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KatRina [158]
3 years ago
8

Alyssa has a goal to read a certain number of pages each week. So far she has read 60 pages, which is 15% of her goal. What is h

er goal?
Mathematics
1 answer:
Arte-miy333 [17]3 years ago
4 0

400 because divide 60 by 15 in order to get 1% of the unknown number, which is 4. Then, just multiply 4 by 100 to get 100% of the number, which equals 400.

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To solve this problem, you can use the compound interest formula. 

Here it is once you plug in the numbers: A = 5,000,000 ( 1 + .04) ^30

Just solve the equation to find the solution.


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3 years ago
Can you help this is my little sisters work
Olin [163]

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There are 15 ounces left in the bucket.

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Starting with 30 ounces.

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Hope this helped. :)

7 0
3 years ago
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
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