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3 years ago

Josh estimate the height of his desk what is reasonable estimate??please help

Mathematics

2 answers:

julsineya [31]3 years ago

6 0

The height will approximately be 3 to 3.6 feet

valina [46]3 years ago

5 0

A reasonable estimate could be any number that makes sense. Say the average elementary student is 4' 10" then a reasonable estimate could be 3' to 3' 6" because the desk would be about a 1' 6" shorter than the person.

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Help with number 12 and 13 And please show work so I can understand how to do it

Len [333]

13: multiply the amount by the tax rate (to make a percent a decimal move the decimal point 2 places to the left) and add the original amount and the taxed amount together. For 12 do above but then multiply the total by the tip rate and add together (basically the same as the tax again) :) I hope this helps

7 0

3 years ago

1. 2 months ago you had 3 mice, you now have 18.

Sphinxa [80]

Answer:

k= 15 every two months

Step-by-step explanation: so you started with 3 and you gained fifth teen in two months so every two months you will gain fifth teen mice.

7 0

3 years ago

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

What makes apparent magnitude different from absolute brightness? Give an example in your response.

Lapatulllka [165]

Apparent magnitude isn’t the actual brightness, it’s what the brightness appears to be from earth, absolute brightness is the actual brightness an example is our sun it appears bright on earth but in space it is so much more brighter

8 0

3 years ago

What do you predict the solution for the system 5x-6y=7 and 6x-7y=8 will be ?<br>

Alenkinab [10]

System of Linear Equations entered :

[1] 5x - 6y = 7
[2] 6x - 7y = 8
Graphic Representation of the Equations :

-6y + 5x = 7 -7y + 6x = 8

Solve equation [2] for the variable x

[2] 6x = 7y + 8

[2] x = 7y/6 + 4/3
// Plug this in for variable x in equation [1]

[1] 5•(7y/6+4/3) - 6y = 7
[1] - y/6 = 1/3
[1] - y = 2
// Solve equation [1] for the variable y

[1] y = - 2
// By now we know this much :

x = 7y/6+4/3
y = -2
// Use the y value to solve for x

x = (7/6)(-2)+4/3 = -1

8 0

3 years ago