There are 3 arms for every 2 eyes. If there are 6 eyes, how many arms are there?

Two pumps connected in parallel fail independently of one another on any given day. The probability that only the older pump wil

Bogdan [553]

Answer : Probability that the pumping system will fail on any given day = 0.005

Explanation:

Probability that only older pump fail = P(A) = 0.10

Probability that only newer pump fail = P(B) = 0.05

Since they are independent events ,

then the conditions will be

P(A∩B) = P(A). P(B)

= 0.10 ×0.05

= 0.005

Hence, the probability will fail on any given day = 0.005

8 0

3 years ago

Find the slope between (-3,1) and (-17,2)

lapo4ka [179]

Answer:

-1/14

Step-by-step explanation:

7 0

3 years ago

Which of the following represents a function?

Svetlanka [38]

Is there a none option? if not then it has to be B.

7 0

3 years ago

Read 2 more answers

Please please help me no links

Slav-nsk [51]

pretty sure it’s 108 because each angle of a pentagon is 108

4 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