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Pani-rosa [81]
1 year ago
15

There are four steps for converting the equation x2 + y2 + 12x + 2y – 1 = 0 into standard form by completing the square. complet

e the last step. group the x terms together and the y terms together, and move the constant term to the other side of the equation. x²+ 12x + y²+ 2y = 1 determine (b ÷ 2)2 for the x and y terms. (12 ÷ 2)2 = 36 and (2 ÷ 2)2 = 1 add the values to both sides of the equation. x2 + 12x + 36 + y2 + 2y + 1 = 1 + 36 + 1 write each trinomial as a binomial squared, and simplify the right side. (x + )2 + (y + )2 =
Mathematics
1 answer:
goblinko [34]1 year ago
3 0
Let's first write each step of the procedure:
 Step 1: 
 group the x terms together and the terms and together, and move the constant term to the other side of the equation:
 x² + 12x + y² + 2y = 1
 Step 2:
 
determine (b ÷ 2) 2 for the x and y terms.
 (12 ÷ 2) 2 = 36
 and
 (2 ÷ 2) 2 = 1
 Step 3:
 add the values to both sides of the equation.
 x2 + 12x + 36 + y2 + 2y + 1 = 1 + 36 + 1
 Step 4:
 write each trinomial to binomial squared, and simplify the right side.
 (x + 6) 2 + (y + 1) 2 = 38
 Answer:
 the last step is:
 (x + 6) 2 + (y + 1) 2 = 38
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A botanist collected one leaf at random from each of 10 randomly selected mature maple trees of the same species. The mean and t
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Answer:

One sample t-test for population mean would be the most appropriate method.

Step-by-step explanation:

Following is the data which botanist collected and can use:

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Answer:

(75.1 -72.1) -1.66 \sqrt{\frac{12.8^2}{35} +\frac{14.6^2}{50}}= -1.96

(75.1 -72.1) +1.66 \sqrt{\frac{12.8^2}{35} +\frac{14.6^2}{50}}= 7.96

And the confidence interval would be given by:

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And since the confidence interval contains the value 0 we have enough evidence to conclude that we don't have significant differences between the two means at 10% of significance.

Step-by-step explanation:

For this case we have the following info given :

\bar X_1= 75.1 represent the sample mean for the scores of the undergraduate students

s_1 = 12.8 represent the standard deviation for the undergraduate students

n_1 =35 the sample size for the undergraduate

\bar X_2= 72.1 represent the sample mean for the scores of the high school students

s_2 = 14.6 represent the standard deviation for the high school students

n_2 =50 the sample size for the high school

The confidence interval for the true difference of means is given by:

(\bar X_1 -\bar X_2) \pm t_{\alpha/2} \sqrt{\frac{s^2_1}{n_1} +\frac{s^2_2}{n_2}}

The degrees of freedom are given by:

df=n_1 +n_2 -2= 35+50-2=83

The confidence level is 90% and the significance level is \alpha=0.1 and \alpha/2 =0.05 then the critical value would be:

t_{\alpha/2}= 1.99

And replacing the info we got:

(75.1 -72.1) -1.66 \sqrt{\frac{12.8^2}{35} +\frac{14.6^2}{50}}= -1.96

(75.1 -72.1) +1.66 \sqrt{\frac{12.8^2}{35} +\frac{14.6^2}{50}}= 7.96

And the confidence interval would be given by:

-1.96 \leq \mu_1 -\mu_2 \leq 7.96

And since the confidence interval contains the value 0 we have enough evidence to conclude that we don't have significant differences between the two means at 10% of significance.

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Answer:

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