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kvasek [131]
3 years ago
14

Jordan spent a total of 14.85 on a trip to the zoo 2.85 on snacks and the rest on bus fares. How much did she spend on the bus f

ares to and from the zoo
Mathematics
2 answers:
Alika [10]3 years ago
7 0

Answer:

Jordan spent $12 to and from the zoo.

Step-by-step explanation:

since he only used $2.85 on snacks the the rest was for the bus fares so it would be 14.85-2.85. Which equals 12.

Liono4ka [1.6K]3 years ago
6 0

Answer: Hello there!

we know that Jordan spent a total of $14.85 on the trip.

He spends $2.85 on snacks and the rest on bus fares.

the amount spend on bus fares is the difference between the total amount spend and the amount that he spends on snacks, this is:

$14.85 - $2.85 = $12

now we also know that he takes two buses ( one to the zoo, and other from the zoo)

then in two buses, he spended $12, this means that he spends $12/6 = $6 on each bus.

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The slope of AB and the slope of DC based on the information about the coordinate will be -2 and -2 respectively.

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From the information given, we are told to use the slope formula along with the coordinates of points A, B, C, and D to find the slope.

The slope of AB based on the information will be:

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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.

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Pls. see attachment. 

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Area of a triangle = ½ * base * height

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