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Dima020 [189]
3 years ago
15

Someone help me asap?

Mathematics
1 answer:
laila [671]3 years ago
6 0
4) 8n
5) x=27
6) 2n=22
7)5
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bonufazy [111]

1. 1000 2. 0.001. hope this helps.

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3 years ago
What is 8 divided by 2 2/3
babymother [125]

Answer:

3

Step-by-step explanation:

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3 years ago
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What is 3/10 as a percent
timama [110]
3/10 is represented as 30%.
7 0
3 years ago
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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
Y is such that 4y _ 7&lt;= 3y and 3y &lt;= 5y +8. What range of values of y satisfies both inequalities
fgiga [73]

Answer:

<h3>-4≤y≤7</h3>

Step-by-step explanation:

Given the inequality expressions

4y - 7 ≤ 3y and 3y≤5y+8

For 4y - 7 ≤ 3y

Collect like terms

4y - 3y ≤ 7

y ≤ 7

For  3y≤5y+8

Collect like terms

3y - 5y ≤ 8

-2y ≤ 8

y ≥ 8/-2

y ≥ -4

Combining both solutions

-4≤y≤7

<em>Hence the range of values of y that satisfies both inequalities is -4≤y≤7</em>

6 0
3 years ago
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