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SVETLANKA909090 [29]
3 years ago
13

Write an equation for the nth term of the arithmetic sequence. 2, 5, 8, 11, ...

Mathematics
1 answer:
ivanzaharov [21]3 years ago
4 0

Answer:

An equation for the nth term of the arithmetic sequence.

a_n=3n-1

Step-by-step explanation:

Given the sequence

2,5,8,11,...

An arithmetic sequence has a constant difference 'd' and is defined by  

a_n=a_1+\left(n-1\right)d

here

  • a₁ = 2

computing the differences of all the adjacent terms

  • d = 5-2 = 3, d = 8-5=3, d=11-8=3

Using the nth term formula

a_n=a_1+\left(n-1\right)d

substituting a₁ = 2, d = 3

a_n=2+\left(n-1\right)3

     =2+3n-3

     =3n-1

Thus, an equation for the nth term of the arithmetic sequence.

a_n=3n-1

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Answer:

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p_v =P(z>1.837)=1-P(Z  

Comparing the p value with a significance level for example \alpha=0.05 we see that p_v so we can conclude that we have enough evidence to reject the null hypothesis, and we can say that the average IQ on city A is signficantly higher than city B at 5% of singificance.  

Step-by-step explanation:

\bar X_{A}=120 represent the mean for sample 1

\bar X_{B}=116 represent the mean for sample 2

s_{A}=18 represent the sample standard deviation for 1  

s_{B}=15 represent the sample standard deviation for 2  

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n_{B}=150 sample size for the group 2  

\alpha Significance level provided

z would represent the statistic (variable of interest)  

Concepts and formulas to use  

We need to conduct a hypothesis in order to check if residents of City A smarter on average, the system of hypothesis would be:  

Null hypothesis:\mu_{A}-\mu_{B}\leq 0  

Alternative hypothesis:\mu_{A} - \mu_{B}> 0  

We don't have the population standard deviation's, but the sample sizes are large enough we can apply a z test to compare means, and the statistic is given by:  

z=\frac{(\bar X_{A}-\bar X_{B})-\Delta}{\sqrt{\frac{s^2_{A}}{n_{A}}+\frac{s^2_{B}}{n_{B}}}} (1)

z-test: Is used to compare group means. Is one of the most common tests and is used to determine whether the means of two groups are equal to each other.

With the info given we can replace in formula (1) like this:  

z=\frac{(120-116)-0}{\sqrt{\frac{18^2}{100}+\frac{15^2}{150}}}}=1.837

P value

Since is a one right tailed test the p value would be:  

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