Level 1 A B C D 1 3 2 4 Enter the correct 4 digit code (no spaces) *

The cost of printing a catalog of the products that your company distributes is a flat rate of $52 plus $0.20 per copy. Let C re

marusya05 [52]

The answer is d, just find the equation and then plug in the first few numbers of copies and you’ll find the correct answer

mark me brainliest if the answer is correct, thank you in advance :)

7 0

3 years ago

Ella has 50 stacks of ten pennies in each stack. Describe how to find how many pennies Ella has in all

k0ka [10]

You would do ten times fifty. which would be 500

6 0

3 years ago

Read 2 more answers

Wich is bigger 1.320. 1.302 < > =

elena-14-01-66 [18.8K]

1.320 >1.302 or 1.302<1.320

4 0

3 years ago

Sooooo bored ask me anything even if it´s not on the list

serg [7]

Answer:

What oets do you like? whats your height u noob

Step-by-step explanation:

6 0

3 years ago

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] .

(c) 90% confidence interval estimate of the population standard deviation is given by ;

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

3 years ago