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Aliun [14]
3 years ago
15

Leo invests $2,000 loses money at a rate of 4%. How much is the investment worth in total at the end of 3 years? Round to the ne

arest cent.
Mathematics
1 answer:
barxatty [35]3 years ago
5 0

Answer:

<em>Leo's investment is worth $1,760 at the end of 3 years</em>

Step-by-step explanation:

<em>Simple Interest</em>

Leo invests $2,000 and loses money at a rate of 4% (yearly),

It's required to find the value of his money at the end of 3 years.

Since Leo losses 4% per year, the amount lost per year is:

$2,000*4/100=$80

For three years, he loses 3*$80 = $240

Thus, his money is worth:

$2,000 - $240 = $1,760

Leo's investment is worth $1,760 at the end of 3 years

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(b) The probability of the event (<em>X</em> < 64) is 0.483.

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The random variable <em>X</em> follows a Poisson distribution with parameter <em>λ</em> = 64.

The probability mass function of a Poisson distribution is:

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(a)

Compute the probability of the event (<em>X</em> > 84) as follows:

P (X > 84) = 1 - P (X ≤ 84)

                =1-\sum _{x=0}^{x=84}\frac{e^{-64}(64)^{x}}{x!}\\=1-[e^{-64}\sum _{x=0}^{x=84}\frac{(64)^{x}}{x!}]\\=1-[e^{-64}[\frac{(64)^{0}}{0!}+\frac{(64)^{1}}{1!}+\frac{(64)^{2}}{2!}+...+\frac{(64)^{84}}{84!}]]\\=1-0.99308\\=0.00692\\\approx0.007

Thus, the probability of the event (<em>X</em> > 84) is 0.007.

(b)

Compute the probability of the event (<em>X</em> < 64) as follows:

P (X < 64) = P (X = 0) + P (X = 1) + P (X = 2) + ... + P (X = 63)

                =\sum _{x=0}^{x=63}\frac{e^{-64}(64)^{x}}{x!}\\=e^{-64}\sum _{x=0}^{x=63}\frac{(64)^{x}}{x!}\\=e^{-64}[\frac{(64)^{0}}{0!}+\frac{(64)^{1}}{1!}+\frac{(64)^{2}}{2!}+...+\frac{(64)^{63}}{63!}]\\=0.48338\\\approx0.483

Thus, the probability of the event (<em>X</em> < 64) is 0.483.

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