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elena-s [515]
3 years ago
13

A 26-sided die is rolled 5 times. How many different outcomes are possible?

Mathematics
2 answers:
joja [24]3 years ago
8 0

Answer:130

Step-by-step explanation:If you multiply 26 x 5 it is 130.

Greeley [361]3 years ago
5 0
Answer: 130 I’m pretty sure
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You are challenged to a lucky draw game. If you draw a face card (K, Q, J) from a standard deck of cards, you earn 10 points. If
sveticcg [70]
There are 3 face cards for each of the 4 suits, leading to a total possible gain of 120, there are 40 other cards leading to a total possible loss of 80.
120 - 80 = 40.
40 / 52 (total number of cards) = about .77

Final Answer:
The expected value of a draw is positive 0.77 points.
Hope I helped :)
7 0
3 years ago
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Alex777 [14]

Answer: i cant help you because it don't make sense

Step-by-step explanation:

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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
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dedylja [7]

Substituting L = 7.29,

t = 2.01 \sqrt{7.29} \approx \boxed {5.4}

5 0
2 years ago
This is my last question I need to do what’s the answer
MAXImum [283]

Step-by-step explanation:

According to the question,

l+5=2h

5 0
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