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3 years ago

14

The table below gives the list price and the number of bids received for five randomly selected Items sold through online auctio

ns. Using this data, consider the equation of the regression line hat y =b 0 +b 1 x for predicting the number of bids an item will receive based on the list price. Keep in mind
, the correlation coefficient may or may not be statistically significant for the data givenRemember, practice , it would not be appropriate to use the regression line to make a prediction the correlation coefficient is not statistically significant Price in Dollars 39 40 42 43 45 Number of Bids 1 3 5 6 8 Table Copy Data Step 1 of 6: Find the estimated slope: Round your answer to three decimal places

Advanced Placement (AP)

2 answers:

kobusy [5.1K]3 years ago

3 0

Answer:

14 birds

Explanation:

julsineya [31]3 years ago

3 0

14 birds is the answer

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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

<em>This is proved as follows;</em>

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

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

<em>Assume n = 3</em>

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

<em>Multiply both sides by 2</em>

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

$2T_{3} = T_1 + T_2$

<em>Assume n = 4</em>

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

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

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 $T_{4} = T_3$

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

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

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

<em>Replace 5 with n</em>

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

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

$T_{n} = T_3$

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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