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mrs_skeptik [129]
3 years ago
14

What is the density of an object having a mass of 8.0 g and a volume of 25 cm655-02-05-00-00_files/i0020000.jpg?

Mathematics
1 answer:
gogolik [260]3 years ago
8 0
The answer is A. 0.32g/cm3
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Anton [14]

Answer:

I would go with the second one.

Step-by-step explanation:

3 0
3 years ago
Hey someone who knows this want to give me the answer thanks
rewona [7]
The answer is C. (x+6)(x+8)
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
Answer the two questions below. Show ALL of your work.
Yuliya22 [10]

Answer A: 84π

Answer B: I'm also confused. If anyone else knows can you please help me also. Sorry for not answering B. I hope this helps though.

Step-by-step explanation:

Answer A:

V=1/3π 6^2 (7) Since 12 is a diameter you half it which is 6

V=1/3π 36(7)

V=1/3π 252

V= 252π/3

V=84π

3 0
3 years ago
In a two week period it was sunny 3/7 of the day's. How many day's were sunny?
netineya [11]
\dfrac{3}{7}\cdot14=3\cdot2=6

6 days
4 0
3 years ago
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