Take the equation and distribute the 16 on the right:
64x - 16 = 16(4x - 1)
64x - 16 = 64x - 16
Since what is on both sides is identical, you'll always get a true statement no matter what value you pick for x.
This equation has infinitely many solutions.
This problem lends itself to the binomial probability approach.
Focus on students who are not secretly robots.
Then P(student is not a robot) = 2/6, or 1/3. Here n=6 and x=2.
Then the desired probability is binompdf(6,1/3, 2), which, by calculator, comes out to 0.329.
Do the brackets first as it says in BIDMAS
<span>Answer:
The multiplication factor of increase should be inverse of the multiplication factor of decrease.
e.g. Say you have a number 100.
You increase it by 25%. The multiplication factor is 5/4 i.e. when you multiply 100 by 5/4, you get 100*5/4 = 125. This is 25% more than 100.
Now you want to decrease it by a certain % such that you get 100 back.
Basically, 100*5/4 * x = 100
So x = 4/5 (inverse of 5/4)
Hence, you decrease by 20% (the multiplication factor of 20% is 4/5)
or
Use this formula: cumulative % change = a + b + ab/100
You want the cumulative change to be 0.
a + b + ab/100 = 0
If you know that you are increasing by 25% and want to find the % by which you should decrease to get the same number,
25 + b + 25b/100 = 0
5b/4 = -25
b = -20
So you need to decrease (hence you get the -ve sign) by 20%.</span>
<h3>
Answer: They're all the same</h3>
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Reason:
means we have 6 copies of "2" multiplied out as shown in choice B. That explains how A and B are the same, and we can say

The parenthesis are optional, but I find they're handy to count the '2's easier.
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Now notice that

So,

The last step is possible because we have two copies of
multiplied together.
This shows that choice C is equivalent to A and B.
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Lastly,

The jump to the last step is possible because we have three copies of
multiplied together.
This shows choice D is equivalent to the others.
All four expressions are the same.
They represent different ways to say the same number. That number being 64.