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2 years ago

2.5 X 10^5 divided by 1.0 X 10^10

Mathematics

1 answer:

bekas [8.4K]2 years ago

6 0

Answer:

0.000025

Step-by-step explanation:

PEMDAS (Exponets first.)

10^5 = 100,000

10^10 = 10,000,000,000

rewrite

2.5 x 100,000 divided by 1.0 x 10,000,000,000

PEMDAS (multiplication next.)

2.5 x (10^5) = 250,000

1.0 x (10^10) = 10,000,000,000

rewrite.

250,000 divided by 10,000,000,000 = 0.000025

0.000025

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One root of f(x) = x + 10x2 – 25x – 250 is x = -10. What are all the roots of the function? Use the Remainder Theorem.

Natalka [10]

Answer:

-10

-5

Step-by-step explanation:

From the answers given, you probably mean f(x) = x^3 + 10x2 – 25x – 250

The Remainder Theorem is going to take a bit to solve.

You have to try the factors of 250. One way to make your life a lot easier is to graph the equation. That will give you the roots.

The graph appears below. Since the y intercept is -250 the graph goes down quite a bit and if you show the y intercept then it will not be easy to see the roots.

However just to get the roots, the graph shows that

x = -10

x = - 5

x = 5

The last answer is the right one. To use the remainder theorem, you could show none of the answers will give 0s except the last one. For example, the second one will give

f((10) = 10^3 + 10*10^2 - 25*10 - 250

f(10) = 1000 + 1000 - 250 - 250

f(10) = 2000 - 500

f(10) = 1500 which is not 0.

==================

f(1) = (1)^3 + 10*(1)^2 - 25(1) - 250

f(1) = 1 + 10 - 25 - 250

f(1) = -264 which again is not zero

3 0

3 years ago

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x can do a job in 16 days and y can do the same jovb in 20 days x y and z can do the job in 7.5 days. alone z can do the same jo

MissTica

Answer:

24 days

Step-by-step explanation:

4 0

3 years ago

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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

Solve for Y please help me

spayn [35]

Answer:

y = j - rx/c

Step-by-step explanation:

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3 years ago

What are the best approximations of the solutions to this system?

Scilla [17]

(-1.2,-2.0) and (1.9,2.2) are the best approximations of the solutions to this system.

Option B

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

Here, we have a graph of two functions from which we need to find the approximate value of common solutions. Let's find this:

First look at where we have intersection points, In first quadrant & in third quadrant.

<u>At first quadrant:</u>

Draw perpendicular lines from x-axis & y-axis from this point . After doing this we can clearly see that the perpendicular lines cut x-axis at x=1.9 and y-axis at y=2.2. So, one point is (1.9,2.2)

<u>At Third quadrant:</u>

Draw perpendicular lines from x-axis & y-axis from this point. After doing this we can clearly see that the perpendicular lines cut x-axis at x=-1.2 and y-axis at y= -2.0. So, other point is (-1.2,-2.0).

5 0

3 years ago