Hi Johnpeter,
Welcome to the forum :)
Questions like this can be very confusing.
The question is
What percent of 24 is 80?
You have answered a different question
i.e.
What percent of 80 is 24?
Hectictar's answer is correct.
I have an unlimited supply of standard 6-sided dice. What's the fewest number of dice that I have to simultaneously roll to be at least 95% likely to roll at least one 6?
\(1-\left(\frac{5}{6}\right)^n>0.95\\ 0.05>\left(\frac{5}{6}\right)^n\\ log(0.05)>log\left(\frac{5}{6}\right)^n\\ log(0.05)>nlog\left(\frac{5}{6}\right)\\ nlog\left(\frac{5}{6}\right) \frac{log(0.05)}{log\left(\frac{5}{6}\right)}\\ n>16.43\\~\\ smallest\;\;n=17\)
The points B, C, D and E all lie on the same line segment, in that order, such that the ratio of BC:CD:DEBC:CD:DE is equal to 2:4:5.2:4:5. If BE=44,BE=44, find CD.
our question does not make sense.
I expect you should proof read it.
A die is rolled 5 times. What is the probability that a number greater than 5 is rolled at least once? Write your answer rounded to 3 decimal places.
P(at least one 5 is rolled)
1 - P(no 5 is rolled)
1- (5/6)^5
HOW SOLVE 3POWER ROOT 2
\(3^{\sqrt{2}}\quad?\)
There is nothing to solve.
If a number is divisible by 15 then it must be divisible by both 3 and 5.
If a number is divisible by 5 what does it have to end in?
\(ax^2+bx+c=2(x-4)^2+8\\~\\ 3(ax^2+bx+c)=3[2(x-4)^2+8]\\ 3(ax^2+bx+c)=6(x- {\color{red} 4})^2+24\\ \)
SO
h=4
LaTex:
ax^2+bx+c=2(x-4)^2+8\\~\\ 3(ax^2+bx+c)=3[2(x-4)^2+8]\\ 3(ax^2+bx+c)=6(x -{\color{red} 4})^2+24\\
I like this answer guest.
Heureka has elaborated on it below.... for anyone who doesn't already understand.
It is very nice to see you back again Heureka
Use the calculator
3 divided by 8.8 =
+118725 
