To obtain the probability of obtaining heads at least 3 times out of 5 times we recall that in her simulation <span>she
assigned odd digits to represent heads and even digits to represent
tails. Thus, we count the simulated numbers to see in how many numbers
do we have 3 or more odd digits.
Below, the simulated numbers with 3 or more odd digits are bolded.
</span> <span>32766 53855 34591 27732
47406 31022 25144 72662
03087 35521 26658 81704
56212 72345 44019 <span>65311
</span>
We have 6 simulated numbers having 3 or more odd digits.
Therefore, </span><span>P("heads" at least 3 out of 5 times) = 6 / 16 = 3 / 8.</span>
I'm pretty sure the pattern is ÷-2
We are given the following function:
f(x) = 4/(x+2) - 2
We are asked to determine the inverse of this function. To do that we will first change the "f(x)" for "y":
Now we will switch "x" and "y", like this:
Now we will solve for "y", first by adding 2 to both sides:
Now we multiply both sides by "y+2":
Now we divide both sides by "x+2":
Now we subtract 2 to both sides:
Now we change "y" for the inverse of f(x), that is:
And thus we found the inverse. A similar procedure can be used for function 2.
Each person would get 17 blocks and there will be two left over.If you have a number greater than 30 and you want to divide it by 3 you use the distributive property. 30\3 is 10 plus 23 divided by 3 is 7r2.10 plus 7r2 is 17r2...
