amygdaleon305

Really? What do you think we're bots designed to solve your homework or something? Posting too many questions is unacceptable here clearly it isn't "doubt-clearing" as we can see. 

I'm assuming repetition of digits are allowed. 

No. of 1-digit no.'s = 1 (6)

No. of 2-digit no.'s = 3 (24,42 & 66) 

No. of 3-digit no.'s = 9 (222,246,264,444,426,462,666,624 & 642) 

So total no. of multiples = 1+3+9

                                      = 13

Okay from the question, 

5 members can paint wall in = 27 min

∴ By inverse proportion, 

18 members can paint wall in \(={27 *5 \over 18}\)

                                                = 7 min and 30 sec

Oh yeah! You're right thanks. 

Yes its absolutely right, apart from the fact that I got the mean as 4.8 and when I solved it with this value I got the variance as 2.46. But I guess that minute difference can be ignored right? 

f(x) = 7x + 5

and g(x) = x + 1

⇒ \(h(x)=f(g(x))\)

              \(=f(x+1)\)

              \(=7x+7+5\)

              \(=7x+12\)

In order to find  \(h^{-1}(x)\) 

Let  \(y=h(x)\)

       \(y=7x+12\)

       \(x= {y-12\over 7}\)

⇒ \(h^{-1}(x) = {x-12 \over 7}\)

   Thus, inverse of h(x) is \({x-12 \over 7}\)

~Amy 

Now in this problem, 

Let the no. of  10 dollar bills be x, 5 dollar bills be y, 2 dollar bills be z and 1 dollar bills be w.

According to question, 

\(10x+5y+2z+w=20\)                    \(...(1)\)

In order to find no. of ways, we've to find no. of solutions eq. (1) has.

∴ By the theory of combinatrics, 

No. of solutions \(=\left( \begin{array}{c} 20+4-1 \\ 4-1 \end{array} \right)\)

                          \(=\left( \begin{array}{c} 23 \\ 3 \end{array} \right)\)

                          \(=1771\)

∴ Eq. (1) has 1771 possible set of solutions.

Thus a $20 bill can be changed in 1771  possible ways. 

~Amy 

According to the question, 

All the integers between 500 & 1000 are 3 digit no.s greater than 500. 

∴ First place may have digits 5,6,7,8 & 9 = 5 possibilities

Second & third place may have digits 3 or 4 = 2 possibilities 

⇒Total no. of possibilities \(=5*2\)

                                         \(=10\)

Thus, total no. of such integers are 10. 

~Amy 

Let's consider this graph 

So as you can see the seismographs are placed at \(A(-2,5), B(2,3)\) and  \(C(-2,-4)\).

From the question, 

Epicentre is located 2 miles from A, 4 miles from B & 7 miles from C

∴ Eliminating all possibilities, epicentre is located at \(E(-2,3)\).

~Amy