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2 years ago

Find f(-2) for f(x) = 2(4)^x

Mathematics

2 answers:

emmainna [20.7K]2 years ago

8 0

Answer:

f(- 2) = $\frac{1}{8}$

Step-by-step explanation:

To find f(- 2) substitute x = - 2 into f(x) , that is

f(- 2) = 2 $(4)^{-2}$ = 2 × $\frac{1}{4^2}$ = 2 × $\frac{1}{16}$ = $\frac{2}{16}$ = $\frac{1}{8}$

stepladder [879]2 years ago

3 0

Answer:

2(4)^-2

=2/4^2

2/16

1/8

thanks

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2x^2+5x+3/x^2-3x-4 divided by 4x^2+2x-6/x^2-8x+16

Masja [62]

Answer: $\frac{x-4}{2x-2}$

Step-by-step explanation:

$\frac{2x^2+5x+3}{x^2-3x-4}$ divided by $\frac{4x^2+2x-6}{x^2-8x+16}$ is the same thing as multiplying $\frac{2x^2+5x+3}{x^2-3x-4}$ by $\frac{x^2-8x+16}{4x^2+2x-6}$ .

The equation we get is:

$\frac{2x^2+5x+3}{x^2-3x-4}*\frac{x^2-8x+16}{4x^2+2x-6}$

With this equation, we can factor each equation:

$\frac{(x+1)(2x+3)}{(x+1)(x-4)} *\frac{(x-4)(x-4)}{2(x-1)(2x+3)}$

We can cancel out like terms since they would be dividing each other:

$\frac{x-4}{2x-2}$

8 0

3 years ago

Mrrafil [7]

Answer:

3280

Step-by-step explanation:

52×63=3276

Round 3276 to 3280

Hope this helps! :)

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2 years ago

6. Write an algebraic expression for the word phrase 5 times the product of x and y

Lena [83]

Answer:

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Step-by-step explanation:

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3 years ago

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Ms. Santor divides 32 students into 8 equal groups for a field trip.Draw a tape diagram,and label the number of students in each

Nataly [62]

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3 years ago

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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

2 years ago