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Rus_ich [418]
2 years ago
12

What are prime factors of 1000?

Mathematics
1 answer:
Ede4ka [16]2 years ago
4 0

Answer:

1,2,4,5,8,10,20,25,40,50,10,125,200,250,500,1000

Step-by-step explanation:

prime factorization is

1000=2*2*2*5*5*5

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At an aquarium, a shrimp if fed 1/5 gram of food each feeding and is fed 3 times each day. What is the constant of proportionali
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Answer:

The answer is most likely 5 shrimp eats 5 grams of food im one day,

Step-by-step explanation:

At an aquarium, a shrimp if fed 1/5 gram of food each feeding and is fed 3 times each day. <em>What is the constant of proportionality and what does it tell us about the situation?</em>

So, We know the shrimp is fed 1/5 gram of food each feeding, And we know the shrimp is fed 3 times each day.

Now, We need to answer the question, <u><em>What is the constant of proportionality and what does it tell us about the situation?</em></u>

<u><em /></u>

Then, That means '5, shrimp eats 5 grams of food in one day ' is correct.

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An article contained the following observations on degree of polymerization for paper specimens for which viscosity times concen
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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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