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4 years ago

10. a? = 3,600algebra solve each equation

Mathematics

1 answer:

fenix001 [56]4 years ago

7 0

Answer:

a=360

Step-by-step explanation:

3600 divided by 10 is 360

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X is greater than or equal to -1 and less than or equal to 8

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Answer:

Is their an equation that can be used a reference to find the sum/difference/product/quotient to this equation?

Step-by-step explanation:

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3 years ago

Help this is due in 2 minutes

Alika [10]

Answer:

The percentage change each year is 2%

Step-by-step explanation:

Generally, we can have an exponential equation written as;

f(t) = I( 1 + r)^t

where f(t) is the value at a certain time

I is the initial value at t = 0

r is the growth rate

t is the time we are looking at

Thus, we have it that;

1.02 = 1 + r

r = 1.02 -1

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The percentage change each year is 2%

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3 years ago

The total claim amount for a health insurance policy follows a distribution with density function 1 ( /1000) ( ) 1000 x fx e− =

gizmo_the_mogwai [7]

Answer:

the approximate probability that the insurance company will have claims exceeding the premiums collected is $\mathbf{P(X>1100n) = 0.158655}$

Step-by-step explanation:

The probability of the density function of the total claim amount for the health insurance policy is given as :

$f_x(x) = \dfrac{1}{1000}e^{\frac{-x}{1000}}, \ x> 0$

Thus, the expected total claim amount $\mu$ = 1000

The variance of the total claim amount $\sigma ^2 = 1000^2$

However; the premium for the policy is set at the expected total claim amount plus 100. i.e (1000+100) = 1100

To determine the approximate probability that the insurance company will have claims exceeding the premiums collected if 100 policies are sold; we have :

P(X > 1100 n )

where n = numbers of premium sold

$P (X> 1100n) = P (\dfrac{X - n \mu}{\sqrt{n \sigma ^2 }}> \dfrac{1100n - n \mu }{\sqrt{n \sigma^2}})$

$P(X>1100n) = P(Z> \dfrac{\sqrt{n}(1100-1000}{1000})$

$P(X>1100n) = P(Z> \dfrac{10*100}{1000})$

$P(X>1100n) = P(Z> 1) \\ \\ P(X>1100n) = 1-P ( Z \leq 1) \\ \\ P(X>1100n) =1- 0.841345$

$\mathbf{P(X>1100n) = 0.158655}$

Therefore: the approximate probability that the insurance company will have claims exceeding the premiums collected is $\mathbf{P(X>1100n) = 0.158655}$

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3 years ago

Lesson 10 congruent triangles unit test

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4 years ago

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3 years ago

10. a? = 3,600algebra solve each equation​

10. a? = 3,600algebra solve each equation