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Cold Play, that's an excellent understanding of an ambiguously-stated problem.  Good work.  I interpreted it another way and my answer was not among the choices, so I abandoned the problem.  What I read was "Each subsequent year, he earned an additional $4000 more."  

_?

omg hecticar the 1+6 2+5 3+4 etc explanation  

nobody ever showed it to me that way before

it's so obvious now ... i'm so happy i'm crying

The answer is (B) 1/7.

Find the asymptotes of the hyperbola x^2 - 2x - 4y^2 - 16y - 19 = 0.

The answer is (B) 38%.

1) The answer is (D) 1/6.

2) The answer is (A) 1/2.

(a + b)/c = 2/3.

a^5 divided by a^3         

equals  a²  

Remember that a⁵ means  a • a • a • a • a 

                  and a³ means  a • a • a    

So when you divide a⁵ by a³ ... you simply cancel 3 of the a's ... leaving a • a ... which of course is a²    

_.

A way to solve this, is to find a link between number 15 and 6

List the multiples of 15: (15,30,45,60...)

List the multiples of 6: (6,12,18,24,30...)

Notice 30 in both 

Well if she goes swimming every 6 days and hiking every 15 days

Swimming every 6 days so after 30 days she will go swimming with hiking as she go hiking each 15 days, the next 15 days which will be january 23th she will go only hiking since she will go swimming on 14th jan. and 20 jan, 26 jan. 1 feb. , 7th feb. 

they don't meet there but notice the next swimming will be january 23th + 15 days = 7th feb 

so on 7th february she will go both swimming and hiking 

(Notice it took a while , a faster way would be notice the multiplies between 6 and 15, in 15 to reach 30 we took 30 days, in 6 to reach 30 we took 30 days as well) so we just need to add 30 days + Current date which is january 8th (Assuming january ends at 31th) it will be february 7th! 

\(\frac{a^5}{a^3}\)=\(a^2\)

same base, we subtract the power so, 

5-3=2 

a + b + c = 5.     (1)

Divide (1) by b + c:    a/(b+c) + 1 = 5/(b+c).  (2)

Divide (1) by a + c:    b/(a+c) + 1 = 5/(a+c).   (3)

Divide (1) by a + b:    c/(a+b) + 1 = 5/(a+b).   (4)

Add (2), (3) and (4): a/(b+c) + b/(a+c) + c/(a+b) + 3 = 5*6

Hence  a/(b+c) + b/(a+c) + c/(a+b) = 27

I agree!.

The vectors are [3,-3] and [1,1].

The matrix is \(\begin{pmatrix} 1/9 & -2/9 \\ 2/9 & 4/9 \end{pmatrix}\).

The vectors are [2,-3], [0,1], [5,-2], [3,-1].

(c/a,e/a) = (-3,7/2).

To make this easier, we can find the probability of the numbers being less than or equal to 10, and then subtract that from 1.

Combinations:

We can find that all 1's work.

1 x 1, 1 x 2, 1 x 3, 1 x 4, ... , 1 x 10

Same with the other way

1 x 1, 2 x 1, 3 x 1, 4 x 1, ... , 10 x 1

Now we subtract 1 because we counted 1 x 1 twice, to get a total of 19.

Now for 2's:

2 x 2, 2 x 3, 2 x 4, 2 x 5

2 x 2, 3 x 2, 4 x 2, 5 x 2

Subtract 1 and get 7.

Now 3's:

3 x 3

3 x 3

Subtract 1 and get 1.

4 or higher won't work because it's greater than 10, and we already counted values like 4 x 1 and 4 x 2.

19+7+1 = 27

There are a total of 100 possibilities

So 1-27/100 = 73/100.

You are very welcome!

:P

Hi Guest,

You have assumed that the two must choose different digits.  I do not think that is the case.

To answer I drew a 10 by 10 grid. Like a times table grid.

There are 100 pairs of numbers in the grid. That is the sample space.

Then I ticked of the pairs that met the condition. I found 73 of them

So the answer I go was   73/100

All answers hon this site shold be verified by the asker.

A bag contains five white balls and five black balls. Your goal is to draw two black balls.
You draw two balls at random. Once you have drawn two balls, you put back any white balls, and redraw so that you again have two drawn balls. What is the probability that you now have two black balls? (Include the probability that you chose two black balls on the first draw.)

5 white, 5 black

\(P(BB)=\frac{5*4}{10*9}=\frac{20}{90}=\frac{2}{9}\\ P(BWB)=\frac{5*5*4}{10*9*9}=\frac{100}{810}=\frac{10}{81}\\ P(WBB)=\frac{10}{81}\\ P(WWBB)=\frac{5*4*5*5}{10*9*10*9}=\frac{500}{8100}=\frac{5}{81}\)

2/9+20/81+5/81 = 0.5308641975308642 

You need to check my working.

You have used the wrong sample space.