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12345 [234]
2 years ago
9

PLEASE HELP! GRAPH THE FUNCTION

Mathematics
2 answers:
algol [13]2 years ago
8 0

#8

  • y=4(x-0)²+0

Vertex at (0,0)

#11

  • y=-6(x-0)²+0

Same Vertex

#14

  • y=-(x-0)²+2

Vertex (0,2)

damaskus [11]2 years ago
4 0

Step-by-step explanation:

by order

the 1st one y = 4x^2

vetex (0,0) and upward

the 2nd one y = -6x^2

vertex (0,0) and downward

the 3rd one y = -x^2 + 2

vertex (0,2) and downward

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mixer [17]

There are no real solution

8 0
3 years ago
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What is this?<br><img src="https://tex.z-dn.net/?f=%20%5Cfrac%7Bf%7D%7B4%7D%20%20-%205%20%3D%20%20-%209" id="TexFormula1" title=
tigry1 [53]

Hey there!

<h2>ANSWER: f=-16</h2><h2>EXPLANATION:</h2>

\frac{f}{4}-5=-9

Simplify both sides and you get:

\frac{1}{4}f-5=9

Now add 5 to both sides and you get:

\frac{1}{4}f=-4

Now you have to multiply both sides and you get:

f=-16(ANSWER)

Hope this helps!

\text {-TestedHyperr}

6 0
3 years ago
8x+8(c+8)-6 show work its for my homework i get it but im just making sure i did it right:)
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8x+8c+164-6
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7 0
3 years ago
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Some scientists believe alcoholism is linked to social isolation. One measure of social isolation is marital status. A study of
frez [133]

Answer:

1) H0: There is independence between the marital status and the diagnostic of alcoholic

H1: There is association between the marital status and the diagnostic of alcoholic

2) The statistic to check the hypothesis is given by:

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

3) \chi^2 = \frac{(21-33.143)^2}{33.143}+\frac{(37-41.429)^2}{41.429}+\frac{(58-41.429)^2}{41.429}+\frac{(59-46.857)^2}{46.857}+\frac{(63-58.571)^2}{58.571}+\frac{(42-58.571)^2}{58.571} =19.72

4) 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} >19.72)=5.22x10^{-5}

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

"=1-CHISQ.DIST(19.72,2,TRUE)"

Since the p value is lower than the significance level so then we can reject the null hypothesis at 5% of significance, and we can conclude that we have association between the two variables analyzed.

Step-by-step explanation:

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".

Assume the following dataset:

                    Diag. Alcoholic   Undiagnosed Alcoholic    Not alcoholic    Total

Married                     21                              37                            58                116

Not Married              59                             63                            42                164

Total                          80                             100                          100              280

Part 1

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

H0: There is independence between the marital status and the diagnostic of alcoholic

H1: There is association between the marital status and the diagnostic of alcoholic

The level os significance assumed for this case is \alpha=0.05

Part 2

The statistic to check the hypothesis is given by:

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

Part 3

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{80*116}{280}=33.143

E_{2} =\frac{100*116}{280}=41.429

E_{3} =\frac{100*116}{280}=41.429

E_{4} =\frac{80*164}{280}=46.857

E_{5} =\frac{100*164}{280}=58.571

E_{6} =\frac{100*164}{280}=58.571

And the expected values are given by:

                    Diag. Alcoholic   Undiagnosed Alcoholic    Not alcoholic    Total

Married             33.143                       41.429                        41.429                116

Not Married     46.857                      58.571                        58.571                164

Total                   80                              100                             100                 280

And now we can calculate the statistic:

\chi^2 = \frac{(21-33.143)^2}{33.143}+\frac{(37-41.429)^2}{41.429}+\frac{(58-41.429)^2}{41.429}+\frac{(59-46.857)^2}{46.857}+\frac{(63-58.571)^2}{58.571}+\frac{(42-58.571)^2}{58.571} =19.72

Part 4

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} >19.72)=5.22x10^{-5}

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

"=1-CHISQ.DIST(19.72,2,TRUE)"

Since the p value is lower than the significance level so then we can reject the null hypothesis at 5% of significance, and we can conclude that we have association between the two variables analyzed.

7 0
3 years ago
What is the volume of a rectangular prism with a height of 22 1/2 centimeters and a base area of 562 1/2 square centimeters?
umka2103 [35]

The correct answer is D. You can solve this by multiplying the values.

7 0
2 years ago
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