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2 years ago

Express 0.484848..... in p/q form

Mathematics

1 answer:

cupoosta [38]2 years ago

7 0

<h3>Answer: 16/33</h3>

It's in p/q form where p = 16 and q = 33.

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Work Shown:

x = 0.484848.....

100x = 48.4848.....

I multiplied both sides by 100 to move the decimal over 2 spots. Both decimal values for x and 100x have an infinite string of "48"s repeated after the decimal point. When we subtract, those infinite strings will cancel out

100x - x = 99x

48.4848..... - 0.484848..... = 48

So after subtracting straight down, we have the new equation 99x = 48 which solves to x = 48/99

Divide both parts by the GCF 3 to fully reduce

48/3 = 16

99/3 = 33

Therefore, x = 48/99 = 16/33 = 0.484848...

I recommend using a calculator to confirm that 16/33 = 0.484848...

Side note: your calculator may round the last digit, but this is of course rounding error

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3 years ago

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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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2 years ago

Express 0.484848..... in p/q form​

Express 0.484848..... in p/q form