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WITCHER [35]
3 years ago
10

What is 12/25 as a percent? (Step by step) Sorry I don’t pay attention :p

Mathematics
2 answers:
ikadub [295]3 years ago
8 0

Answer:

48%

Step-by-step explanation:

explanation in picture

kiruha [24]3 years ago
5 0
12/25 as a percentage is 48%

So, since our denominator in 12/25 is 25, we could adjust the fraction to make the denominator 100. To do that, we divide 100 by the denominator: Now we can see that our fraction is 48/100, which means that 12/25 as a percentage is 48%. Once we have the answer to that division, we can multiply the answer by 100 to make it a percentage: And there you have it! Two different ways to convert 12/25 to a percentage. Both are pretty straightforward and easy to do, but I personally prefer the convert to decimal method as it takes less steps.
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What percent of 60 is 88
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Answer:

95% confidence interval for the true mean score is [4.6 , 6.6].

Step-by-step explanation:

We are given that a sample of 81 tobacco smokers who recently completed a new smoking-cessation program were asked to rate the effectiveness of the program on a scale of 1 to 10.

The average rating was 5.6 and the standard deviation was 4.6.

Firstly, the pivotal quantity for 95% confidence interval for the true mean is given by;

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

where, \bar X = sample average rating = 5.6

            s = sample standard deviation = 4.6

            n = sample of tobacco smokers = 81

            \mu = population mean score

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

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

P(-1.993 < t_8_0 < 1.993) = 0.95  {As the critical value of t at 80 degree of

                                            freedom are -1.993 & 1.993 with P = 2.5%}  

P(-1.993 < \frac{\bar X-\mu}{\frac{s}{\sqrt{n} } } < 1.993) = 0.95

P( -1.993 \times {\frac{s}{\sqrt{n} } } < {\bar X-\mu} < 1.993 \times {\frac{s}{\sqrt{n} } } ) = 0.95

P( \bar X-1.993 \times {\frac{s}{\sqrt{n} } } < \mu < \bar X+1.993 \times {\frac{s}{\sqrt{n} } } ) = 0.95

<u>95% confidence interval for</u> \mu = [ \bar X-1.993 \times {\frac{s}{\sqrt{n} } } , \bar X+1.993 \times {\frac{s}{\sqrt{n} } } ]

                                            = [ 5.6-1.993 \times {\frac{4.6}{\sqrt{81} } } , 5.6+1.993 \times {\frac{4.6}{\sqrt{81} } } ]

                                            = [4.6 , 6.6]

Therefore, 95% confidence interval for the true mean score is [4.6 , 6.6].

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