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ycow [4]
3 years ago
15

Length of a rectangle is 5 cm longer than the width. Four squares are constructed outside the rectangle such that each of the sq

uares share one side with the rectangle. The total area of the constructed figure is 120 cm². What is the perimeter of the rectangle?
Mathematics
1 answer:
stiv31 [10]3 years ago
8 0

Answer:

18

Step-by-step explanation:

Remark

This is one of those questions that can throw you. The problem is that do you include the original rectangle or not. The way it is written it sounds like you shouldn't

However if you don't the question gives you 2 complex answers. (answers with the sqrt( - 1) in them.

Solution

Let the width = x

Let the length = x + 5

Area of the rectangle: L * w = x * (x + 5)

Area of the smaller squares (there are 2)

Area = 2*s^2

x = s

Area = 2 * x^2

Area of the larger squares = 2 * (x+5)^2

Total Area

x*(x + 5) + 2x^2 + 2(x + 5)^2 = 120                 Expand

x^2 + 5x + 2x^2 + 2(x^2 + 10x + 25) = 120     Remove the brackets

x^2 + 5x + 2x^2 + 2x^2 + 20x + 50 = 120      collect the like terms on the left

5x^2 + 25x +  50 = 120                                   Subtract 120 from both sides.

5x^2 + 25x - 70 = 0                                         Divide through by 5

x^2 + 5x - 14 = 0                                               Factor

(x + 7)(x - 2) = 0                                                 x + 7 has no meaning  

x - 2 =  0

x = 2                                                              

Perimeter

P = 2*w + 2*L

w = 2

L = 2 + 5

L = 7

P = 2*2 + 2 * 7

P = 4 + 14

P = 18

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Answer:

a) There is a 1.21% probability that both contain diet soda.

b) There is a 79.21% probability that both contain diet soda.

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Step-by-step explanation:

There are only two possible outcomes. Either the can has diet soda, or it hasn't. So we use the binomial probability distribution.

Binomial probability distribution

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P(X = x) = C_{n,x}.\pi^{x}.(1-\pi)^{n-x}

In which C_{n,x} is the number of different combinatios of x objects from a set of n elements, given by the following formula.

C_{n,x} = \frac{n!}{x!(n-x)!}

And \pi is the probability of X happening.

A number of sucesses x is considered unusually low if P(X \leq x) \leq 0.05 and unusually high if P(X \geq x) \geq 0.05

In this problem, we have that:

Two cans are randomly chosen, so n = 2

Two out of 18 cans are filled with diet coke, so \pi = \frac{2}{18} = 0.11

a) Determine the probability that both contain diet soda. P(both diet soda)

That is P(X = 2).

P(X = x) = C_{n,x}.\pi^{x}.(1-\pi)^{n-x}

P(X = 2) = C_{2,2}(0.11)^{2}(0.89)^{0} = 0.0121

There is a 1.21% probability that both contain diet soda.

b)Determine the probability that both contain regular soda. P(both regular)

That is P(X = 0).

P(X = x) = C_{n,x}.\pi^{x}.(1-\pi)^{n-x}

P(X = 0) = C_{2,0}(0.11)^{0}(0.89)^{2} = 0.7921

There is a 79.21% probability that both contain diet soda.

c) Would this be unusual?

We have that P(X = 2) is unusual, since P(X \geq 2) = P(X = 2) = 0.0121 \leq 0.05

For P(X = 0), it is not unusually high nor unusually low.

d) Determine the probability that exactly one is diet and exactly one is regular. P(one diet and one regular)

That is P(X = 1).

P(X = x) = C_{n,x}.\pi^{x}.(1-\pi)^{n-x}

P(X = 1) = C_{2,1}(0.11)^{1}(0.89)^{1} = 0.1958

There is a 19.58% probability that exactly one is diet and exactly one is regular.

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