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3 years ago

7

Write code which takes inputs and creates two Rectangle objects (using the edhesive.shapes.Rectangle class) and compares them us

ing the equals method.
JAVA!!

Computers and Technology

1 answer:

iogann1982 [59]3 years ago

6 0

Answer:

System.out.println("Enter length:");

Scanner scan = new Scanner(System.in);

Double x = scan.nextDouble();

System.out.println("Enter 2 lengths:");

Double a = scan.nextDouble();

Double b = scan.nextDouble();

Rectangle rect = new Rectangle(x);

Rectangle rect2 = new Rectangle (a,b);

if (rect2.equals(rect)){

System.out.print("Congruent Rectangles");

}

else {

System.out.print("Different Rectangles");

}

}

}

Explanation:

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<em />

Explanation:

Given

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Required

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Such that

$p = 95\%\\$

It should be noted that probabilities always add up to 1;

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Subtract p from both sides

$p - p + q = 1 - p$

$q = 1 - p$

Substitute $p = 95\%\\$

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Convert % to fraction

$q = 1 - \frac{95}{100}$

Convert fraction to decimal

$q = 1 - 0.95$

$q = 0.05$

------------------------------------------------------------------------------------------------

<em>To get an expression for one server</em>

The probabilities of 1 servers having 99.98% reliability is as follows;

$p = 99.98\%$

Recall that probabilities always add up to 1;

So,

$p + q = 1$

Subtract q from both sides

$p + q - q = 1 - q$

$p = 1 - q$

So,

$p = 1 - q = 99.98\%$

$1 - q = 99.98\%$

Let the number of servers be represented with n

The above expression becomes

$1 - q^n = 99.98\%$

Convert percent to fraction

$1 - q^n = \frac{9998}{10000}$

Convert fraction to decimal

$1 - q^n = 0.9998$

Add $q^n$ to both sides

$1 - q^n + q^n= 0.9998 + q^n$

$1 = 0.9998 + q^n$

Subtract 0.9998 from both sides

$1 - 0.9998 = 0.9998 - 0.9998 + q^n$

$1 - 0.9998 = q^n$

$0.0002 = q^n$

Recall that $q = 0.05$

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$0.0002 = 0.05^n$

Take Log of both sides

$Log(0.0002) = Log(0.05^n)$

From laws of logarithm $Loga^b = bLoga$

So,

$Log(0.0002) = Log(0.05^n)$ becomes

$Log(0.0002) = nLog(0.05)$

Divide both sides by $Log0.05$

$\frac{Log(0.0002)}{Log(0.05)} = \frac{nLog(0.05)}{Log(0.05)}$

$\frac{Log(0.0002)}{Log(0.05)} = n$

$n = \frac{Log(0.0002)}{Log(0.05)}$

$n = \frac{-3.69897000434}{-1.30102999566}$

$n = 2.84310893421$

$n = 3 (Approximated)$

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