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3 years ago

10

12 ducks fly overhead. Each of 6 hunters picks one duck at random to aim at and kills it with probability 0.6. What's the expect

ed number of hunters who hit the duck they aim at?

1 answer:

nexus9112 [7]3 years ago

6 0

Answer:

The expected number of hunters who hit the duck they aim at is 3.6

Step-by-step explanation:

Given;

number of hunters, n = 6

the probability of killing a duck, p = 0.6

The expected number of hunters who hit the duck they aim at?

In binomial distribution, the expected value is equal to the product of the number of trials and the probability of success.

The expected number of hunters who hit the duck they aim at is calculated as follows;

E = np

E = 0.6 x 6

E = 3.6

Therefore, the expected number of hunters who hit the duck they aim at is 3.6

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The mass of radioactive material remaining after 50 years would be 48.79 kilograms

<h3>How to determine the amount</h3>

It is important to note that half - life is the time it takes for the amount of a substance to reduce by half its original size.

Given the radioactive decay formula as

m(t)=120e−0.018t

Where

t= 50 years

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Substitute the value of t

$m(t) = 120e^-^0^.^0^1^8^(^5^0^)$

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m(t) = 48.788399

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Thus, the mass of radioactive material remaining after 50 years would be 48.79 kilograms

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2 years ago

Determine formula of the nth term 2, 6, 12 20 30,42

nalin [4]

Check the forward differences of the sequence.

If $\{a_n\} = \{2,6,12,20,30,42,\ldots\}$ , then let $\{b_n\}$ be the sequence of first-order differences of $\{a_n\}$ . That is, for n ≥ 1,

$b_n = a_{n+1} - a_n$

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and we can solve for $b_n$ in terms of $b_1=4$ :

$b_{n+1} = b_n + 2$

$b_{n+1} = (b_{n-1}+2) + 2 = b_{n-1} + 2\times2$

$b_{n+1} = (b_{n-2}+2) + 2\times2 = b_{n-2} + 3\times2$

and so on down to

$b_{n+1} = b_1 + 2n \implies b_{n+1} = 2n + 4 \implies b_n = 2(n-1)+4 = 2(n + 1)$

We solve for $a_n$ in the same way.

$2(n+1) = a_{n+1} - a_n \implies a_{n+1} = a_n + 2(n + 1)$

Then

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$a_{n+1} = (a_{n-2} + 2(n-1)) + 2((n+1)+n) \\ ~~~~~~~ = a_{n-2} + 2 ((n+1) + n + (n-1))$

$a_{n+1} = (a_{n-3} + 2(n-2)) + 2((n+1)+n+(n-1)) \\ ~~~~~~~= a_{n-3} + 2 ((n+1) + n + (n-1) + (n-2))$

and so on down to

$a_{n+1} = a_1 + 2 \displaystyle \sum_{k=2}^{n+1} k = 2 + 2 \times \frac{n(n+3)}2$

$\implies a_{n+1} = n^2 + 3n + 2 \implies \boxed{a_n = n^2 + n}$

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2 years ago

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Answer:

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Step-by-step explanation:

took the test hope it helps =\/

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2 years ago

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Answer:

Think about it

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