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3 years ago

PLSS HELP look at pic

Mathematics

1 answer:

gavmur [86]3 years ago

6 0

We can’t really see the picture

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use a graphing calculator or other technology to answer the question which quadratic regression equation best fits the data set

Greeley [361]

Answer:

Option C is the correct option.

In other words, the quadratic regression $y=32.86\:\left(x\right)^2-379.14\left(x\right)+1369.14\:$ best fits the data set, as it gets very much close to the data values given in the data table.

The graph of the equation $y=32.86\:\left(x\right)^2-379.14\left(x\right)+1369.14\:$ is also attached.

Step-by-step explanation:

x y

3 470

4 416

5 403

Analyzing Option A:

Considering the equation

$y=32.86\:\left(x\right)^2+379.14\left(x\right)-1369.14\:$

From (3, 470), putting x = 3

$y=32.86\:\left(3\right)^2+379.14\left(3\right)-1369.14\:$

$y=64.02$

From (4, 470), putting x = 4

$y=32.86\:\left(4\right)^2+379.14\left(4\right)-1369.14\:\:$

$y=673.18$

From (5, 403), putting x = 5

$y=32.86\:\left(5\right)^2+379.14\left(5\right)-1369.14\:$

$y=1348.06$

Analyzing Option B:

$y=32.86\:\left(x\right)^2-379.14\left(x\right)$

From (3, 470), putting x = 3

$y=32.86\:\left(3\right)^2-379.14\left(3\right)$

$y=-841.68$

From (4, 470), putting x = 4

$y=32.86\:\left(4\right)^2-379.14\left(4\right)$

$\:y=-990.8$

From (5, 403), putting x = 5

$y=32.86\:\left(5\right)^2-379.14\left(5\right)$

$\:y=-1074.2$

Analyzing Option C:

Considering the equation

$y=32.86\:\left(x\right)^2-379.14\left(x\right)+1369.14\:$

From (3, 470), putting x = 3

$y=32.86\:\left(3\right)^2-379.14\left(3\right)+1369.14\:$

$y=527.46$

So, the approximately result is (3, 527)

From (4, 470), putting x = 4

$y=32.86\:\left(4\right)^2-379.14\left(4\right)+1369.14\:$

$y=378.34$

So, the approximately result is (4, 378)

From (5, 403), putting x = 5

$y=32.86\:\left(5\right)^2-379.14\left(5\right)+1369.14\:\:\:$

$y=294.94$

So, the approximately result is (5, 295)

Analyzing Option D:

Considering the equation

$y=-1369.14\:\left(x\right)^2-379.14\left(x\right)+32.86$

From (3, 470), putting x = 3

$y=-1369.14\:\left(3\right)^2-379.14\left(3\right)+32.86\:\:$

$y=-13426.82$

From (4, 470), putting x = 4

$y=-1369.14\:\left(4\right)^2-379.14\left(4\right)+32.86$

$y=-23389.94$

From (5, 403), putting x = 5

$y=-1369.14\:\left(5\right)^2-379.14\left(5\right)+32.86\:\:$

$y=-36091.34$

Therefore, from the above calculations and analysis, we conclude that Option C is the correct option.

In other words, the quadratic regression $y=32.86\:\left(x\right)^2-379.14\left(x\right)+1369.14\:$ best fits the data set, as it gets very much close to the data values given in the data table.

The graph of the equation $y=32.86\:\left(x\right)^2-379.14\left(x\right)+1369.14\:$ is also attached.

3 0

3 years ago

Mr. Taylor filled out a bracket for the NCAA National Tournament. Based on his knowledge of college basketball, he has a 0.54 pr

boyakko [2]

Answer:

The probability is $2.7327 \times 10^{-9}$

Step-by-step explanation:

The probability of guessing correctly, P = 0.54

Probability of not guessing correctly, q = 1 – P

q = 1 – 0.54 = 0.46

Number of trials, n = 32

Now calculate the probability that Mr. Taylor will pick 32 correctly in first round of the game.

Below is the calculation using binomial distribution.

$Probability = \left ( _{k}^{n}\textrm{} \right )P^{k}(1-P)^{(n-k)} \\= \left ( _{32}^{32}\textrm{} \right )0.54^{32}(0.46)^{(32-32)} \\= 0.54^{32} \\= 2.7327 \times 10^{-9}$