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2 years ago

PLZZZ HELP ASAP

Mathematics

1 answer:

taurus [48]2 years ago

5 0

This is a bar graph.

I was chosen to show difference in time and money and profit.

Upward trend

Hoped this helped <3

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What will be the sum of an integer and its additive inverse

klemol [59]

Answer:

Step-by-step explanation:

4 is an integer

-4is its additive inverse

4+-4=0

5 1

3 years ago

Help marking brainly

Aneli [31]

Answer:

9 + 25

Step-by-step explanation:

Step 1: Define expression

a² + b²

a = 3

b = 5

Step 2: Substitute

3² + 5²

Step 3: Evaluate

9 + 25

5 0

3 years ago

Read 2 more answers

Explain the steps necessary to convert a quadratic function in standard form to vertex form

SIZIF [17.4K]

Answer:

See below.

Step-by-step explanation:

Here's an example to illustrate the method:

f(x) = 3x^2 - 6x + 10

First divide the first 2 terms by the coefficient of x^2 , which is 3:

= 3(x^2 - 2x) + 10

Now divide the -2 ( in -2x) by 2 and write the x^2 - 2x in the form

(x - b/2)^2 - b/2)^2 (where b = 2) , which will be equal to x^2 - 2x in a different form.

= 3[ (x - 1)^2 - 1^2 ] + 10 (Note: we have to subtract the 1^2 because (x - 1)^2 = x^2 - 2x + 1^2 and we have to make it equal to x^2 - 2x)

= 3 [(x - 1)^2 -1 ] + 10

= 3(x - 1)^2 - 3 + 10

= <u>3(x - 1)^2 + 7 </u><------- Vertex form.

In general form the vertex form of:

ax^2 + bx + c = a [(x - b/2a)^2 - (b/2a)^2] + c .

This is not easy to commit to memory so I suggest the best way to do these conversions is to remember the general method.

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

levacccp [35]

Easy hijdbggffafaysuwus

6 0

2 years ago