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3 years ago

Find the 67th term of the arithmetic sequence –17, -23, -29,...

Mathematics

2 answers:

Maksim231197 [3]3 years ago

7 0

Answer:

T₆₇ = -413

Step-by-step explanation:

Olin [163]3 years ago

6 0

Answer:

- 413

Step-by-step explanation:

The n th term of an arithmetic sequence is

$a_{n}$ = a₁ + (n - 1)d

where a₁ is the first term and d the common difference

Here a₁ = - 17 and d = - 23 - (- 17) = - 23 + 17 = - 6

$a_{67}$ = - 17 + (66 × - 6)

= - 17 - 396

= - 413

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3 years ago

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Answer:

The Decision Rule

Fail to reject the null hypothesis

The conclusion

There is no sufficient evidence to support the claim that the mean age of the cars is greater than that of taxi

Step-by-step explanation:

From the question we are told that

The data is

Car Ages 4 0 8 11 14 3 4 4 3 5 8 3 3 7 4 6 6 1 8 2 15 11 4 1 6 1 8

Taxi Ages 8 8 0 3 8 4 3 3 6 11 7 7 6 9 5 10 8 4 3 4

The level of significance $\alpha = 0.05$

Generally the null hypothesis is $H_o : \mu_1 - \mu_2 = 0$

the alternative hypothesis is $H_a : \mu_1 - \mu_2 > 0$

Generally the sample mean for the age of cars is mathematically represented as

$\= x_1 = \frac{\sum x_i }{n}$

=> $\= x_1 = \frac{4+ 0+ 8 +11 + \cdots + 8
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=> $\= x_1 = 5.56$

Generally the standard deviation of age of cars

$\sigma _1 = \sqrt{\frac{\sum (x_i - \= x)^2}{n_1} }$

=> $\sigma _1 = \sqrt{\frac{(4 - 5.56)^2 + (0 - 5.56)^2+ (8 - 5.56)^2 + \cdots + 8}{ 27} }$

=> $\sigma _1 = 3.88$

Generally the sample mean for the age of taxi is mathematically represented as

$\= x_2 = \frac{\sum x_i }{n}$

=> $\= x_2 = \frac{8 +8 +0 + \cdots + 4
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=> $\= x_2 = 5.85$

Generally the standard deviation of age of taxi

$\sigma _2 = \sqrt{\frac{\sum (x_i - \= x)^2}{n_1} }$

=> $\sigma _2 = \sqrt{\frac{(8 - 5.85)^2 + (8 - 5.85)^2+ (0 - 5.85)^2 + \cdots + 8}{ 20} }$

=> $\sigma _2 = 2.83$

Generally the test statistics is mathematically represented as

$t = \frac{(\= x_ 1 - \= x_2 ) - 0}{\sqrt{\frac{\sigma^2_1}{n_1} + \frac{\sigma^2_2}{n_2} } }$

=> $t = \frac{(5.56 - 5.85 ) - 0}{\sqrt{\frac{(3.88)^2}{27} + \frac{(2.83)}{20} }}$

=> $t = -0.30$

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$df = n_1 + n_2 -2$

$df = 27 + 20 -2$

$df = 45$

From the t distribution table the $P(t > t )$ at the obtained degree of freedom = 45 is

$P(t > -0.30 ) = 0.61722067$

So the p-value is

$p-value = P(t > T) = 0.61722067$

From the obtained values we see that the p-value > $\alpha$ hence we fail to reject the null hypothesis

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