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larisa [96]
3 years ago
8

Mguel sells nsurence. He earns a 6% commision on each sell he makes. Miguel has a goal of earning $600 in commisions. Miguel nee

ds to sell what Worth of nsurence to meet his goal.
Mathematics
1 answer:
dexar [7]3 years ago
5 0

Answer:

$10,000= sales

Step-by-step explanation:

Giving the following information:

Commission rate= 6%

Desired total earnings= $600

<u>To calculate the sales level required to reach the objective, we need to use the following formula:</u>

Desired earning= commission rate*sales

600= 0.06*sales

600/0.06= sales

$10,000= sales

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In 1898 L. J. Bortkiewicz published a book entitled The Law of Small Numbers. He used data collected over 20 years to show that
attashe74 [19]

Answer:

(a) The probability of more than one death in a corps in a year is 0.1252.

(b) The probability of no deaths in a corps over 7 years is 0.0130.

Step-by-step explanation:

Let <em>X</em> = number of soldiers killed by horse kicks in 1 year.

The random variable X\sim Poisson(\lambda = 0.62).

The probability function of a Poisson distribution is:

P(X=x)=\frac{e^{-\lambda}\lambda^{x}}{x!};\ x=0,1,2,...

(a)

Compute the probability of more than one death in a corps in a year as follows:

P (X > 1) = 1 - P (X ≤ 1)

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

             =1-\frac{e^{-0.62}(0.62)^{0}}{0!}-\frac{e^{-0.62}(0.62)^{1}}{1!}\\=1-0.54335-0.33144\\=0.12521\\\approx0.1252

Thus, the probability of more than one death in a corps in a year is 0.1252.

(b)

The average deaths over 7 year period is: \lambda=7\times0.62=4.34

Compute the probability of no deaths in a corps over 7 years as follows:

P(X=0)=\frac{e^{-4.34}(4.34)^{0}}{0!}=0.01304\approx0.0130

Thus, the probability of no deaths in a corps over 7 years is 0.0130.

6 0
3 years ago
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yarga [219]
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3 years ago
Use the binomial expression (p+q)n
zvonat [6]

Answer:

{0.16807, 0.36015, 0.3087, 0.1323, 0.02835, 0.00243}

Step-by-step explanation:

The expansion of (p+q)^n for n = 5 is ...

(p+q)^5 = p^5 +5·p^4·q +10·p^3·q^2 +10·p^2·q^3 +5·p·q^4 +q^5

When the probability p=0.3 and q = 1-p = 0.7 the terms of this series correspond to the probabilities of 5, 4, 3, 2, 1, and 0 favorable outcomes out of 5 trials.

For example, p^5 = 0.3^5 = 0.00243 is the probability of 5 favorable outcomes in 5 trials where the probability of each favorable outcome is 0.3.

___

The attachment shows the calculation of these numbers using a graphing calculator. It lists them in reverse order of the expansion of (p+q)^5 shown above, so that they are the probabilities of 0–5 favorable outcomes in the order 0–5.

3 0
3 years ago
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