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Vesna [10]
3 years ago
9

You have ​$23 comma 000 to​ invest, part at 8 % and the rest at 9 %. If x is the amount invested at 8 %​, write an algebraic exp

ression that represents the total annual income from both investments. Simplify the expression.
Mathematics
1 answer:
Vinvika [58]3 years ago
3 0

<em>x</em> = amount invested at 8 %. Then 23 000 - <em>x</em> = amount invested at 9 %

Annual income (<em>I</em>) = interest at 8 % + interest at 9 %

<em>I</em> = 0.08<em>x</em> + 0.09(23 000 – <em>x</em>)

<em>I</em> = 0.08<em>x</em> + 2070 – 0.09<em>x</em>

<em>I</em> = 2070 – 0.01<em>x</em>

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HACTEHA [7]

In this question, the Poisson distribution is used.

Poisson distribution:

In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by the following formula:

P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}

In which

x is the number of sucesses

e = 2.71828 is the Euler number

\mu is the mean in the given interval.

Parameter of 5.2 per square yard:

This means that \mu = 5.2r, in which r is the radius.

How large should the radius R of a circular sampling region be taken so that the probability of finding at least one in the region equals 0.99?

We want:

P(X \geq 1) = 1 - P(X = 0) = 0.99

Thus:

P(X = 0) = 1 - 0.99 = 0.01

We have that:

P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}

P(X = 0) = \frac{e^{-5.2r}*(5.2r)^{0}}{(0)!} = e^{-5.2r}

Then

e^{-5.2r} = 0.01

\ln{e^{-5.2r}} = \ln{0.01}

-5.2r = \ln{0.01}

r = -\frac{\ln{0.01}}{5.2}

r = 0.89

Thus, the radius should be of at least 0.89.

Another example of a Poisson distribution is found at brainly.com/question/24098004

3 0
3 years ago
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Answer:

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