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andreev551 [17]
2 years ago
5

The lengths of the sides of a triangle are in the extended ratio 5 ​: 8 ​: 10. The perimeter of the triangle is 69 cm. What are

the lengths of the​ sides?
Mathematics
1 answer:
pshichka [43]2 years ago
7 0

Answer:

The lengths of the triangle are 15 cm, 24 cm and 30 cm.

Step-by-step explanation:

The side lengths of a triangle are in the extended ratio 5 : 8 : 10.

Let the side lengths of the triangle are 5x, 8x and 10x.

Now, the perimeter of the triangle is given to be 69 cm.

So, 5x + 8x + 10x = 69

⇒ 23x = 69

⇒ x = 3

So, the lengths of the triangle are 5x = 5 × 3 = 15 cm, 8x = 8 × 3 = 24 cm and 10x = 10 × 3 = 30 cm. (Answer)

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gogolik [260]

First, order the number set from least to greatest:

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Mean: You find the mean by combining all the terms, and dividing by the amount of the terms there are in the number set:

(3 + 4 + 4 + 7 + 8 + 9 + 12 + 14 + 16 + 20)/10

(97)/10 = 9.7

Mean: 9.7

Median: You find the median by first ordering the number set from least to greatest, and finding the middle number. Note that if there is a even number of numbers in the set, you find the mean with the two given median digits:

8 & 9 are the median numbers:

(8 + 9)/2 = (17)/2 = 8.5

Median: 8.5

Mode: The mode is the number(s) in the set that shows up the most:

Mode: 4 (shows up one more time than all other numbers)

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Range: 20 - 3 = 17

Range: 17

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5 0
2 years ago
Which function represents exponential decay
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4 0
3 years ago
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

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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

3 0
3 years ago
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So I believe you will have to do 10% x 7 as it has been seven years which will be 70% and 70% of 8000 is 5600 so the answer will be £5600
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Answer:

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The purpose of this question is to determine the research objective:

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