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tatyana61 [14]
3 years ago
11

A​ 12-sided die is rolled. The set of equally likely outcomes is {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12} . Find the probability

of rolling a number greater than 1.
Mathematics
2 answers:
Free_Kalibri [48]3 years ago
7 0

Answer:

11/12

Step-by-step explanation:

Greater than 1: 2,3.... 12

P(greater than 1) = 11/12

morpeh [17]3 years ago
3 0

Answer:

11/12 or 0.91666 or 91.6%

Step-by-step explanation:

Any number above one makes up 11 out of the 12 options.

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2 years ago
20 POINTS
chubhunter [2.5K]
Answer: 2 inches, 3 inches, or 3.125 and 2.083


Explanations:
The simplest way is to take 20% of the 2.5 inches and go that much above & below 2.5 inches.

2.5 x 20% = 2.5 x 0.20 = 0.5

So 2.5 - 0.5 = 2 inches was predicted
And 2.5 + 0.5 = 3 inches was predicted.

The more complicated way is to see number + 20% of that number = 2.5, and what number - 20% = 2.5.

Which solution sounds more like what you’re doing in class right now?

If it’s the more complicated way:

0.8x = 2.5 (80% of the predicted rain value equals 2.5)
x = 3.125 inches was predicted

1.2x = 2.5 (120% of the predicted rain value equals 2.5)
x = 2.083 inches was predicted

Sorry, this is probably confusing. Let me know what questions you have.
6 0
3 years ago
An article contained the following observations on degree of polymerization for paper specimens for which viscosity times concen
devlian [24]

Answer:

(a) A 95% confidence interval for the population mean is [433.36 , 448.64].

(b) A 95% upper confidence bound for the population mean is 448.64.

Step-by-step explanation:

We are given that article contained the following observations on degrees of polymerization for paper specimens for which viscosity times concentration fell in a certain middle range:

420, 425, 427, 427, 432, 433, 434, 437, 439, 446, 447, 448, 453, 454, 465, 469.

Firstly, the pivotal quantity for finding the confidence interval for the population mean is given by;

                              P.Q.  =  \frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }  ~ t_n_-_1

where, \bar X = sample mean = \frac{\sum X}{n} = 441

            s = sample standard deviation = \sqrt{\frac{\sum (X-\bar X)^{2} }{n-1} }  = 14.34

            n = sample size = 16

            \mu = population mean

<em>Here for constructing a 95% confidence interval we have used One-sample t-test statistics as we don't know about population standard deviation.</em>

<em />

<u>So, 95% confidence interval for the population mean, </u>\mu<u> is ;</u>

P(-2.131 < t_1_5 < 2.131) = 0.95  {As the critical value of t at 15 degrees of

                                             freedom are -2.131 & 2.131 with P = 2.5%}  

P(-2.131 < \frac{\bar X-\mu}{\frac{s}{\sqrt{n} } } < 2.131) = 0.95

P( -2.131 \times {\frac{s}{\sqrt{n} } } < {\bar X-\mu} < 2.131 \times {\frac{s}{\sqrt{n} } } ) = 0.95

P( \bar X-2.131 \times {\frac{s}{\sqrt{n} } } < \mu < \bar X+2.131 \times {\frac{s}{\sqrt{n} } } ) = 0.95

<u>95% confidence interval for</u> \mu = [ \bar X -2.131 \times {\frac{s}{\sqrt{n} } } , \bar X +2.131 \times {\frac{s}{\sqrt{n} } } ]

                                      = [ 441-2.131 \times {\frac{14.34}{\sqrt{16} } } , 441+2.131 \times {\frac{14.34}{\sqrt{16} } } ]

                                      = [433.36 , 448.64]

(a) Therefore, a 95% confidence interval for the population mean is [433.36 , 448.64].

The interpretation of the above interval is that we are 95% confident that the population mean will lie between 433.36 and 448.64.

(b) A 95% upper confidence bound for the population mean is 448.64 which means that we are 95% confident that the population mean will not be more than 448.64.

6 0
3 years ago
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