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Stels [109]
2 years ago
10

Bilet ulgowy do cyrku jesto 10 zł tańszy od biletu normalnego za 3 bilety normalne i 2 ulgowe zapłacono 160 zł ile należało by z

apłacić za 2 bilety normalnei 1 bilet ulgowy
Mathematics
1 answer:
kipiarov [429]2 years ago
3 0

Answer:

Kwota zapłacona za 2 bilety normalne i 1 bilet ulgowy wynosi 98 złotych

Step-by-step explanation:

Po pierwsze, zadeklarujmy zmienne. Niech kwota zapłacona za normalne bilety wyniesie n złotych, a kwota zapłacona za bilety zniżkowe będzie d złotych

Od pierwszej części pytania powiedziano nam, że kwota za bilet ulgowy jest o 10 złotych mniejsza niż w przypadku biletu normalnego.

Stąd matematycznie n = d + 10 ......... (i)

Mówi się nam również, że kwota zapłacona za 3 normalne bilety i 2 bilety ze zniżką wynosi 160 złotych

Matematycznie można to zapisać jako;

3n + 2d = 160 .................... (ii)

Teraz, aby uzyskać kwotę zapłaconą za dwa bilety normalne i jeden bilet zniżkowy, musimy najpierw znaleźć kwotę zapłaconą za każdy bilet, korzystając z dwóch powyższych równań. Teraz zamień (i) na (ii)

mamy; 3 (d + 10) + 2d = 160

3d + 30 + 2d = 160

5d + 30 = 160

5d + 160-30

5d = 130

d = 130/5

d = 26 złotych

Kwota zapłacona za normalny bilet wynosi zatem = 26 + 10 = 36 złotych

Znajdując kwotę zapłaconą za 2 bilety normalne i 1 bilet zniżkowy, mamy 2 (36) + 26 = 72 + 26 = 98 złotych

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Last year, 46% of business owners gave a holiday gift to their employees. A survey of business owners indicated that 35% plan to
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Answer:

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b) Since is a left tailed test the p value would be:  

p_v =P(Z  

c) If we compare the p value obtained and using the significance level given \alpha=0.05 we have p_v so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 5% the proportion of business owners providing holiday gifts had decreased from the 2008 level.  

We can use as smallest significance level 0.044 and we got the same conclusion.  

Step-by-step explanation:

Data given and notation  

n=60 represent the random sample taken

X represent the business owners plan to provide a holiday gift to their employees

\hat p=0.35 estimated proportion of business owners plan to provide a holiday gift to their employees

p_o=0.46 is the value that we want to test

\alpha=0.05 represent the significance level

Confidence=95% or 0.95

z would represent the statistic (variable of interest)

p_v represent the p value (variable of interest)  

Part a

On this case w ejust need to multiply the value of th sample size by the proportion given like this:

Number = 60 *0.35=21

2) Part b

We need to conduct a hypothesis in order to test the claim that the proportion of business owners providing holiday gifts had decreased from the 2008 level:  

Null hypothesis:p\geq 0.46  

Alternative hypothesis:p < 0.46  

When we conduct a proportion test we need to use the z statisitc, and the is given by:  

z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}} (1)  

The One-Sample Proportion Test is used to assess whether a population proportion \hat p is significantly different from a hypothesized value p_o.

Calculate the statistic  

Since we have all the info requires we can replace in formula (1) like this:  

z=\frac{0.35 -0.46}{\sqrt{\frac{0.46(1-0.46)}{60}}}=-1.710  

Statistical decision  

It's important to refresh the p value method or p value approach . "This method is about determining "likely" or "unlikely" by determining the probability assuming the null hypothesis were true of observing a more extreme test statistic in the direction of the alternative hypothesis than the one observed". Or in other words is just a method to have an statistical decision to fail to reject or reject the null hypothesis.  

The significance level provided \alpha=0.05. The next step would be calculate the p value for this test.  

Since is a left tailed test the p value would be:  

p_v =P(Z  

Part c

If we compare the p value obtained and using the significance level given \alpha=0.05 we have p_v so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 5% the proportion of business owners providing holiday gifts had decreased from the 2008 level.  

We can use as smallest significance level 0.044 and we got the same conclusion.  

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