<span>There are several possible events that lead to the eighth mouse tested being the second mouse poisoned. There must be only a single mouse poisoned before the eighth is tested, but this first poisoning could occur with the first, second, third, fourth, fifth, sixth, or seventh mouse. Thus there are seven events that describe the scenario we are concerned with. With each event, we want two particular mice to become diseased (1/6 chance) and the remaining six mice to remain undiseased (5/6 chance). Thus, for each of the seven events, the probability of this event occurring among all events is (1/6)^2(5/6)^6. Since there are seven of these events which are mutually exclusive, we sum the probabilities: our desired probability is 7(1/6)^2(5/6)^6 = (7*5^6)/(6^8).</span>
I think it's 12.50
Im not sure so sorry if im wrong but im 99% sure
Answer:
16^25
Step-by-step explanation:
The applicable relation is ...
(a^b)^c = a^(bc)
2^100 = 16^x = (2^4)^x = 2^(4x)
Equating exponents, we have ...
100 = 4x
25 = x
Then 2^100 = 16^25.
Answer:
Rectangle= lw (length times width)
Square= a²
Parallelogram=bh (base times height)
The point that the graphs of f and g have in common are (1,0)
<h3>How to get the points?</h3>
The given functions are:
f(x) = log₂x
and
g(x) = log₁₀x
We know that logarithm of 1 is always zero.
This means that irrespective of the base, the y-values of both functions will be equal to 0 at x=1
Therefore the point the graphs of f and g have in common is (1,0).
Learn more about graph on:
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