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Ugo [173]
3 years ago
14

Which statement is true if the degree of the numerator polynomial function is equal to the degree of the denominator polynomial?

Mathematics
2 answers:
tatiyna3 years ago
8 0

Answer:

degree of the denominator`s polynomial function than:

  lim f ( x ) =  ∞

x→∞

Answer: it has no horizontal asymptote.

Step-by-step explanation:

We know that the horizontal asymptote of a function is obtained as follows:

1)

if the degree of the numerator and denominator are equal then we divide the coefficients of the higher degree term.

2)

If the degree of denominator term is greater than the numerator than the horizontal asymptote is the x-axis.

3)

Also if the numerator is of greater degree as compared to the denominator there are no horizontal asymptote.

Hence, in the given question we are given 3) case.

     Hence, there are no horizontal asymptote.

poizon [28]3 years ago
3 0

Answer: C. It has only one horizontal asymptote.

Step-by-step explanation:

We know that if the degree of the numerator polynomial function is equal to the degree of the denominator polynomial then , there is only one horizontal asymptote at:

y=\dfrac{\text{Leading coefficient of numerator }}{\text{Leading coefficient of deniominator}}

Hence, the statement is true if the degree of the numerator polynomial function is equal to the degree of the denominator polynomial is " It has only one horizontal asymptote."

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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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aleksklad [387]

Answer:

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Answer:

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Step-by-step explanation:

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Answer:

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Step-by-step explanation:

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