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Wewaii [24]
4 years ago
10

Find the 11th term of the sequence 11, 19, 27, 35

Mathematics
1 answer:
iragen [17]4 years ago
7 0

Answer:

  91

Step-by-step explanation:

Each term is 8 more than the last, so you can write the terms of the sequence until you get there:

  11, 19, 27, 35, 43, 51, 59, 67, 75, 83, 91

__

Or, you can use the formula for the general term of an arithmetic sequence. You have first term a1=11 and common difference d=8.

  an = a1 +d(n -1)

  a11 = 11 + 8(11 -1) = 11 + 80

  a11 = 91

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2a) md= 24 b) Q1=20.5 c) Q3= 30 3) Q3-Q1 =9.5 b) 19/48

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To answer this question the 1st and the 2nd we need to order the data entries. So from ordering from the lowest to the highest value:

1. {18, 18 , 19 , 19 , 19 , 20 , 21 , 21 , 21 , 21 , 23 , 24 , 24 , 26 , 27 , 27 , 29 , 30 , 30 , 30 , 33 , 33, 34 , 35 , 38 }

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{18, 18 , 19 , 19 , 19 , 20 , 21 , 21 , 21 , 21 , 23 , 24 , 24 , 26 , 27 , 27 , 29 , 30 , 30 , 30 , 33 , 33, 34 , 35 , 38 }

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md=24

b) To find out the 1st quartile, we can use this way:

Q_{1}=\frac{i}{4}(n+1)\\ Q_{1}=\frac{1}{4}(25+1)\\ Q_{1}=\frac{1}{4}(26)=6.5

Then 6.5 is between the 6th and 7th position. Let's find the mean of them, now:  

Q_1=\frac{20+21}{2}= 20.5

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Q_{3}=\frac{i}{4}(n+1)\\ Q_{3}=\frac{3}{4}(25+1)\\ Q_{1}=\frac{3}{4}(26)=19.5

The 19th position and 20th position average is:\frac{30+30}{2} =30

3)

a) To find the Interquartile Range, we just need to find out the difference of the upper quartile and the lower one:[tex](Q_3-Q_1)Q_3-Q_1[/tex]

(30-20.5)=9.5

b) Interquartile Ratio is given by the quotient of the Interquartile Range over the Median

\frac{IQR}{md}=\frac{9.5}{24}=\frac{19}{48}

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k=1+3.32logn

7=1+3.32logn

6=3.32logn

n≈66

Each class must have an interval of 10 ages, for (91-18)/7≈ 10. Notice the orange line intercepts the midpoint of each interval.

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