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maw [93]
2 years ago
7

3(-x - 4) = 3x + 5(x + 2)

Mathematics
2 answers:
Ket [755]2 years ago
7 0
3(-x - 4) = 3x + 5(x + 2)
x=-2
snow_lady [41]2 years ago
7 0
There you go!!! Hope this helps

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Gina has 5 2/8 feet of silver ribbon and 3 6/8 of gold ribbon. How much more silver ribbon does Gina have than gold ribbon.
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Answer: we are simply to subtract the length of the gold ribbon which is 2 4/6 ft from the length of the silver ribbon, 5 2/6 feet. Mathematically,

                               5 2/6 feet - 2 4/6 feet = 8/3 feet

Therefore, Gina has 8/3 feet more of the silver ribbon than the golden ribbon.

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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}

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