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Radda [10]
3 years ago
5

What is the fractional equivalent of 0.2323? 23/99 23/999 2/9

Mathematics
1 answer:
morpeh [17]3 years ago
7 0
It should be 23/99

23/99= 0.232323...
23/999= 0.023023023....
2/9= 0.222222....

Final answer: 23/99
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When using pi, do you use 3.14 or 3.141592653?
sergey [27]

Answer:

You would usually use 3.14

Step-by-step explanation:

It kind of depends on where you are using it. In an equation, pi would be used as 3.14 and the answer would be approximated. To write it anywhere else (I dont know where else you would though), you would do the pi symbol. I hope this helps.

4 0
3 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

#SPJ1

<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
In circle P with mNRQ = 60°, find the angle measure of minor arc NQ
Galina-37 [17]

Given:

m∠NRQ = 60°

To find:

The angle measure of minor arc NQ

Solution:

The inscribed angle is half of the intercepted arc.

$\Rightarrow m\angle NRQ =\frac{1}{2} m(ar NQ)

Multiply by 2 on both sides.

$\Rightarrow 2 \times m\angle NRQ =2 \times \frac{1}{2} m(ar NQ)

$\Rightarrow 2 \ m\angle NRQ =m(ar NQ)

Substitute m∠NRQ = 60°.

$\Rightarrow 2\times 60^\circ=m(ar NQ)

$\Rightarrow 120^\circ=m(ar NQ)

The measure of minor arc NQ is 120°.

8 0
3 years ago
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