+4)+6 = 56k+38\\ 56k+38+3\pmod{5}=0\\ k + 1\pmod{5} = 0\\ k \pmod{5} = 4\\ N=56(5j+4)+38 = 280j+262\)

\(\text{we can't have negative candies so the smallest }N\\ \text{occurs when }j=0 \text{ i.e. when }N=262\)

#5)

\(N\pmod{9}=5\\ N\pmod{7}=3\\ N\pmod{5}=1\)

\(9m+5=7n+3\\ n=\dfrac{9m+2}{7} \in \mathbb{Z}\\ 9m+2 \pmod{7} = 0\\ 2m \pmod{7}=5\\ m \pmod{7} = 20 \pmod{7} = 6\\ m = 7k+6, ~k \in \mathbb{Z}\)

\(N=9(7k+6)+5 = 63k + 59\\ 63k+59 = 5j + 1\\ j = \dfrac{63k+58}{5}\in \mathbb{Z}\\ 63k+58 \pmod{5} = 0\\ 3k+3 \pmod{5} = 0\\ 3k \pmod{5} = 2\\ k \pmod{5} = 4\\ k = 5i + 4\)

\(N = 63(5i+4)+59\\ N = 315i + 311\\ i = 3 \text{ gets us the smallest 4 digit number}\\ N = 945 + 311 = 1256\)

Thanks!

\(\#3)\\ N = 3m + 2 = 4n+3\\ m = \dfrac{4n+1}{3} \in \mathbb{Z}\\ 4n+1 \pmod{3} = 0\\ n \pmod{3} = -1\\ n\pmod{3} = 2\\ n = 2+3k\\ N = 4n+3 = 4(2+3k)+3 = 12k+11\\ N \pmod{12} = 11\)

This is a slight variation on Rom's excellent answer above.

Put three coins on one side of the balance, and three on the other side.

One side will be heavier. 

Take two of the three from the heavier side and put them on the balance. 

If they balance, the odd coin you did not put on the balance is the fake.

If they don't balance, the heavier is the fake. 

Two weighings

thank you very much!

\(R=F_1+F_2 = (p-6)\hat{i} + (5-2p)\hat{j}\\ \text{This is parallel to the vector }\hat{i} - 3\hat{j}\\ \text{The easiest way to do this is just to equate the slopes}\\ \dfrac{5-2p}{p-6}=\dfrac{-3}{1} \\ 5-2p = 18-3p\\ p = 13\)

\(R = (7,-21) \text{ which is clearly parallel to }(1,-3)\)

The common difference is second term - first term =   - 9 - 6 =   - 15

So the third term is  -9 - 15  = -24

And the common ratio   is  third term / second term =   -24 /-9 =  8/3

So....the fourth term is   -24 (8/3) =  -64

So   x , y   =      -24, -64  

\(\text{in two dimensions the ratio of the x and y coordinates}\\ \text{of parallel vectors will be equal}\\ \dfrac{q}{12} = \dfrac{-1}{9}\\ 9q = -12\\ q = -\dfrac 4 3 \)

Requires sign in

OK!!!

80 km / hr   x   1 hr / 60 min  x  18 min   =   24 km

Thank you so much CPhill

Let's look at this again

The exterior angles sum to 360

The smallest exterior angle will be supplemental to the largest  interior angle

So....the smallestt exterior angle is  (2/9)(360) = 80°

So...the largest interior angle = 180 - 80   =  100°  = y

EDIT TO CORRECT AN ERROR

Link requires a sign in.....

180°  x  pi / 180    =    pi  rads

Minor arc AB   =  360 - 286  =  74°

The minute hand moves 2pi/60 =  pi/30 rads  every minute  = 6° every minute

The hour hand moves pi/6 rads every hour so it moves pi/360 rads every miniute = .5° every  minute

So...in 18 minutes....

The minute hand moves  18* 6 = 108°

And the hour hand moves   .5 * 18  =  9°

So....

The minute hand is    108 - ( 9 + 20)  =  108 - 29 = 79°  ahead of the hour hand after 18 mnutes

The ratio of the circumference of a circle to its diameter is same for every circle.

To see why this is true.....note that the circumference = 2pi^r = (2r)*pi  = (diameter)* pi

So

circumference / diameter  =   (diameter) * pi  / diameter  =   pi    (which is a constant)

Dilate circle Q by a scale factor of p/q.

To see this.....let the radius of P = 6   and the radius of Q = 3

Then.....dilating Q  we  get the new radius of   3 * 6/3  =  6 = the same radius as P

Dilation with center B and scale factor 3 on circle A 

Dilate Circle B about B by a scale factor of 1/2 followed by a translation of 8 right and 2 down.

5x^{2}+19x+16 > 20    subtract  20 from both sides

5x^2 + 19x - 4 > 0        (1)   

Let's solve this.....and then we can find the interval that works

5x^2 + 19x  - 4 = 0      factor

(5x  - 1) ( x + 4)   =  0   

Setting each factor to 0 and solving for x we get that

x = 1/5      or      x  = 4

We have three possible intervals that might solve (1)

(-infinity, -4)   or (-4, 1/5)   or ( 1/5, infinity)

Usually, either the middle interval works or  the two outside intervals work

If we pick a point in the middle interval - I'll choose 0 -    note that this value makes (1) false

So....the intervals that solve this  are

(-infinity, 4)   U  ( 1/5 , infinity)