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Nadusha1986 [10]
3 years ago
8

Can u help me with this

Mathematics
1 answer:
daser333 [38]3 years ago
6 0

Answer:

acc to ques....perimeter of figure= 3a^2+15a+9

let missing dimension be x

= x+10a-2+a^2-7+5a=3a^2+15a+9

x+15a+a^2-9=3a^2+15a+9

x=2a^2+18

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Answer: 22.5 ; 5 ; 14

Step-by-step explanation:

Given the dataset:

{20,22,23,24,26,26,28,29,30}

The lower quartile (Q1) = 1/4(n + 1)th term

Where n = number of observations, n = 9

Q1 = 1/4 (9 + 1)th term

Q1 = 1/4(10) = 2.5

We average the 2nd and 3rd term:

(22 + 23) / 2

45 / 2 = 22.5

B) The interquartile range(IQR) of the dataset :

{62,63,64,65,67,68,68,68,69,74}

IQR = Q3 - Q1

The lower quartile (Q1) = 1/4(n + 1)th term

Where n = number of observations, n = 10

Q1 = 1/4 (10 + 1)th term

Q1 = 1/4(11) = 2.75 term

We take the average of the 2nd and 3rd term:

(63 + 64) / 2

45 / 2 = 63.5

The upper quartile (Q3) = 3/4(n + 1)th term

Where n = number of observations, n = 10

Q3 = 3/4 (10 + 1)th term

Q3 = 3/4(11) = 8.25 term

We take the average of the 8th and 9th term:

(68 + 69) / 2

137 / 2 = 68.5

IQR = Q3 - Q1

IQR = 68.5 - 63.5

IQR = 5

C) give the dataset :

{7,8,8,9,10,12,13,15,16}

The upper quartile (Q3) = 3/4(n + 1)th term

Where n = number of observations, n = 9

Q3 = 3/4 (9 + 1)th term

Q3 = 3/4(10) = 7.5 term

We take the average of the 7th and 8th term:

(13 + 15) / 2

28 / 2 = 14

7 0
3 years ago
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