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xxMikexx [17]
3 years ago
4

What is the answer to this question ?

Mathematics
1 answer:
FrozenT [24]3 years ago
8 0
approximate. multiply by 400. answer is C. you do not need a calc for this
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The function f(x) is shown below.
JulsSmile [24]

Answer:

f(g(2)) = 2

Step-by-step explanation:

Given

x \to f(x)

-6 \to 1

-3 \to 2

g(x) = f^{-1}(x) --- inverse

Required

f(g(2))

For two functions f(x) and g(x) where f(x) and g(x)are inverse;

f(g(x)) = x

So, by comparison:

f(g(2)) = 2

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What is the slope-intercept equation of the line below? y - i ntercept = (0, - 4) slope = 3
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Answer:

y = 3x -4

Step-by-step explanation:

The slope-intercept equation looks like this: y = mx +b. Where m is the slope of the function and b is the y-coordinate of the vertical intercept (where the function crosses the y-axis). Let's apply the formula:

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Step-by-step explanation:

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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}

4 0
3 years ago
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