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Eddi Din [679]
3 years ago
5

What is the solution to set the inequality k-8<32

Mathematics
1 answer:
Orlov [11]3 years ago
3 0

Answer:

k <40

Step-by-step explanation:

k-8<32

Add 8 to each side

k-8+8<32+8

k < 40

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A sign is 4 1/2 feet wide and 5 1/4 feet long . Find the area of the surface of the sign.
ZanzabumX [31]

Answer: 94.5

Step-by-step explanation:

9/2 × 21/4

Get common denominator

9 × 2 =18

2 × 2 = 4

18/4 × 21/4

21 × 18 = 378

378/4 = 94.5

7 0
3 years ago
Graph the funxtion y=|x+2|-3
zhenek [66]
First you need to pick a number for y. Lets say you pick 0
Then you need to +3 to each side
Then you can say 3=x+2 or -3=x+2.
To find x you need to subtract 2.
You answer would  be (1,3) and (-5.-3)
5 0
3 years ago
Read 2 more answers
a) Does the Range Rule of Thumb estimate more often underestimate or overestimate the actual standard deviation?
qwelly [4]
Yes it dose hope this helped :)
8 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
Which ordered pair is in the solution set of the system of linear inequalities? y &gt; x – 1 y &lt; x – 1
suter [353]
There are no possible answers to this. 
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3 years ago
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