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forsale [732]
3 years ago
10

Which point to is the solution set of the given system of inequalities?

Mathematics
2 answers:
ryzh [129]3 years ago
8 0
<h2>Answer:</h2>

The point that lie in the solution set of the given system of inequalities is:

                          (0,0)

<h2>Step-by-step explanation:</h2>

We are given a system of inequality as:

          3x+y ≥ -3---------------(1)

and    x+2y ≤ 4----------------(2)

From the given points we will check which satisfies both the inequality and hence the one which satisfies will be a solution.

a)

                (5,0)

on putting in inequality (1) we get:

  15≥  -3

and from inequality (2) we get:

             5≤4

which is incorrect.

Hence, option: a is not a solution.

b)

                 (-2,0)

on putting in inequality (1) we get:

        -6 ≥ -3

which is incorrect( since -6<-3)

Hence, option: b is incorrect.

c)

                          (0,3)

on putting in inequality (1) we get:

                    3 ≥ -3

which is true

on putting in inequality (2) we get:

         6 ≤ 4

which is incorrect.

Hence, option: c is false.

d)

             (0,0)

on putting in inequality (1) we get:

   0 ≥ -3

which is  a true expression.

on putting in inequality (2) we get:

              0 ≤ 4

which is a true expression.

Hence, the point (0,0) will lie in the solution set.

MatroZZZ [7]3 years ago
4 0

Answer:the second one is the answer


Step-by-step explanation:I did this last year


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MrRissso [65]

Answer:

We conclude that the true average percentage of organic matter in such soil is something other than 3% at 10% significance level.

We conclude that the true average percentage of organic matter in such soil is 3% at 5% significance level.

Step-by-step explanation:

We are given a random sample of soil specimens was obtained, and the amount of organic matter (%) in the soil was determined for each specimen;

1.10, 5.09, 0.97, 1.59, 4.60, 0.32, 0.55, 1.45, 0.14, 4.47, 1.20, 3.50, 5.02, 4.67, 5.22, 2.69, 3.98, 3.17, 3.03, 2.21, 0.69, 4.47, 3.31, 1.17, 0.76, 1.17, 1.57, 2.62, 1.66, 2.05.

Let \mu = <u><em>true average percentage of organic matter</em></u>

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The test statistics that will be used here is <u>One-sample t-test statistics</u> because we don't know about the population standard deviation;

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

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             s = sample standard deviation = 1.616%

            n = sample of soil specimens = 30

So, <u><em>the test statistics</em></u> =  \frac{2.481-3}{\frac{1.616}{\sqrt{30} } }  ~ t_2_9

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(a) Now, at 10% level of significance the t table gives a critical value of -1.699 and 1.699 at 29 degrees of freedom for the two-tailed test.

Since the value of our test statistics doesn't lie within the range of critical values of t, so we have <u><em>sufficient evidence to reject our null hypothesis</em></u> as it will fall in the rejection region.

Therefore, we conclude that the true average percentage of organic matter in such soil is something other than 3% at 10% significance level.

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Therefore, we conclude that the true average percentage of organic matter in such soil is 3% at 5% significance level.

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