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Volgvan
2 years ago
6

First five term for An=n^2-2n

Mathematics
1 answer:
densk [106]2 years ago
6 0

First 5 terms are: -1,0,3,8,15

Step-by-step explanation:

we will put n=1,2,3,4,5 to find first 5 terms of the sequence

Explicit formula for sequence is:

a_n = n^2-2n

Putting n=1

a_1 = (1)^2-2(1)\\= 1-2\\= -1

Putting n=2

a_2 = (2)^2-2(2)\\= 4 - 4 \\=0

Putting n=3

a_3 = (3)^2-2(3)\\= 9-6\\= 3

Putting n=4

a_4 = (4)^2 - 2(4)\\= 16-8 \\= 8

Putting n=5

a_5 = (5)^2-2(5)\\= 25 - 10\\= 15

Hence,

First 5 terms are: -1,0,3,8,15

Keywords: Arithmetic sequence, Common difference

Learn more about arithmetic sequence at:

  • brainly.com/question/1349456
  • brainly.com/question/1332667

#LearnwithBrainly

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Answer:

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4.One attorney claims that more than 25% of all the lawyers in Boston advertise for their business. A sample of 200 lawyers in B
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Answer:

z=\frac{0.315 -0.25}{\sqrt{\frac{0.25(1-0.25)}{200}}}=2.123  

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The p value obtained was a very low value 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% of significance the proportion of lawyers had used some form of advertising for their business is significantly higher than 0.25 or 25% .  

Step-by-step explanation:

1) Data given and notation  

n=200 represent the random sample taken

X=63 represent the lawyers had used some form of advertising for their business

\hat p=\frac{63}{200}=0.315 estimated proportion of lawyers had used some form of advertising for their business

p_o=0.25 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 more than 25% of all the lawyers in Boston advertise for their business:  

Null hypothesis:p\leq 0.25  

Alternative hypothesis:p > 0.25  

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{0.315 -0.25}{\sqrt{\frac{0.25(1-0.25)}{200}}}=2.123  

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 right tailed test the p value would be:  

p_v =P(Z>2.123)=0.0169  

The p value obtained was a very low value 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% of significance the proportion of lawyers had used some form of advertising for their business is significantly higher than 0.25 or 25% .  

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