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3 years ago

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In the month of October, the population of a species of fish in a river increased by a factor of 1.05 every day. The function be

low shows the number of fish in the river f(x) after x days:
f(x) = 100(1.05)x

Which of the following is a reasonable domain for the function?

1 answer:

Solnce55 [7]3 years ago

7 0

I don’t know but I hope you have a good day

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Evaluate the piecewise at x=-4 and x=4

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nikitadnepr [17]

Answer:

The distribution is $\frac{\lambda^{n}e^{- \lambda t}t^{n - 1}}{(n - 1)!}$

Solution:

As per the question:

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A matrix is diagonal if

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4 years ago

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Answer:

The value of P (A ∩ B) is 0.25.

Step-by-step explanation:

Independent events are those events that do not effect the occurrence of each other, i.e. if event <em>X </em>and <em>Y</em> are independent then the occurrence of <em>X</em> and <em>Y </em>are not influenced by each other.

For independent events <em>X</em> and <em>Y</em> the joint probability of <em>X</em> and <em>Y</em> is:

$P (X\cap Y)=P(X)\times P(Y)$

It is provided that events <em>A</em> and <em>B </em>are independent of each other.

And P (A) = P (B) = 0.50.

Compute the value of P (A ∩ B) as follows:

$P (A\cap B)=P(A)\times P(B) \\= 0.50\times0.50\\=0.25$

Thus, the value of P (A ∩ B) is 0.25.

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3 years ago