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Lostsunrise [7]
3 years ago
6

Find the quotient of n(x) = 6x^5y^7 and m(x) = 2x^3y

Mathematics
1 answer:
likoan [24]3 years ago
6 0

Answer:

a.  Yes

b.  Yes

c. Yes

d. Degree 8

Step-by-step explanation:

a. Yes, n(x) is a polynomial of one single term (also called monomial) because it contains variables raised to positive integers.

b. Yes, m(x) is a polynomial of also one single term (also called monomial) because it contains variables raised to positive integers.

c. The quotient of n(x) / m(x) can be reduced to a polynomial of one single term as follows:

\frac{n(x)}{m(x)} =\frac{6\,x^5\,y^7}{2\,x^3\,y} =3\,x^2\,y^6

which as can be seen, also contains variables raised to positive integers.

d. The degree of the polynomial resultant is the addition of the powers of all variables present (x and y) which results in: 2 + 6 = 8

Therefore the degree of this polynomial is 8.

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Complete the chart to find the mean, variance, and standard deviation. Remember to use commas and round numbers to the nearest t
vitfil [10]

Answer: The mean of the data is 433.75, variance of the data is 99667.19 and the standard deviation of the data is 315.7011.

Explanation:

The given data is 900, 35, 500 and 300.

The number of observation is 4.

Formula of mean is,

\text{Mean}=\frac{\text{Sum of observations}}{\text{Number of observations}}

\bar{X}=\frac{900+35+50+300}{4} =433.75

The formula of variance is given below,

\sigma^2=\frac{1}{n}\sum{(X-\bar{X})^2}

\sigma^2=\frac{398668.75}{4}

\sigma^2=99667.19

The variance of the data is 99667.19

\sigma=\sqrt{99667.19}

\sigma=315.7011

The standard deviation of the data is 315.7011.

The other information or values of given chart is shown in the attached table.

8 0
3 years ago
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Convert –2y2 + x – 4y + 6 = 0 into standard form.
Rufina [12.5K]

Answer:A

Step-by-step explanation:

X=2(y+1)^2-6

4 0
3 years ago
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Mrs. McCarthy bought 1.5 pounds of chicken from Smiths. She paid a total of $6.75 for the chicken. What was the price per pound?
REY [17]

$ 6.75/ 1.5 lbs

$4.50 per lb

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For this question you will need to analyze data given in file seed Emergence. For your convenience, the data is posted as a text
Mrac [35]

Answer:

A) You will get the Mean, Sample variance and standard deviation, Minimum, maximum, range, and Boxplot.

B) The two-way ANOVA (analysis of variance) model is used for the two factors.

C) Outputs are obtained

D) Here, P-value (= 0.0377) < α (= 0.05). There is evidence that the treatments differ with respect to emerging plants.

Step-by-step explanation:

(A)  

Use the MegaStat add-in in Excel to draw the boxplot and to find; the group counts, means, standard deviations, variances, minimums, maximums, and ranges for the 5 treatment groups.

In another Excel worksheet. Enter the data in 5 different columns, each column representing a treatment, and the first row holding the treatment names.

Go to Add-Ins > MegaStat > Descriptive Statistics.

Enter Sheet1!$A$1:$E$5 in Input range.

Tick on Mean, Sample variance and standard deviation, Minimum, maximum, range, and Boxplot.

Click OK.

(B)

Since there are two factors affecting the outcomes- the five treatments (Control, Arasan, Spergon, Semesan, and Fermate), and the four blocks, the two-way ANOVA (analysis of variance) model must be used.

(C)

We have used the Data Analysis tool-pack in Excel to construct the analysis of variance table.

We have arranged the data and entered it as follows:

              Control,    Arasan,    Spergon,    Semesan,    Fermate

Block 1:        86           98             96               97              91

Block 2:       90           94             90               95              93

Block 3:       88           93             91                91               95

Block 4:       87           89             92               92              95

Go to Data > Data Analysis > Anova: Two-Factor Without Replication > OK.

In Input Range, enter $A$1:$E$6; tick on labels, enter Alpha as 0.05, and click OK.

The following output is obtained. Note that the analysis of variance table is given under “ANOVA” in the output.

(D)

In the analysis of variance table above, the “Rows” under “Sources of Variation” correspond to the treatments (as the observations under each treatment are noted along a row), whereas the “Columns” title relates to the block effects.

The P-value for Rows, hat is, for the treatments is 0.0377 (4 decimal places).

The level of significance is given as α = 0.05.

In this case, the null hypothesis is that, there is no significant difference between the 5 treatments, and the alternative hypothesis should be that, not all the treatments have the same effect.

The rejection rule for a test using the P-value is: Reject the null hypothesis, if P-value ≤ α. Otherwise, fail to reject the null hypothesis.

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3 years ago
Please help i am confused
son4ous [18]

Answer: (4)

Step-by-step explanation:

The parts listed in the congruence statements don't correspond, so they aren't necessarily congruent.

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