Write the sentence as an equation. 323 decreased by d equals 369

How is the Recursive Formula used in real life

OlgaM077 [116]

Answer:

People often sort stacks of documents using a recursive method. For example, imagine you are sorting 100 documents with names on them. First place documents into piles by the first letter, then sort each pile. Looking up words in the dictionary is often performed by a binary-search-like technique, which is recursive.

Step-by-step explanation:

This is just an example.

If this helps please mark as brainliest

4 0

2 years ago

What is the volume of the portion of the sphere in which the students' favorite color is red? Round your answer to the nearest t

Finger [1]

Answer:

.28

Step-by-step explanation:

4 0

3 years ago

Read 2 more answers

Help, quick! Worth 100 points

finlep [7]

Answer:

the answer is (C) 2(1-2x)=x-3

4 0

2 years ago

Solve the proportion x/8= 3/12. Please help.

galben [10]

Answer:

2

Step-by-step explanation:

$\frac{x}{8} = \frac{3}{12} \\ \\ x = \frac{8 \times 3}{12} \\ \\ x = \frac{24}{12}\\ \\ x = 2$

5 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