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3 years ago

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The length of Rebecca’s rectangular garden was two times the width “w”. She increased the length and width of the garden so that

the area of the new garden is (2w2+7w+6)square yards. By how much did Rebecca increase the length and width of the garden. Show your complete working.

1 answer:

Damm [24]3 years ago

6 0

Answer:

rebecca increased the length by 3 and the width by 2

Step-by-step explanation:

First we need to find the length and width by factorizing the expressio ofr the area;

A = 2w^2+7w+6

A = 2w^2+4w+3w+6

A = 2w(w+2)+3(w+2)

A = (2w+3)(w+2)

Since l = 2w

Length = 2w+3

width = w+2

This shows that rebecca increased the length by 3 and the width by 2

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What is the area of Shaded part?

svetlana [45]

Answer:

1π

Step-by-step explanation:

suppose the radius of semicircle P is r,

then the radius of semicircle Q = (r+d)/2 ... d≤r

radius of semicircle R = (r-d)/2

area P = 1/2 (r)²π

area Q = 1/2 ((r+d)/2)² π = 1/8 (r² + 2rd + d²)π

area R = 1/2 ((r-d)/2)² π = 1/8 (r² - 2rd + d²)π

shaded area = P-Q-R = 1/2 r²π - 1/4 (r² + d²)π

= ((r² - d²)/4) * π

because there is no constant r value in the question and d value changes with the r change, when the vertical segment length equal the semicircle P radius (r), r=2 and d = 0

therefore the shaded area = ((2² -0²)/4)*π = 1π

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3 years ago

Given the original number n. Multiply the number by 9. Add 99. Divide this sum by 9. Subtract the original number, n, from the q

tangare [24]

Answer:

11

Step-by-step explanation:

Hello,

Given the original number n.

$\large \boxed{n}$

Multiply the number by 9.

$\large \boxed{9n}$

Add 99.

$\large \boxed{9n+99}$

Divide this sum by 9.

$\large \boxed{\dfrac{9n+99}{9}=n+11}$

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Thank you.

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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

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2 years ago

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ANTONII [103]

Answer:

The easiest way to do this is to reflect Figure I over the Y axis. Then move it up 3 and left 2.

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Hunter-Best [27]

Well let's see first we have to divide
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Now to check and prove your answer we should use the multiplication.
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So now you have a solid answer with proof behind it.
If u need a sentence you might want it to go something like this;
Matt brought 11 packs of baseball cards since 132 divided by 12 is 11. 12 times 11 is 132 cards.
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