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2 years ago

Describe what happens to the exponential function when the value of "k" is positive. *

Mathematics

1 answer:

Schach [20]2 years ago

5 0

Answer:

Step-by-step explanation:It isn’t. There are really 4 different versions of the exponential function. Positive (like the growth of money invested at 5%), negative (like the US debt constantly becoming more negative), positive decay (like the half life of a radioactive element), or negative decay ( pay half your loan amount in year 1, 1/4 in year two, 1/8 in year three, you in theory will never pay your loan completely but the debt is growing closer to zero.

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here is the base of a prism if the height of the prism is 5cm what is it’s surface area? what is it’s volume

denpristay [2]

<h2>Volume Of Prism</h2>

Here is the base of a prism if the height of the prism is 5cm what is it’s surface area? what is it’s volume.

<h3>Solution:</h3>

Volume = s²
V = 5cm
V = 5cm x 5cm
V = 25cm²

- So The Surface area of Prism is 25cm².

#ProblemSolve

3 0

2 years ago

Read 2 more answers

A random variable X with a probability density function () = {^-x > 0

Sliva [168]

The solutions to the questions are

The probability that X is between 2 and 4 is 0.314
The probability that X exceeds 3 is 0.199
The expected value of X is 2
The variance of X is 2

<h3>Find the probability that X is between 2 and 4</h3>

The probability density function is given as:

f(x)= xe^ -x for x>0

The probability is represented as:

$P(x) = \int\limits^a_b {f(x) \, dx$

So, we have:

$P(2 < x < 4) = \int\limits^4_2 {xe^{-x} \, dx$

Using an integral calculator, we have:

$P(2 < x < 4) =-(x + 1)e^{-x} |\limits^4_2$

Expand the expression

$P(2 < x < 4) =-(4 + 1)e^{-4} +(2 + 1)e^{-2}$

Evaluate the expressions

P(2 < x < 4) =-0.092 +0.406

Evaluate the sum

P(2 < x < 4) = 0.314

Hence, the probability that X is between 2 and 4 is 0.314

<h3>Find the probability that the value of X exceeds 3</h3>

This is represented as:

$P(x > 3) = \int\limits^{\infty}_3 {xe^{-x} \, dx$

Using an integral calculator, we have:

$P(x > 3) =-(x + 1)e^{-x} |\limits^{\infty}_3$

Expand the expression

$P(x > 3) =-(\infty + 1)e^{-\infty}+(3+ 1)e^{-3}$

Evaluate the expressions

P(x > 3) =0 + 0.199

Evaluate the sum

P(x > 3) = 0.199

Hence, the probability that X exceeds 3 is 0.199

<h3>Find the expected value of X</h3>

This is calculated as:

$E(x) = \int\limits^a_b {x * f(x) \, dx$

So, we have:

$E(x) = \int\limits^{\infty}_0 {x * xe^{-x} \, dx$

This gives

$E(x) = \int\limits^{\infty}_0 {x^2e^{-x} \, dx$

Using an integral calculator, we have:

$E(x) = -(x^2+2x+2)e^{-x}|\limits^{\infty}_0$

Expand the expression

$E(x) = -(\infty^2+2(\infty)+2)e^{-\infty} +(0^2+2(0)+2)e^{0}$

Evaluate the expressions

E(x) = 0 + 2

Evaluate

E(x) = 2

Hence, the expected value of X is 2

<h3>Find the Variance of X</h3>

This is calculated as:

V(x) = E(x^2) - (E(x))^2

Where:

$E(x^2) = \int\limits^{\infty}_0 {x^2 * xe^{-x} \, dx$

This gives

$E(x^2) = \int\limits^{\infty}_0 {x^3e^{-x} \, dx$

Using an integral calculator, we have:

$E(x^2) = -(x^3+3x^2 +6x+6)e^{-x}|\limits^{\infty}_0$

Expand the expression

$E(x^2) = -((\infty)^3+3(\infty)^2 +6(\infty)+6)e^{-\infty} +((0)^3+3(0)^2 +6(0)+6)e^{0}$

Evaluate the expressions

E(x^2) = -0 + 6

This gives

E(x^2) = 6

Recall that:

V(x) = E(x^2) - (E(x))^2

So, we have:

V(x) = 6 - 2^2

Evaluate

V(x) = 2

Hence, the variance of X is 2

Read more about probability density function at:

brainly.com/question/15318348

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<u>Complete question</u>

A random variable X with a probability density function f(x)= xe^ -x for x>0\\ 0& else

a. Find the probability that X is between 2 and 4

b. Find the probability that the value of X exceeds 3

c. Find the expected value of X

d. Find the Variance of X

7 0

2 years ago

Share £405 in the ratio 4:11

Advocard [28]

I think the answer to your question is 15 hope this helps

6 0

3 years ago

Which of the following tables is a relation that represents a function?

uysha [10]

The first one is not a function due to the rule that there can not be more than one x, such as there is a repeated number 1 , 2 , 1 , 4 There should not be two
x = 1 terms.

Same the same rule applies to the second option as well as the last.

Your answer is C. or the third option
x= 6, 5, 4, 1
y=6, 4, 6, 2

Hope this Helps

3 0

3 years ago

Which Triangles are congruent by ASA

PolarNik [594]

Triangles congruent by ASA have two pairs of congruent sides and an included congruent angle.

The graph indicates that sides TV, HG, and AB are congruent, and that sides TU, FG, and BC are congruent. It also indicates that angles U, F, and C are congruent, and that angles G and B are congruent. Notice that angle U of triangle TUV is not an included angle; this eliminates triangle TUV as it can't be congruent to another triangle by ASA with the information provided.

That leaves triangles FGH and ABC. Evidently, angles G and B are included angles, so these triangles are congruent by ASA.

Answer:
b. ΔHGF and ΔABC

8 0

3 years ago

Read 2 more answers