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Setler79 [48]
3 years ago
13

I WILL GIVE A LOT OF POINTS AND BRAINLIEST

Mathematics
1 answer:
nasty-shy [4]3 years ago
4 0

Hello there!

First of all, I like your profile picture!

1) There's a lot of different methods for solving a system of equation. But here are my three favorite.

  • Graphing
  • Substitution (Which I preferred to use and I think it is easier).
  • Elimination

2) When solving a system of equations, you can get;

  • One solution
  • No solution
  • Infinite number of solution

3) The graphing method help solving a system of equation by knowing that the solution to a system of equations is the point of intersection of the two lines.

I hope this answer helps. As always, it is my pleasure helping student like you!

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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] .

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