How do you calculate quadratic regression by hand

A polynomial function has a root of -4 with multiplicity of 4, a root of -1 with multiplicity of 3, and a root of 5 with multipl

Dmitrij [34]

Answer:

To draw this graph, we start from the left in quadrant 3 drawing the curve to -4 on the x-axis to touch it but not cross. We continue back down and curve back around to cross the x-axis at -1. We continue up past -1 and curve back down to 5 on the x-axis. We touch here without crossing and draw the rest of our function heading back up. It should form a sideways s shape.

Step-by-step explanation:

A polynomials is an equation with many terms whose leading term is the highest exponent known as degree. The degree or exponent tells how many roots exist. These roots are the x-intercepts.

This polynomial has roots -4, -1, and 5. This means the graph must touch or cross through the x-axis at these x-values. What determines if it crosses the x-axis or the simple touch it and bounce back? The even or odd multiplicity - how many times the root occurs.

In this polynomial:

Root -4 has even multiplicity of 4 so it only touches and does not cross through.

Root -1 has odd multiplicity of 3 so crosses through.

Root 5 has even multiplicity of 6 so it only touches and does not cross through.

Lastly, what determines the facing of the graph (up or down) is the leading coefficient. If positive, the graph ends point up. If negative, the graph ends point down. All even degree graphs will have this shape.

To draw this graph, we start from the left in quadrant 3 drawing the curve to -4 on the x-axis to touch it but not cross. We continue back down and curve back around to cross the x-axis at -1. We continue up past -1 and curve back down to 5 on the x-axis. We touch here without crossing and draw the rest of our function heading back up. It should form a sideways s shape.

7 0

3 years ago

Read 2 more answers

Average movie prices in the United States are, in general, lower than in other countries. It would cost $81.61 to buy three t

777dan777 [17]

3 Japan tickets cost 27.20
3 Switzerland tickets cost 25.88
Hope this helps :)

5 0

3 years ago

Health insurance benefits vary by the size of the company (the Henry J. Kaiser Family Foundation website, June 23, 2016). The sa

xxMikexx [17]

Answer:

$\chi^2 = \frac{(32-42)^2}{42}+\frac{(18-8)^2}{8}+\frac{(68-63)^2}{63}+\frac{(7-12)^2}{12}+\frac{(89-84)^2}{84}+\frac{(11-16)^2}{16}=19.221$

Now we can calculate the degrees of freedom for the statistic given by:

$df=(rows-1)(cols-1)=(3-1)(2-1)=2$

And we can calculate the p value given by:

$p_v = P(\chi^2_{2} >19.221)=0.000067$

And we can find the p value using the following excel code:

"=1-CHISQ.DIST(19.221,2,TRUE)"

Since the p values is higher than a significance level for example $\alpha=0.05$ , we can reject the null hypothesis at 5% of significance, and we can conclude that the two variables are dependent at 5% of significance.

Step-by-step explanation:

Previous concepts

A chi-square goodness of fit test "determines if a sample data matches a population".

A chi-square test for independence "compares two variables in a contingency table to see if they are related. In a more general sense, it tests to see whether distributions of categorical variables differ from each another".

Solution to the problem

Assume the following dataset:

Size Company/ Heal. Ins. Yes No Total

Small 32 18 50

Medium 68 7 75

Large 89 11 100

_____________________________________

Total 189 36 225

We need to conduct a chi square test in order to check the following hypothesis:

H0: independence between heath insurance coverage and size of the company

H1: NO independence between heath insurance coverage and size of the company

The statistic to check the hypothesis is given by:

$\sum_{i=1}^n \frac{(O_i -E_i)^2}{E_i}$

The table given represent the observed values, we just need to calculate the expected values with the following formula $E_i = \frac{total col * total row}{grand total}$

And the calculations are given by:

$E_{1} =\frac{50*189}{225}=42$

$E_{2} =\frac{50*36}{225}=8$

$E_{3} =\frac{75*189}{225}=63$

$E_{4} =\frac{75*36}{225}=12$

$E_{5} =\frac{100*189}{225}=84$

$E_{6} =\frac{100*36}{225}=16$

And the expected values are given by:

Size Company/ Heal. Ins. Yes No Total

Small 42 8 50

Medium 63 12 75

Large 84 16 100

_____________________________________

Total 189 36 225

And now we can calculate the statistic:

$\chi^2 = \frac{(32-42)^2}{42}+\frac{(18-8)^2}{8}+\frac{(68-63)^2}{63}+\frac{(7-12)^2}{12}+\frac{(89-84)^2}{84}+\frac{(11-16)^2}{16}=19.221$

Now we can calculate the degrees of freedom for the statistic given by:

$df=(rows-1)(cols-1)=(3-1)(2-1)=2$

And we can calculate the p value given by:

$p_v = P(\chi^2_{2} >19.221)=0.000067$

And we can find the p value using the following excel code:

"=1-CHISQ.DIST(19.221,2,TRUE)"

Since the p values is higher than a significance level for example $\alpha=0.05$ , we can reject the null hypothesis at 5% of significance, and we can conclude that the two variables are dependent at 5% of significance.

3 0

3 years ago

3/7 to 6/7. what is the percent of change

stepan [7]

Percent change = (new - old)/old * 100
(6/7 - 3/7)/(3/7) * 100
3/7 * 7/3 * 100
1 * 100
100%

4 0

3 years ago

Read 2 more answers

1-1=............... HELP!!!!!!!!!!!!!!!!!!

Verizon [17]

1-1=0 lol how do u not know this ???

7 0