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Luda [366]
3 years ago
9

Stanford form to y=Mx+b What is 6x-8y=32 in y=Mx+b form?

Mathematics
1 answer:
OverLord2011 [107]3 years ago
5 0

Answer:

y=3/4x-4 and ignore this stuff ejdndjsienensmsk

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What is the solution set
Free_Kalibri [48]

We graphically solve f(x)=g(x) by graphing f(x) and g(x) and eyeballing the points where they intersect each other.  

Really none of our choices are the solution set listing all the (x,y) pairs where the two functions meet because none of them is a set.

Anyway we count from the origin five to the right and three up to the meet of the lines, which is the point (5,3), last choice.
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3 years ago
A certain city covers an area of 42.5 mi? How many square kilometers is this?
klasskru [66]

Answer:

go in creative and count how many blocks there are

One of the many points of contention between Federalists, who advocated a strong national government, and Anti-Federalists, who wanted power to remain with state and local governments, was the Constitution’s lack of a bill of rights that would place specific limits on government power. Federalists argued that the Constitution did not need a bill of rights, because the people and the states kept any powers not given to the federal government. Anti-Federalists held that a bill of rights was necessary to safeguard individual liberty.

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The House approved 17 amendments. Of these, the Senate approved 12, which were sent

Step-by-step explanation:

6 0
3 years ago
Read 2 more answers
A college infirmary conducted an experiment to determine the degree of relief provided by three cough remedies. Each cough remed
KiRa [710]

Answer:

p_v = P(\chi^2_{4,0.05} >3.81)=0.43233

Since the p values is higher than the significance level we FAIL to reject the null hypothesis at 5% of significance, and we can conclude that we don't have significant differences between the 3 remedies analyzed. So we can say that the 3 remedies ar approximately equally effective.

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:

                               NyQuil          Robitussin       Triaminic     Total

No relief                    11                     13                      9               33

Some relief               32                   28                     27              87

Total relief                7                       9                      14               30

Total                         50                    50                    50              150

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

H0: There is no difference in the three remedies

H1: There is a difference in the three remedies

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

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*33}{150}=11

E_{2} =\frac{50*33}{150}=11

E_{3} =\frac{50*33}{150}=11

E_{4} =\frac{50*87}{150}=29

E_{5} =\frac{50*87}{150}=29

E_{6} =\frac{50*87}{150}=29

E_{7} =\frac{50*30}{150}=10

E_{8} =\frac{50*30}{150}=10

E_{9} =\frac{50*30}{150}=10

And the expected values are given by:

                               NyQuil          Robitussin       Triaminic     Total

No relief                    11                     11                       11               33

Some relief               29                   29                     29              87

Total relief                10                     10                     10               30

Total                         50                    50                    50              150

And now we can calculate the statistic:

\chi^2 = \frac{(11-11)^2}{11}+\frac{(13-11)^2}{11}+\frac{(9-11)^2}{11}+\frac{(32-29)^2}{29}+\frac{(28-29)^2}{29}+\frac{(27-29)^2}{29}+\frac{(7-10)^2}{10}+\frac{(9-10)^2}{10}+\frac{(14-10)^2}{10} =3.81

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

df=(rows-1)(cols-1)=(3-1)(3-1)=4

And we can calculate the p value given by:

p_v = P(\chi^2_{4,0.05} >3.81)=0.43233

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

"=1-CHISQ.DIST(3.81,4,TRUE)"

Since the p values is higher than the significance level we FAIL to reject the null hypothesis at 5% of significance, and we can conclude that we don't have significant differences between the 3 remedies analyzed.

4 0
3 years ago
Help please, look at picture for question
alexgriva [62]

Answer:

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Step-by-step explanation:

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3 years ago
Plz help me out a bit :(<br> which one do you think it is?
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C. 9^4
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8 0
3 years ago
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