A is the answer i think
if not then c
By "density" I assume you mean "probability density function". For this to be the case for
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, we require

Since
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you have

which means
The answer:
by definition, an exponential function with base c is defined by <span>h (x) = ac^x</span><span>
where a ≠0, c > 0 , b ≠1, and x is any real number.</span>
The base, c, is a constant and the exponent, x<span>, is a variable.
</span>so if we have f(x)=3(3\8)^2x, this equivalent to f(x)=3(3\8)^y(x),
where y (x)=2x, <span>
therefore, the base is 3/8, and the variable is the function </span>y (x)=2x,
Answer:
answer is A
Step-by-step explanation: