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Gnoma [55]
3 years ago
13

Let f(x) = 8x3 − 28x + 61 and g(x) = 2x + 5. Find f of x over g of x.

Mathematics
1 answer:
antoniya [11.8K]3 years ago
8 0

Answer:

4x^2-10x+11+\frac{6}{2x+5}

Step-by-step explanation:

The given functions are:

f(x)=8x^3-28x+61 and g(x)=2x+5

By algebraic properties of functions;

\frac{f(x)}{g(x)}=\frac{8x^3-28x+61}{2x+5}

We perform the long division of the two polynomials as shown in the attachment.

Therefore:

\frac{8x^3-28x+61}{2x+5}=4x^2-10x+11+\frac{6}{2x+5}

The last choice is correct

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What is the ratio of the calories Robyn burned on Thursday to the total number of calories she burned for the week?
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6 divided by 3,724? I really need to know this............
ziro4ka [17]

Answer: 620 r4 (620.6, the decimal 6 has a line over it because it is repeating)

<u>Step-by-step explanation:</u>

Using long division:

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6  ) 3 7 2 4

  <u> - 36</u>  ↓

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8 0
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}

4 0
3 years ago
Solve this problem. What is x?
DedPeter [7]

a^2 + b^2 = c^2

a^2 = 6^2 + √117^2

a^2 + 36 = 117

a^2 = 117 - 36

a^2 = 81

a = √81

a = 9

x = 9

5 0
3 years ago
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