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3 years ago

Which of the following is NOT a risk to research integrity?

Advanced Placement (AP)

1 answer:

nikdorinn [45]3 years ago

4 0

Answer: D- repeating the same research design multiple times

Explanation: I can use the same research design multiple times only when design is according to my investigation.

the rest of answers are wrong

A- overclaiming, I can perform a claim only when is necessary but I can not do it overclaiming.

B- rushing prematurely to publication, I must follow each step of the research one by one. I can not be premature.

C- the incorrect use of statistics, this can give me wrong results during my investigation, so it can give me bad conclusions.

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Explain in detail, how you solved the following problem: The first two terms of a sequence are 10 and 20. If each term after the

KIM [24]

Answer:

$T_{2020} = 15$

Explanation:

Given

$T_1 = 10$

$T_2 = 20$

Each term after the second term is the average of all of the preceding terms

Required:

Explain how to solve the 2020th term

Solve the 2020th term

Solving the 2020th term of a sequence using conventional method may be a little bit difficult but in questions like this, it's not.

The very first thing to do is to solve for the third term;

The value of the third term is the value of every other term after the second term of the sequence; So, what I'll do is that I'll assign the value of the third term to the 2020th term

This is proved as follows;

From the question, we have that "..... each term after the second term is the average of all of the preceding terms", in other words the MEAN

$T_{n} = \frac{\sum T{k}}{n-1} ; where: k = 1 .... n -1$

Assume n = 3

$T_{3} = \frac{T_1 + T_2}{2}$

Multiply both sides by 2

$2 * T_{3} = \frac{T_1 + T_2}{2} * 2$

$2T_{3} = T_1 + T_2$

Assume n = 4

$T_{4} = \frac{T_1 + T_2 + T_3}{3}$

$T_{4} = \frac{(T_1 + T_2) + T_3}{3}$

Substitute $2T_{3} = T_1 + T_2$

$T_{4} = \frac{2T_3 + T_3}{3}$

$T_{4} = \frac{3T_3}{3}$

$T_{4} = T_3$

Assume n = 5

$T_{5} = \frac{T_1 + T_2 + T_3 +T_4}{4}$

$T_{5} = \frac{(T_1 + T_2) + T_3 +(T_4)}{4}$

Substitute $2T_{3} = T_1 + T_2$ and $T_{4} = T_3$

$T_{5} = \frac{2T_3 + T_3 +T_3}{4}$

$T_{5} = \frac{4T_3}{4}$

$T_{5} = \frac{(5-1)T_3}{5-1}$

Replace 5 with n

$T_{n} = \frac{(n-1)T_3}{n-1}$

(n-1) will definitely cancel out (n-1); So, we're left with

$T_{n} = T_3$

Hence,

$T_{2020} = T_3$

Calculating $T_3$

$T_{3} = \frac{10 + 20}{2}$

$T_{3} = \frac{30}{2}$

$T_{3} = 15$

Recall that $T_{2020} = T_3$

$T_{2020} = 15$

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Answer:

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