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cestrela7 [59]
3 years ago
5

What is the factored form of polynomial 6xy-8x^2

Mathematics
2 answers:
frutty [35]3 years ago
6 0
Well we first have to find the GCF of this polynomial, the GCF is the maximum of numbers or variables that the polinomial share, in this case this two sets of terms only share 1 x so lets take it out x(6y-8x) this cant be simplyfied any further so your answer should be x(6y-8x)

Hope this helps
gogolik [260]3 years ago
4 0
I believe your answer should be x(6y-8x), I believe so :) I hope this helps! :) 
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Answer : 2/5, 1/2, 6/10
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A random sample of soil specimens was obtained, and the amount of organic matter (%) in the soil was determined for each specime
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>

So, Null Hypothesis, H_0 : \mu = 3%      {means that the true average percentage of organic matter in such soil is 3%}

Alternate Hypothesis, H_A : \mu \neq 3%      {means that the true average percentage of organic matter in such soil is something other than 3%}

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

where, \bar X = sample mean percentage of organic matter = 2.481%

             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

                                     =  -1.76

The value of t-test statistics is -1.76.

(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.

(b) Now, at 5% level of significance the t table gives a critical value of -2.045 and 2.045 at 29 degrees of freedom for the two-tailed test.

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

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

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Answer:

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Step-by-step explanation:

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Paladinen [302]

Answer:

a) 117.4

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Step-by-step explanation:

We are given the following data set in the question:

108.6, 117.4, 128.4, 120.0, 103.7, 112.0, 98.3, 121.5, 123.2

Median:\\\text{If n is odd, then}\\\\Median = \displaystyle\frac{n+1}{2}th ~term \\\\\text{If n is even, then}\\\\Median = \displaystyle\frac{\frac{n}{2}th~term + (\frac{n}{2}+1)th~term}{2}

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Sorted Values: 98.3, 103.7, 108.6, 112.0, 117.9, 120.0, 121.5, 123.2, 128.4

New Median =

\dfrac{9 + 1}{2}^{th}\text{ term} = 5^{th}\text{ term} = 117.9

c) Since median is a position based descriptive statistics, a small change in values can bring a change in the median value as the order of the data may change.

Option A)  When there is rounding or grouping, the median can be highly sensitive to small change

7 0
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