Melody

Hello Mr Patel :)

1. There are five $5 and three $10 bills in a bag. A person reaches in and pulls out two bills. What is the probability of the sum of the bills will be $10, $15 or $20?

5,5,5,5,5,  10,10,10           8 notes altogether.

$10=$5+$5

\(P(5,5)=  \frac{5}{\not{8}^2}\times \frac{\not{4}^1}{7}=\frac{5}{14}\)

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$15    =   $5+$10    or     $10+$5

\(P(5,10)=  \frac{5}{8}\times \frac{3}{7}=\frac{15}{56}\\ P(10,5)=  \frac{3}{8}\times \frac{5}{7}=\frac{15}{56}\\ \text{P(sum of 15)}=\frac{30}{56}=\frac{15}{28} \)

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You can do the $20 yourself I think :)

150 = 10 times log x/10⁻¹²

I will assume that you mean

150 = 10 times log[ x/10⁻¹²]

\(150=10\times log(\frac{x}{10^{-12}})\\ 150=10\times log(x*10^{12})\\ 150=10\times[ log(x)+log(10^{12})]\\ 150=10\times[ log(x)+12log(10)]\\ 15=log(x)+12*1\\ 3=log(x)\\ 10^3=10^{log(x)}\\ 10^{log(x)}=10^3\\ x=1000 \)

or maybe you mean:

\(150 = 10 \times \frac{log (x)   }{   10^{-12}}\\ 15 = \frac{log (x)   }{   10^{-12}}\\ 15\times 10^{12} = log (x)  \\ x=10^{15\times 10^{12} }\\ x=10^{1.5\times 10^{13} }\\~\\ \text{WOW that is a big number !}\)

in an obstacle course race how many ways can five finalists be ordered?

5 choices for 1st, 4 choices for second etc

5*4*3*2*1=5!=120 ways

potatos  :)

Appologies all around :)

Yes I see :)

Chris I do not think your answer is correct either becasue Wolfram Alpha says the answer is approx 0.0023

Integrate: ∫x^4[1 - x]^4 / [1 + x^2]dx,from x=0 to 1

\(\displaystyle\int_0^1\;\frac{x^4(1-x)^4}{1-x^2}\;dx\\ =\displaystyle\int_0^1\;\frac{x^4(1-x)^4}{(1-x)(1+x)}\;dx\\ =-\displaystyle\int_0^1\;\frac{x^4(x-1)^3}{x+1}\;dx\\ =-\displaystyle\int_0^1\;\frac{x^4(x^3-3x^2+3x-1)}{x+1}\;dx\\ =-\displaystyle\int_0^1\;\frac{x^7-3x^6+3x^5-x^4}{x+1}\;dx\\ \text{Do the algebraic division}\\ =-\displaystyle\int_0^1\;x^6-4x^5+7x^4-6x^3+6x^2-6x+6-\frac{6}{x+1}\;dx\\ \)

\( =-\displaystyle\int_0^1\;x^6-4x^5+7x^4-6x^3+6x^2-6x+6-\frac{6}{x+1}\;dx\\ =-\left[\; \frac{x^7}{7}-\frac{4x^6}{6}+\frac{7x^5}{5}-\frac{6x^4}{4}+\frac{6x^3}{3}-\frac{6x^2}{2}+6x-6ln(x+1)\;\right]_0^1\\ =-\left[\; \frac{x^7}{7}-\frac{2x^6}{3}+\frac{7x^5}{5}-\frac{3x^4}{2}+2x^3-3x^2+6x-6ln(x+1)\;\right]_0^1\\ =-\left[\; \frac{1}{7}-\frac{2}{3}+\frac{7}{5}-\frac{3}{2}+2-3+6-6ln(2)\;\right]\\ =-\left[\; \frac{1}{7}-\frac{2}{3}+\frac{7}{5}-\frac{3}{2}+5-6ln(2)\;\right]\)

1/7-2/3+7/5-3/2+5 = 4.3761904761904762 = 919/210

\(=6ln2-\frac{919}{210}\)

-4.3761904761904762+6*log(2) = -2.570010502206589

This is not correct, I have made some silly mistake somewhere, but the method is correct :)

Wolfram Alpha says the answer is approx 0.0023

Thanks Heureka and Alan :))

x to the second power - 5.3x = -8.6

\(x^2-5.3x=-8.6\\ x^2-5.3x+8.6=0\\ \triangle=5.3^2-4*8.6=-6.31\\ \triangle <0\\\therefore \text{No real roots}\)