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Papessa [141]
4 years ago
6

A student reader is making $8.10 per hour and gets a $0.66 raise. What is the percent increase? (Round to the nearest tenth of a

percent.)
Mathematics
1 answer:
kicyunya [14]4 years ago
4 0
The answer is 8.2 percent
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Annual starting salaries for college graduates with degrees in business administration are generally expected to be between $30,
saveliy_v [14]

Answer:

The value is   n =1351

Step-by-step explanation:

From the question we are told that

  The standard deviation is  \sigma  =  \$3750

   The margin of error is  E = \$ 200

From the question we are told the confidence level is  95% , hence the level of significance is    

      \alpha = (100 - 95 ) \%

=>   \alpha = 0.05

Generally from the normal distribution table the critical value  of  \frac{\alpha }{2} is  

   Z_{\frac{\alpha }{2} } =  1.96

Generally the sample size is mathematically represented as

         n = [\frac{Z_{\frac{\alpha }{2} } *  \sigma }{E} ] ^2

=>      n = [1.96  *  3750 }{200} ] ^2

=>      n =1351

6 0
3 years ago
Value for m: 10m=10-5(m-7
Nataly [62]
I think the answer is 3 
6 0
3 years ago
Read 2 more answers
a consumer magazine counts the number of tissues per box in a random sample of 15 boxes of No- Rasp facial tissues. The sample s
Alex

Answer:

95% confidence interval for the population variance of the number of tissues per box is [5043.11 , 23401.31].

Step-by-step explanation:

We are given that a consumer magazine counts the number of tissues per box in a random sample of 15 boxes of No- Rasp facial tissues. The sample standard deviation of the number of tissues per box is 97.

Firstly, the pivotal quantity for 95% confidence interval for the population variance is given by;

                         P.Q. = \frac{(n-1)s^{2} }{\sigma^{2} }  ~ \chi^{2}__n_-_1

where, s^{2}  = sample variance = 97^{2} = 9409

              n = sample of boxes = 15

           \sigma^{2}  = population variance

<em>Here for constructing 95% confidence interval we have used chi-square test statistics.</em>

So, 95% confidence interval for the population variance, \sigma^{2} is ;

P(5.629 < \chi^{2}__1_4 < 26.12) = 0.95  {As the critical value of chi-square at 14

                                         degree of freedom are 5.629 & 26.12}  

P(5.629 < \frac{(n-1)s^{2} }{\sigma^{2} } < 26.12) = 0.95

P( \frac{5.629 }{(n-1)s^{2} } < \frac{1}{\sigma^{2} } < \frac{26.12 }{(n-1)s^{2} } ) = 0.95

P( \frac{(n-1)s^{2} }{26.12 } < \sigma^{2} < \frac{(n-1)s^{2} }{5.629 } ) = 0.95

<u><em>95% confidence interval for</em></u> \sigma^{2} = [ \frac{(n-1)s^{2} }{26.12 } , \frac{(n-1)s^{2} }{5.629 } ]

                                                  = [ \frac{14 \times 9409 }{26.12 } , \frac{14 \times 9409 }{5.629 } ]

                                                  = [5043.11 , 23401.31]

Therefore, 95% confidence interval for the population variance of the number of tissues per box is [5043.11 , 23401.31].

6 0
3 years ago
39 is 84 percent of which number ​
tamaranim1 [39]

X= (39 * 100)/84

X= 46.42

8 0
3 years ago
Determine the value of x.<br><br><br><br> 20°<br> 30°<br> 45°<br> 60°
mars1129 [50]

30 is the value of x given of above quertion

4 0
3 years ago
Read 2 more answers
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