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1 year ago

Given the data shown below, which of the following is the best approximation of the coefficient of determination (r^2)

Mathematics

1 answer:

HACTEHA [7]1 year ago

8 0

The coefficient of determination can be found using the following formula:

$r^2=\mleft(\frac{n(\sum ^{}_{}xy)-(\sum ^{}_{}x)(\sum ^{}_{}y)}{\sqrt[]{(n\sum ^{}_{}x^2-(\sum ^{}_{}x)^2)(n\sum ^{}_{}y^2-(\sum ^{}_{}y)^2}^{}}\mright)^2$

Using a Statistics calculator or an online tool to work with the data we were given, we get

So the best aproximation of r² is 0.861

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3 years ago

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Consider the following quadratic function Part 3 of 6: Find the x-intercepts. Express it in ordered pairs.Part 4 of 6: Find the

maksim [4K]

Answer:

The line of symmetry is x = -3

Explanation:

Given a quadratic function, we know that the graph is a parabola. The general form of a parabola is:

$y=ax^2+bx+c$

The line of symmetry coincides with the x-axis of the vertex. To find the x-coordinate of the vertex, we can use the formula:

$x_v=-\frac{b}{2a}$

In this problem, we have:

$y=-x^2-6x-13$

Then:

a = -1

b = -6

We write now:

$x_v=-\frac{-6}{2(-1)}=-\frac{-6}{-2}=-\frac{6}{2}=-3$

Part 3:

For this part, we need to find the x-intercepts. This is, when y = 0:

$-x^2-6x-13=0$

To solve this, we can use the quadratic formula:

$x_{1,2}=\frac{-(-6)\pm\sqrt{(-6)^2-4\cdot(-1)\cdot(-13)}}{2(-1)}$

And solve:

$x_{1,2}=\frac{6\pm\sqrt{36-52}}{-2}$ $x_{1,2}=\frac{-6\pm\sqrt{-16}}{2}$

Since there is no solution to the square root of a negative number, the function does not have any x-intercept. The correct option is ZERO x-intercepts.

Part 4:

To find the y intercept, we need to find the value of y when x = 0:

$y=-0^2-6\cdot0-13=-13$

The y-intercept is at (0, -13)

Part 5:

Now we need to find two points in the parabola. Let-s evaluate the function when x = 1 and x = -1:

$x=1\Rightarrow y=-1^2-6\cdot1-13=-1-6-13=-20$ $x=-1\Rightarrow y=-(-1)^2-6\cdot(-1)-13=-1+6-13=-8$

The two points are:

(1, -20)

(-1, -8)

Part 6:

Now, we can use 3 points to find the graph of the parabola.

We can locate (1, -20) and (-1, -8)

The third could be the vertex. We need to find the y-coordinate of the vertex. We know that the x-coordinate of the vertex is x = -3

Then, y-coordinate of the vertex is:

$y=-(-3)^2-6(-3)-13=-9+18-13=-4$

The third point we can use is (-3, -4)

Now we can locate them in the cartesian plane:

And that's enough to get the full graph:

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1 year ago

What to do do I subtract the corridnates or add them

trapecia [35]

The easiest way to do it is graph it and count.

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3 years ago

Two athletic teams play a series of games; the first team to win 4 games is declared the overall winner. Suppose that one of the

kodGreya [7K]

Answer:

p=0.7103 (4-game series)
p=0.6480 (2-game series)

Step-by-step explanation:

Let X be the random variable equal the the first 4 straight wins. An overall win for the stronger team implies a negative binomial function with the parameters n=4, p=0.6:

$P(X=4)={{i-1}\choose {4-1}}0.6^40.4^{i-4},\ i=4,5,6,7$

#We find probabilities for the different values of i:

$P(X=4)={3\choose 3}0.6^4=0.1296\\\\P(X=5)={4\choose 3}0.6^40.4^1=0.2074\\\\P(X=6)={5\choose 3}0.6^40.4^2=0.2074\\\\P(X=4)={6\choose 3}0.6^40.4^3=0.1659$

Hence, probability of the stronger team winning overall is:

$=P(X=4)+P(X=5)+P(X=6)+P(X=7)\\\\=0.7103$

#Define Y as the random variable for winning 2/3 games.:

$P(Y=2)={1\choose 1}0.6^2=0.3600\\\\P(Y=3)={2\choose3}0.6^20.4=0.2880\\\\P(win)=0.2880+0.3600=0.6480$

Hence, probability of the stronger team winning in 2 out 3 game series is 0.6480

The stronger team has a higher chance of winning in a 4-game series(0.7103>0.6480)

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3 years ago