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jasenka [17]
3 years ago
14

Please help. I’ll mark you as brainliest if correct!

Mathematics
1 answer:
andreev551 [17]3 years ago
8 0

Answer:

The system is dependent:

x=-3t-7

y=-5t-15

z=t

Step-by-step explanation:

I chose to use a matrix to solve this system of equations. Once put into matrix form, you need to row reduce the system into its simplest form (Row Reduced Echelon form). Doing this, we find that the system is dependent on the z variable. And following usual procedures, we let z equal some other letter; which is t in this case. Then we isolate each variable to get the answer.

Check the attachment for the work.

[The arrows indicate a row swap and the parenthesis indicates addition if a constant multiple of one row to another]

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A college student is interested in investigating the TV-watching habits of her classmates and surveys 20 people on the number of
Alla [95]

Answer:

80% confidence interval of the true average number of hours of TV watched per week is [8.28 hours, 11.02 hours].

Step-by-step explanation:

We are given that a college student is interested in investigating the TV-watching habits of her classmates and surveys 20 people on the number of hours they watch per week. The results are provided below;

<u>Hours of TV per week (X)</u>: 6, 14, 13, 6, 16, 10, 19, 4, 5, 5, 18, 8, 7, 14, 8, 8, 9, 12, 6, 5.

Firstly, the Pivotal quantity for 80% confidence interval for the true average is given by;

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

where, \bar X = sample mean number of hours of TV watched per week = \frac{\sum X}{n} = 9.65

            s = sample standard deviation = \sqrt{\frac{\sum (X -\bar X)^{2} }{n-1} }  = 4.61

            n = sample of people = 20

           \mu = true average number of hours of TV watched per week

<em>Here for constructing 80% confidence interval we have used One-sample t-test statistics as we don't know about population standard deviation.</em>

<u>So, 80% confidence interval for the true average, </u>\mu<u> is ;</u>

P(-1.33 < t_1_9 < 1.33) = 0.80  {As the critical value of t at 19 degrees of

                                               freedom are -1.33 & 1.33 with P = 10%}  

P(-1.33 < \frac{\bar X-\mu}{\frac{s}{\sqrt{n} } } < 1.33) = 0.80

P( -1.33 \times {\frac{s}{\sqrt{n} } } < {\bar X-\mu} < 1.33 \times {\frac{s}{\sqrt{n} } } ) = 0.80

P( \bar X-1.33 \times {\frac{s}{\sqrt{n} } } < \mu < \bar X+1.33 \times {\frac{s}{\sqrt{n} } } ) = 0.80

<u>80% confidence interval for</u> \mu = [ \bar X-1.33 \times {\frac{s}{\sqrt{n} } } , \bar X+1.33 \times {\frac{s}{\sqrt{n} } } ]

                                         = [ 9.65-1.33 \times {\frac{4.61}{\sqrt{20} } } , 9.65+1.33 \times {\frac{4.61}{\sqrt{20} } } ]

                                         = [8.28 hours, 11.02 hours]

Therefore, 80% confidence interval of the true average number of hours of TV watched per week is [8.28 hours, 11.02 hours].

7 0
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A. 15
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Step-by-step explanation:

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The simplification form of the provided expression is 2y⁴/x⁴ option (A) 2y⁴/x⁴ is correct after applying the integer exponent properties.

<h3>What is an integer exponent?</h3>

In mathematics, integer exponents are exponents that should be integers. It may be a positive or negative number. In this situation, the positive integer exponents determine the number of times the base number should be multiplied by itself.

The question is incomplete.

The complete question is attached in the picture, please refer picture.

We have an expression:

= \rm \dfrac{\left(14x^4y^6\right)}{\left(7x^8y^2\right)}

\rm =\dfrac{7\cdot \:2x^4y^6}{7x^8y^2}

\rm =\dfrac{2x^4y^6}{x^8y^2}

\rm =\dfrac{2y^4}{x^4}

Thus, the simplification form of the provided expression is 2y⁴/x⁴ option (A) 2y⁴/x⁴ is correct after applying the integer exponent properties.

Learn more about the integer exponent here:

brainly.com/question/4533599

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