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andre [41]
3 years ago
13

What is a3 of the arithmetic sequence for which a10 is 41 and a15 is 61

Mathematics
1 answer:
Alexus [3.1K]3 years ago
3 0
For the arithmetic sequence
a₁, a₂, a₃, ..., 
The n-th term is
a_{n}=a_{1}+(n-1)d
where d = common difference.

Because a₁₀ = 41,
a₁ + 9d = 41                (1)

Because a₁₅ = 61
a₁ + 14d = 61             (2)

Subtract (1) from (2).
5d = 20
  d = 4
From (1), obtain
a₁ = 41 - 9*4 = 5

Therefore
a₃ = 5 + 2*4 = 13

Answer:  a₃ = 13
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•The exam outlier at 60 makes the IQR narrower and the median higher.

•The class data is more evenly spread, which pulls its median down.

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•The class Q3 is higher than the exam Q3.

Step-by-step explanation:

In the question we have two box plots

a) Box plots for class is given as

Minimum at 70, Q1 at 74, median at 83, Q3 at 92,

b) Box plots for exam is given as Minimum at 60, Q1 at 81, median at 87, Q3 at 91, maximum at 95.

When we compare the median of the class which is 83 and the median of the exam plots 87, we can see that the median of the exam is higher than the median of the class, hence the option

"•The class median is lower than the exam median". is wrong.

From the options given, the only correct option which best describes the information about the medians is

"•The class Q3 is higher than the exam Q3".

This is because when we compare the Q3 of the class which is 92 to the Q3 of the box which is 91 , we can see that the class Q3 is higher than the box Q3 hence the option is correct.

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