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Drupady [299]
3 years ago
7

Which of the following polynomials corresponds to the subtraction of the multivariate polynomials 19x 3 + 44x 2 y + 17 and y 3 -

11xy 2 + 2x 2 y - 13x 3? y 3 - 6x 3 + 33x 2y + 2xy 2 + 17 20x3 - y3 + 33x 2y + 2xy 2 + 17 31x 3 - 6x 3 + 44x 2y + 11xy 2 + 17 32x 3 - y 3 + 42x 2y + 11xy 2 + 17
Mathematics
1 answer:
mixer [17]3 years ago
6 0
(19x^3+44x^2y+17)-(y^3-11xy^2+2x^2y-13x^3)

We can use the distributive property to distribute the subtraction sign:
19x^3+44x^2y+17-y^3--11xy^2-2x^2y--13x^3
When we subtract a negative it is the same as adding a positive:
19x^3+44x^2y+17-y^3+11xy^2-2x^2y+13x^3
Gather the like terms:
19x^3+13x^3+44x^2y-2x^2y+17-y^3+11xy^2
\\
\\=32x^3+42x^2y+17-y^3+11xy^2
Rearrange the polynomial in order of powers to x, greatest to least:
32x^3+42x^2y+11xy^2-y^3+17
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Two statistics teachers both believe that each has the smarter class. To put this to the test, they give the same final exam to
SVETLANKA909090 [29]

Answer:

We conclude that there is no difference between the two classes.

Step-by-step explanation:

We are given that two statistics teachers both believe that each has a smarter class.

A summary of the class sizes, class means, and standard deviations is given below:n1 = 47, x-bar1 = 84.4, s1 = 18n2 = 50, x-bar2 = 82.9, s2 = 17

Let \mu_1 = mean age of student cars.

 \mu_2 = mean age of faculty cars.

So, Null Hypothesis, H_0 : \mu_1=\mu_2      {means that there is no difference in the two classes}  

Alternate Hypothesis, H_A : \mu_1\neq \mu_2      {means that there is a difference in the two classes}

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

                         T.S.  =  \frac{(\bar X_1-\bar X_2)-(\mu_1-\mu_2)}{s_p \times \sqrt{\frac{1}{n_1}+\frac{1}{n_2} } }   ~  t_n_1_+_n_2_-_2

where, \bar X_1 = sample mean age of student cars = 8 years

\bar X_2  = sample mean age of faculty cars = 5.3 years  

s_1 = sample standard deviation of student cars = 3.6 years  

s_2 = sample standard deviation of student cars = 3.7 years  

n_1 = sample of student cars = 110  

n_2 = sample of faculty cars = 75  

Also, s_p=\sqrt{\frac{(n_1-1)\times s_1^{2}+(n_2-1)\times s_2^{2} }{n_1+n_2-2} }  = \sqrt{\frac{(47-1)\times 18^{2}+(50-1)\times 17^{2} }{47+50-2} }  = 17.491

So, <u><em>the test statistics</em></u> =  \frac{(84.4-82.9)-(0)}{17.491 \times \sqrt{\frac{1}{47}+\frac{1}{50} } }  ~  t_9_5

                                     =  0.422    

The value of t-test statistics is 0.422.

<u>Now, the P-value of the test statistics is given by;</u>

P-value = P(t_9_5 > 0.422) = From the t table it is clear that the P-value will lie somewhere between 40% and 30%.

Since the P-value of our test statistics is way more than the level of significance of 0.04, so <u><em>we have insufficient evidence to reject our null hypothesis</em></u> as our test statistics will not fall in the rejection region.

Therefore, we conclude that there is no difference between the two classes.

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3 years ago
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Step-by-step explanation:

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