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Basile [38]
2 years ago
10

Please help :( Algebra 2 Question

Mathematics
1 answer:
satela [25.4K]2 years ago
8 0

Answer:

number of action figures = 3 + 6(n-1)

at n = 9: number of action figures = 51

Step-by-step explanation:

first row: 3

second row: 6 more than the row before it (3) = 6 + 3 = 9

third row: 6 + 9 = 15

arithmetic series: a_n = a_1 + (n-1)d, where

a_n is the nth term in the output

a_1 is the first output

n is the input

d is the difference between terms

here, we are given the row, and we want to figure out the number of action figures. thus, row = input and number of action figures = output.

the first output, in the first row, is 3

the difference between the number of action figures in each row is 6

thus, our formula is

a_n = 3 + (n-1)6 = 3 + 6(n-1)

when the row is 9, the number of action figures is equal to

3 + 6(9-1) = 3 + 6 * 8 = 51

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Answer:

z=\frac{0.694 -0.75}{\sqrt{\frac{0.75(1-0.75)}{180}}}=-1.735  

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Step-by-step explanation:

1) Data given and notation  

n=180 represent the random sample taken  

X=125 represent the number of americans between 17 to 24 that not qualify for the military

\hat p=\frac{125}{180}=0.694 estimated proportion of americans between 17 to 24 that not qualify for the military

p_o=0.75 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)  

2) Concepts and formulas to use  

We need to conduct a hypothesis in order to test the claim that less than 75% of Americans between the ages of 17 to 24 do not qualify for the military :  

Null hypothesis: p\geq 0.75  

Alternative hypothesis:p < 0.75  

When we conduct a proportion test we need to use the z statistic, 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.  

3) Calculate the statistic  

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

z=\frac{0.694 -0.75}{\sqrt{\frac{0.75(1-0.75)}{180}}}=-1.735  

4) 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  

If we compare the p value obtained and 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% of significance the proportion of americans between 17 to 24 that not qualify for the military is significantly less than 0.75 or 75% .  

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