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oee [108]
3 years ago
12

During a baseball season, team a scores 9 runs for every 7 runs that team b scores. The total number of runs scored by both team

s is 1440. How many run does each team score?
Mathematics
1 answer:
mrs_skeptik [129]3 years ago
5 0

Answer:

∴ Team b scores is 630

And team a scores is 810

Step-by-step explanation:

Given;

Team a scores 9 runs for every  7 runs that team b scores.

Lets,

Think b is scores  x runs;

Then a is scores x+(2\times\frac{x}{7} );

Because every 7 runs of b , a is scores more 2 runs.

Total number of runs scored by both teams is 1440

Runs score by team a + Runs score by team b=1440

x+(2\times\frac{x}{7} )+  x =1440

2\times x+ \frac{2\times x}{7}=1440

\frac{(14\times x)+(2\times x)}{7} =1440

\frac{16\times x}{7} =1440

16\times x=1440\times 7

x=\frac{10080}{16}

x=630

∴ Team b scores is 630

And team a scores is 630+(2\times \frac{630}{7} )=810

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g(x) = 2x-6

f(x) = -4x +7

(g•f)(x) = g(f(x))

= 2(f(x)) - 6

= 2 ( -4x+7) -6

= -8x + 14 -6

= -8x +8

now

(g•f)(1) = -8(1) + 8= -8+8

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so option a is answer

4 0
3 years ago
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:

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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
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Answer:

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Step-by-step explanation:

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Answer:

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Andre45 [30]

Answer:

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Step-by-step explanation:

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8 0
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