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Yuliya22 [10]
2 years ago
13

(b) The area of a rectangular painting is 7719 cm2.

Mathematics
2 answers:
Nataliya [291]2 years ago
6 0

Given:

  • The area of a rectangular painting is 7719 cm²
  • Width of painting = 83cm

To Find?

  • Length of painting.

Solution:

  • Let length of painting be x cm

Using formula:

  • Area of rectangle = L × B

Where,

  • Length = x cm
  • Breadth = 73 cm

→ 7719 = x × 83

→ x = 7719/83

→ x = 93 cm

Hence,

  • Length of painting is 93cm
Salsk061 [2.6K]2 years ago
3 0

We are given that the width of of a rectangular painting is 83 cm and it's area is 7719 cm² and we are asked to find it's length , as we know that the area of a rectangular shape is given by <em><u>L×</u></em><em><u>B</u></em>. Where <em><u>L</u></em> is the length of the rectangular shape and <em><u>B</u></em> is the width/Breadth .Now , applying the same concept in our question we have ;

{:\implies \quad \sf L\times 83=7719}

{:\implies \quad \sf L=\dfrac{7719}{83}}

{:\implies \quad \sf L=93}

{:\implies \quad \bf \therefore \quad \underline{\underline{Length=93\:\: cm}}}

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An equation has solutions of m = –5 and m = 9. Which could be the equation?
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When circuit boards used in the manufacture of compact disc players are tested, the long-run percentage of defectives is 5%. Let
vladimir1956 [14]

Answer:

a) P(X ≤ 2) = 0.87

b)  P(X ≥ 5) = 0.01

c) P(1 ≤ X ≤ 4) = 0.71

d) P ( X = 0 ) = 0.28

e) σ(X) = 1.09 , E(X) = 1.25

Step-by-step explanation:

Given:

- Let X = the number of defective boards in a random sample of size n = 25, so X ~ Bin(25, 0.05)

Where, n = 25 and p = 0.05

Find:

(a) Determine P(X ≤ 2).(b) Determine P(X ≥ 5).(c) Determine P(1 ≤ X ≤ 4).(d) What is the probability that none of the 25 boards is defective?(e) Calculate the expected value and standard deviation of X.

Solution:

- The probability mass function for a binomial distribution is given by:

                         P ( X = x ) = nCr * (p)^r * ( 1 - p )^(n-r)

a) P(X ≤ 2):

                         P(X ≤ 2) = P ( X = 0 ) + P ( X = 1 ) + P ( X = 2 )

                         = (0.95)^25 + 25*0.05*0.95^24 + 25C2*0.05^2*0.95^23

                         = 0.87

b) P(X ≥ 5):

            P(X ≥ 5) = 1 - [P ( X = 0 ) + P ( X = 1 ) + P ( X = 2 ) + P ( X = 3 ) + P(X=4)

            = 1 - [ 0.87 + 25C3*0.05^3*0.95^22 + 25C4*0.05^4*0.95^21]

            = 1 - 0.98994

            = 0.01

c) P(1 ≤ X ≤ 4):

            P(1 ≤ X ≤ 4) = P ( X = 1 ) + P ( X = 2 ) + P ( X = 3 ) + P(X=4)

            = ( 0.87 - 0.95^25) + 0.11994

           = 0.71

d) P( X = 0 )

            P ( X = 0 ) = 0.95^25 = 0.28

e) E(X) & σ(X):

            E(X) = n*p

            E(X) = 25*0.05 = 1.25

            σ(X) = sqrt ( Var (X) )

            σ(X) = sqrt ( n*p*(1-p) ) = sqrt ( 25*0.05*0.95 )

            σ(X) = 1.09

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