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UNO [17]
3 years ago
13

At the beginning of the school year, Jamie had

Mathematics
1 answer:
VARVARA [1.3K]3 years ago
4 0

Answer:

Hmmm let me solve then I will answer

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What should be multiplied with -25/36 to get -5/9​
Ierofanga [76]

Answer:

\frac{4}{5}

Step-by-step explanation:

\frac{-\frac{5}{9}}{-\frac{25}{36}}=\\\\-\frac{5}{9}*(-\frac{36}{25})=\\\\\frac{4}{5}

6 0
2 years ago
The next model of a sports car will cost 5.1% more than the current model. The current model costs $45,000. How much will the pr
alukav5142 [94]

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Step-by-step explanation:

43091

8 0
3 years ago
How many times do you need to multiply by ten to get from 19.797 to 1979.7
borishaifa [10]
You need to multiply 19.797 by 10 two times to reach 1979.7
5 0
2 years ago
Read 2 more answers
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
If the airplane had to return to the airport, how far would it have to travel?
valentina_108 [34]

Step-by-step explanation:

Draw a line connecting the plane to the airport

then using Pythagoras theorem u can get d distance from the plane to d airport

4 0
2 years ago
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