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steposvetlana [31]
3 years ago
11

The perimeter of rectangle MNPQ is 80 in and the ratio MN : MQ = 3:5. Find the area of MNPQ

Mathematics
2 answers:
sesenic [268]3 years ago
7 0
Let's write 3:5 = 3k:5k. Then using perimeter formula, we can write that P=(3k+5k)2=80 and k=5 cm. The sides are 15 cm and 25 cm. The area is A=15*25=375 cm^{2}
kolezko [41]3 years ago
7 0

Answer:

The area of MNPQ is 375 square inches.

Step-by-step explanation:

It is given that the ratio MN : MQ = 3:5. It means the ratio of dimensions is 3:5.

Let the dimensions of the rectangle be 3x by 5x.

The perimeter of a rectangle is

P=2(length+width)

It is given that the perimeter of rectangle MNPQ is 80.

80=2(3x+5x)

80=2(8x)

80=16x

Divide both sides by 16.

5=x

The value of x is 5.

The dimensions of the rectangle are 15 and 25.

The area of rectangle is

A=length\times breadth

A=15\times 25

A=375

Therefore the area of MNPQ is 375 square inches.

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15. Suppose a box of 30 light bulbs contains 4 defective ones. If 5 bulbs are to be removed out of the box.
Naddika [18.5K]

Answer:

1) Probability that all five are good = 0.46

2) P(at most 2 defective) = 0.99

3) Pr(at least 1 defective) = 0.54

Step-by-step explanation:

The total number of bulbs = 30

Number of defective bulbs = 4

Number of good bulbs = 30 - 4 = 26

Number of ways of selecting 5 bulbs from 30 bulbs = 30C5 = \frac{30 !}{(30-5)!5!} \\

30C5 = 142506 ways

Number of ways of selecting 5 good bulbs  from 26 bulbs = 26C5 = \frac{26 !}{(26-5)!5!} \\

26C5 = 65780 ways

Probability that all five are good = 65780/142506

Probability that all five are good = 0.46

2) probability that at most two bulbs are defective = Pr(no defective) + Pr(1 defective) + Pr(2 defective)

Pr(no defective) has been calculated above = 0.46

Pr(1 defective) = \frac{26C4  * 4C1}{30C5}

Pr(1 defective) = (14950*4)/142506

Pr(1 defective) =0.42

Pr(2 defective) =  \frac{26C3  * 4C2}{30C5}

Pr(2 defective) = (2600 *6)/142506

Pr(2 defective) = 0.11

P(at most 2 defective) = 0.46 + 0.42 + 0.11

P(at most 2 defective) = 0.99

3) Probability that at least one bulb is defective = Pr(1 defective) +  Pr(2 defective) +  Pr(3 defective) +  Pr(4 defective)

Pr(1 defective) =0.42

Pr(2 defective) = 0.11

Pr(3 defective) =  \frac{26C2  * 4C3}{30C5}

Pr(3 defective) = 0.009

Pr(4 defective) =  \frac{26C1  * 4C4}{30C5}

Pr(4 defective) = 0.00018

Pr(at least 1 defective) = 0.42 + 0.11 + 0.009 + 0.00018

Pr(at least 1 defective) = 0.54

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3 years ago
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Step-by-step explanation:

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