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Scrat [10]
3 years ago
11

Lisa sold $75 worth of play tickets.Adult tickets cost $4 each and children’s ticket cost $4 each.How many children’s tickets co

uld she have sold?Is there more than one answer
Mathematics
1 answer:
Helga [31]3 years ago
3 0
Yes, there are many possible answers. An equation that can model your situation is:
4a+4b=75, with a being the amount of adult tickets sold, and b being the amount of children’s tickets sold.
Because both tickets cost 4 dollars, you would multiply the amount of tickets sold, by the price (in dollars) of the ticket.
So, you can find possible answers by substituting the amount of adult tickets sold. Say, 4 adults bought tickets. You could then substitute a for 4. So the equation would look like this.
4(4)+4b=75
16 + 4b = 75
4b = 75 - 16
4b = 59
Divide each side by four
B = 14.75

The reason you got a decimal number is because 75 cannot be divided by 4 evenly. So, in this context, you could say that 15 children bought tickets. One child had an issue, and Lisa gave them a 25% discount. This would explain why there are only 75 dollars worth in tickets. I hope I helped, this is my first question I have ever answered so I’m not entirely sure if this was helpful or not. Hope you figure this out!
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Answer:

\chi^2 = \frac{(27-33.33)^2}{33.33}+\frac{(31-33.33)^2}{33.33}+\frac{(42-33.33)^2}{33.33}+\frac{(40-33.33)^2}{33.33}+\frac{(28-33.33)^2}{33.33}+\frac{(32-33.33)^2}{33.33} =5.860

p_v = P(\chi^2_{5} >5.860)=0.32

Since the p value is higher than the significance level assumed 0.05 we FAIL to reject the null hypothesis at 5% of significance, and we can conclude that we can assume that we have equally like results.

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:

Number:      1,    2 ,   3 ,  4 , 5    ,6

Frequency: 27, 31, 42, 40, 28, 32

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

H0: The outcomes are equally likely.

H1: The outcomes are not equally likely.

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

The statistic to check the hypothesis is given by:

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

The observed values are given:

O_{1}=27   O_{2}=31

O_{3}=42  O_{4}=40

O_{5}=28  O_{6}=32

The expected values are given by:

E_{1} =\frac{1}{6}*200=33.33   E_{2} =\frac{1}{6}*200=33.33

E_{3} =\frac{1}{6}*200=33.33   E_{4} =\frac{1}{6}*200=33.33

E_{5} =\frac{1}{6}*200=33.33   E_{6} =\frac{1}{6}*200=33.33

And now we can calculate the statistic:

\chi^2 = \frac{(27-33.33)^2}{33.33}+\frac{(31-33.33)^2}{33.33}+\frac{(42-33.33)^2}{33.33}+\frac{(40-33.33)^2}{33.33}+\frac{(28-33.33)^2}{33.33}+\frac{(32-33.33)^2}{33.33} =5.860

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

df=Categories-1=6-1=5

And we can calculate the p value given by:

p_v = P(\chi^2_{5} >5.860)=0.32

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

"=1-CHISQ.DIST(5.860,5,TRUE)"

Since the p value is higher than the significance level assumed 0.05 we FAIL to reject the null hypothesis at 5% of significance, and we can conclude that we can assume that we have equally like results.

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Vlad1618 [11]
<span>2x squared - 5 
= 2(x)^2 - 5
= 2(4)^2 - 5
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= 32 - 5
= 27</span>
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Answer:

Option three

Step-by-step explanation:

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