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3 years ago

Activity

Mathematics

2 answers:

Ipatiy [6.2K]3 years ago

8 0

Answer:

2d + 3

Step-by-step explanation:

Dylan’s weight is d.

Twice Dylan’s weight is 2d.

Three pounds more than twice Dylan’s weight is 2d + 3.

So, the expression for Alan’s weight in terms of Dylan’s weight is 2d + 3.

alina1380 [7]3 years ago

4 0

Answer:

Step-by-step explanation:

a=2d+3

b=3d-12

c=a+b so

c=2d+3+3d-12

c=5d-9

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soldi70 [24.7K]

Answer:

$0.33

Step-by-step explanation:

Unit rate=1

if 3 equal to $0.99, then

1 will equal $0.33.

0.99/3=$0.33

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3 years ago

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The marks obtained by students are 10, 13, 12, 15, 25, 30, 20, 35, 40, 18. Calculate the median mark.

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Answer:

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Step-by-step explanation:

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2 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

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2 years ago

Based on the extreme value theorem, what is the maximum value of f(x) = –x2 + 6x over the interval [1, 4]?

Tju [1.3M]

Answer:

$Maximum\ value\ =9 ,at\ x=3$

Step-by-step explanation:

From the question we are told that:

Function given

$f(x) = -x^2 + 6x$

Co-ordinates

(x,y)=[1, 4]

Generally the second differentiation of function is mathematically given by

$-2x+6$

Therefore critical point

$x=3$

Generally the substitutions of co-ordinate into function is mathematically given by

For 1

$F(1)=-(1)^2 + 6(1)\\F(1)=5$

For 4

$F(4)=-(4)^2 + 6(4)\\F(4)=8$

For critical point 3

$F(3)=-(3)^2 + 6(3)\\F(3)=9$

Therefore the maximum value of f(x) = –x2 + 6x over the interval [1, 4] is given by

$Maximum\ value\ =9 ,at\ x=3$

3 0

2 years ago

Read 2 more answers

Please help thank you

Veronika [31]

Answer:

2.75 · $10^{5}$

Step-by-step explanation:

Hope this helps! :)

6 0

3 years ago

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