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2 years ago

What is the nth term rule of the linear sequence below? 15 , 7 , − 1 , − 9 , − 17 , . . .

Mathematics

1 answer:

Doss [256]2 years ago

7 0

You can see that you're constantly subtracting 8 to get the next number. So, we have

$a_1 = 15$

$a_2 = a_1-8$

$a_3 = a_2-8 = (a_1-8)-8=a_1-2\cdot 8$

$a_4=a_1-3\cdot 8$

$a_5=a_1-4\cdot 8$

So, as you can see, to get the n-th term we must subtract (n-1) times 8 from the original number.

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2 years ago

The decimal .67 can be written as which fraction? A) 67 10,000 B) 67 1,000 C) 67 100 D) 67 10

ss7ja [257]

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2 years ago

Read 2 more answers

Because of staffing decisions, managers of the Gibson-Marimont Hotel are interested in

Law Incorporation [45]

Answer:

(a) 900

(b) [567.35 , 1689.72]

(c) [23.82 , 41.11]

Step-by-step explanation:

We are given that a sample of 20 days of operation shows a sample mean of 290 rooms occupied per day and a sample standard deviation of 30 rooms i.e.;

Sample mean, $xbar$ = 290 Sample standard deviation, s = 30 and Sample size, n = 20

(a) Point estimate of the population variance is equal to sample variance, which is the square of Sample standard deviation ;

$\sigma^{2}$ = $s^{2}$ = $30^{2}$

$\sigma^{2}$ = 900

(b) 90% confidence interval estimate of the population variance is given by the pivotal quantity of $\frac{(n-1)s^{2} }{\sigma^{2} }$ ~ $\chi^{2} __n_-_1$

P(10.12 < $\chi^{2}__1_9$ < 30.14) = 0.90 {At 10% significance level chi square has critical

values of 10.12 and 30.14 at 19 degree of freedom}

P(10.12 < $\frac{(n-1)s^{2} }{\sigma^{2} }$ < 30.14) = 0.90

P( $\frac{10.12}{(n-1)s^{2} }$ < $\frac{1 }{\sigma^{2} }$ < $\frac{30.14}{(n-1)s^{2} }$ ) = 0.90

P( $\frac{(n-1)s^{2} }{30.14}$ < $\sigma^{2}$ < $\frac{(n-1)s^{2} }{10.12}$ ) = 0.90

90% confidence interval for $\sigma^{2}$ = $[\frac{19s^{2} }{30.14} , \frac{19s^{2} }{10.12}]$

= $[\frac{19*900 }{30.14} , \frac{19*900 }{10.12}]$

= [567.35 , 1689.72]

Therefore, 90% confidence interval estimate of the population variance is [567.35 , 1689.72] .

P( $\sqrt{\frac{(n-1)s^{2} }{30.14}}$ < $\sigma$ < $\sqrt{\frac{(n-1)s^{2} }{10.12}}$ ) = 0.90

90% confidence interval for $\sigma$ = $[\sqrt{\frac{19s^{2} }{30.14}} , \sqrt{\frac{19s^{2} }{10.12}} ]$

= [23.82 , 41.11]

Therefore, 90% confidence interval estimate of the population standard deviation is [23.82 , 41.11] .

7 0

2 years ago