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AURORKA [14]
2 years ago
8

The length of a rectangle is 5cm more than its breadth . if the perimeter of the rectangle is 40 cm , find the dimensions f the

rectangle ​
Mathematics
2 answers:
PolarNik [594]2 years ago
8 0

Step-by-step explanation:

find

length and the breath

<h2> given</h2>

<h2>solution </h2>

length = x +5

<h2>perimeter 2 L+ B</h2>

<h3>2 x + x = 40</h3>

2 x + 5 = 40 / 2

2x + 5 = 20

2 x = 20 -5

2 x + 15

x = 15 /2

please mark me as brainlist

mezya [45]2 years ago
5 0
<h3>Answer :- </h3>

  • 7.5 cm

<h3>Solution :- </h3>

  • Let the breadth of the rectangle be x cm

  • length of the rectangle = ( x + 5 ) cm
  • perimeter of the rectangle= 2( l + b )

  • 2 ( x + x + 5 ) cm
  • 2 ( 2x + 5 ) cm
  • ( 4x + 10 ) cm

<em><u>Transpose </u></em><em><u>1</u></em><em><u>0</u></em><em><u> </u></em><em><u>to </u></em><em><u>RHS </u></em>

  • <em>therefore</em><em> </em><em>,</em><em> </em>
  • 4x +<em> </em>10 = 40
  • 4 x = 40 - 10
  • x = 30 / 4 = 67.5

  • Length of the rectangle = x + 5 = 7.5 + 5 = 12.5
  • Breadth of the rectangle = x = 7.5 cm

<em>Hope </em><em>it </em><em>helps </em><em>!</em><em>!</em><em> </em>

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Step-by-step explanation:

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Step-by-step explanation:

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Answer:

a) P(X=2)=(10C2)(0.32)^2 (1-0.32)^{10-2}=0.211

b) P(X> 2)=1-P(X\leq 2)=1-[0.0211+0.0995+0.211]=0.668

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Step-by-step explanation:

1) Previous concepts

The binomial distribution is a "DISCRETE probability distribution that summarizes the probability that a value will take one of two independent values under a given set of parameters. The assumptions for the binomial distribution are that there is only one outcome for each trial, each trial has the same probability of success, and each trial is mutually exclusive, or independent of each other".

2) Solution to the problem

Let X the random variable of interest, on this case we now that:

X \sim Binom(n=10, p=0.32)

The probability mass function for the Binomial distribution is given as:

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Where (nCx) means combinatory and it's given by this formula:

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Part a

P(X=2)=(10C2)(0.32)^2 (1-0.32)^{10-2}=0.211

Part b

P(X> 2)=1-P(X\leq 2)=1-[P(X=0)+P(X=1)+P(X=2)]

P(X=0)=(10C0)(0.32)^0 (1-0.32)^{10-0}=0.0211  

P(X=1)=(10C1)(0.32)^1 (1-0.32)^{10-1}=0.0995

P(X=2)=(10C2)(0.32)^2 (1-0.32)^{10-2}=0.211

P(X> 2)=1-P(X\leq 2)=1-[0.0211+0.0995+0.211]=0.668

Part c

P(2 \leq x \leq 5)=P(X=2)+P(X=3)+P(X=4)+P(X=5)

P(X=2)=(10C2)(0.32)^2 (1-0.32)^{10-2}=0.211

P(X=3)=(10C3)(0.32)^3 (1-0.32)^{10-3}=0.264

P(X=4)=(10C4)(0.32)^4 (1-0.32)^{10-4}=0.218

P(X=5)=(10C5)(0.32)^5 (1-0.32)^{10-5}=0.123

P(2 \leq x \leq 5)=0.211+0.264+0.218+0.123=0.816

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