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3 years ago

The area of a rectangle is x^2−4/2x in^2 and its length is (x+2)^2/2 in. What is the width in inches?

Mathematics

1 answer:

lara31 [8.8K]3 years ago

8 0

Answer:

The answer to your question is Width = $\frac{x - 2}{x + 2}$

Step-by-step explanation:

Data

Area = (x² - 4)/2 x in²

Length = (x + 2)² / 2 in

Width = ?

Process

1.- Write the formula of the area of a rectangle

Area = length x width

-Solve for width

Width = Area / length

2.- Substitute the values

Width = (x² - 4)/2 / (x + 2)² / 2

-Simplify

Width = (x² - 4) / (x + 2)²

-Factor both numerator and denominator

Width = (x - 2)(x + 2) / (x + 2)(x + 2)

- Simplify

Width = (x - 2) /(x + 2) or Width = $\frac{x - 2}{x + 2}$

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Answer:

a) Binomial.

b) n=20, p=0.01, k≥2

The probability hat a package sold will be refunded is P=0.0169.

Step-by-step explanation:

a) We know that

the defective probability is constant and independent.
the sample size is bigger than one subject.

The most appropiate distribution to represent this random variable is the binomial.

b) The parameters are:

Sample size (amount of clips in the package): n=20
Probability of defective clips: p=0.01.
number of defective clips that trigger the money-back guarantee: k≥2

The probability of the package being refunded can be calculated as:

$P(x\geq2)=1-(P(x=0)+P(x=1))\\\\\\P(x=k) = \dbinom{n}{k} p^{k}q^{n-k}\\\\\\P(x=0) = \dbinom{20}{0} p^{0}q^{20}=1*1*0.8179=0.8179\\\\\\P(x=1) = \dbinom{20}{1} p^{1}q^{19}=20*0.01*0.8262=0.1652\\\\\\P(x\geq2)=1-(0.8179+0.1652)=1-0.9831=0.0169$