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2 years ago

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A life insurance salesman sells on the average 4 life insurance policies per week.

1 answer:

Lorico [155]2 years ago

7 0

Answer:

0.195

0.191

Step-by-step explanation:

Given that :

μ = 4

Using poisson :

P(x) = (e^-μ * μ^x) ÷ x!

P(x = 3) = (e^-4 * 4^3) / 3!

= (0.0183156 * 64) / 6

= 0.1953664

= 0.195

b. What is the probability that he will sell at most 5 policies in two weeks period?

Mean per week = 4

Mean in two weeks = 4 * 2 = 8

P(x ≤ 5) = p(5) + p(4) +... + p(0)

Using the poisson distribution calculator :

P(x ≤ 5) = 0.19124

= 0.191

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Solve using quadratic formula for 2x^(2)-10x+5=0

sergij07 [2.7K]

Answer:

$x=\frac{5+\sqrt{15}}{2},\:x=\frac{5-\sqrt{15}}{2}$

Step-by-step explanation:

$\mathrm{For\:a\:quadratic\:equation\:of\:the\:form\:}ax^2+bx+c=0\mathrm{\:the\:solutions\:are\:}$

$x_{1,\:2}=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

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$x_{1,\:2}=\frac{-\left(-10\right)\pm \sqrt{\left(-10\right)^2-4\cdot \:2\cdot \:5}}{2\cdot \:2}$

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$=\sqrt{100-40}$

$\mathrm{Subtract\:the\:numbers:}\:100-40=60$

$=\sqrt{60}$

$60\:\mathrm{divides\:by}\:2\quad \:60=30\cdot \:2$

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$=2\cdot \:2\cdot \:3\cdot \:5$

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Apply Radical Rule:

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$\mathrm{Separate\:the\:solutions}$

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$\frac{-\left(-10\right)+2\sqrt{15}}{2\cdot \:2}$

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$=\frac{2\left(5+\sqrt{15}\right)}{4}$

$\mathrm{Cancel\:the\:common\:factor:}\:2$

$=\frac{5+\sqrt{15}}{2}$

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$=\frac{10-2\sqrt{15}}{2\cdot \:2}$

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$=\frac{10-2\sqrt{15}}{4}$

$=\frac{2\left(5-\sqrt{15}\right)}{4}$

$\mathrm{Cancel\:the\:common\:factor:}\:2$

$=\frac{5-\sqrt{15}}{2}$

$\mathrm{The\:solutions\:to\:the\:quadratic\:equation\:are:}$

$x=\frac{5+\sqrt{15}}{2},\:x=\frac{5-\sqrt{15}}{2}$

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