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pashok25 [27]
3 years ago
7

Given: ∠AOC, ∠BOC - complementary angles

Mathematics
1 answer:
34kurt3 years ago
3 0

Answer:

  • 30
  • COB . . . . or . . . . BOC

Step-by-step explanation:

The reason given on the line of interest is "substitution," so the problem boils down to determining what was substituted for what. The previous statement says ...

  AOC + COB = 90

and the first part of the statement we're to complete has BOC + ___.

We see from the given statements that m∠AOC = m∠BOC + 30°, so it appears that is the substitution that has been made: AOC has been replaced by BOC + 30.

This means the first blank is filled with 30.

__

The second part of the previous statement is ...

  AOC + COB = 90

so we believe this (COB) should go in the second blank.

__

Then the line of interest would read ...

  BOC + 30 + COB = 90

_____

<em>Comment on the problem</em>

There is a curious mix of notations here. Usually, (as in the beginning of this problem) we refer to the measure of an angle using "m∠" in front of the angle designator, and we use a degree symbol to indicate the units of that measure. Part-way through the problem statement written here, those notations were dropped, and we're to assume they are intended. IMO, this is a poor way to demonstrate careful problem solving.

The substitution given for AOC is BOC+30, but the line into which that is substituted has AOC +COB = 90. This means the equation after substitution is ...

  BOC +30 +COB = 90

Since BOC and COB are the same angle, we can sort of fudge the "algebra" to get to  BOC=30, but if the problem were more carefully written, the angle would be referred to by consistent nomenclature:

  m∠AOC + m∠BOC = 90° . . . . . . . . . preferred angle designations

  (m∠BOC + 30°) + m∠BOC = 90° . . . . substitution for m∠AOC

  2(m∠BOC) = 60° . . . . . . algebra (subtract 30°, collect terms)

  m∠BOC = 30° . . . . . . . . algebra (divide by 2)

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A college infirmary conducted an experiment to determine the degree of relief provided by three cough remedies. Each cough remed
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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
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