EvanWei123

Hi guest, think of the question as 4 parts.

First part is heads. The probability of that occurance is 1/2 because it is either heads or tails.

Second part is tails. The probability of that occurance is also 1/2 because it is either heads or tails.

Third part is heads. The probability is 1/2 same reasoning of the First and Second part

Fourth part is tails. The probability is 1/2 same reasoning.

Multiply them together, 

Your welcome, Evan

Hi guest, think of the question as 4 parts.

First part is heads. The probability of that occurance is 1/2 because it is either heads or tails.

Second part is tails. The probability of that occurance is also 1/2 because it is either heads or tails.

Third part is heads. The probability is 1/2 same reasoning of the First and Second part

Fourth part is tails. The probability is 1/2 same reasoning.

Multiply them together, \(\frac{1}{2}*\frac{1}{2}*\frac{1}{2}*\frac{1}{2}=\frac{1}{2^4}=\boxed{\frac{1}{16}}\)

Your welcome, Evan

Thanks rom :)

Hectictar thank you so much! Sorry if i wasted your time. Your solution is clear, smart, and correct!

Thanks travisio the graph helps other problem solvers!

Thank you melody :D

I'm how to do this but this is my best shot. So first, you can think of it as a quadratic. So simplifying the expression you get \(x^2-kx-3x-k+6>0\)  

Using the quadratic formula, you get \(\frac{-(-k+3) \sqrt {(-k-3)^2-4(6-k)}}{2} > 0\)

simplify, \(\frac{k-3\sqrt{k^2+6x+9-24+4k}}{2} = \frac{k-3\sqrt{k^2+10x-13}}{2} > 0\)

I hope that helped, i just kind of bashed it out, now i hope you can solve it. I dont really know how to myself. Sorry if this post was no help at all :( 

-Evan 

This is not a off topic post, i just accidently clicked it.

Thanks CPhill!

You know that \(\sqrt{16}=4\) and \(\sqrt{100}=10\) so you need to find how many numbers in 17,18,19....99

My way of doing this is just making the first number 1 and the last number is the number of numbers in the set. So you need to subtract 16 from all the numbers to make the first one 1. The set becomes 1,2,3....83. Therefore there is 83 positive integers that satisfy the inequality.