Use the method of completing the square to transform the quadratic equation into the equation form (x + p)2 = q.

1 answer:

5 0

$\text{Use:}(*)\ (a+b)^2=a^2+2ab+b^2\\\\2x^2+2x+1=0\ \ \ \ |:2\\\\x^2+x+0.5=0\\\\x^2+2\cdot x\cdot0.5+0.5=0\\\\\underbrace{x^2+2\cdot x\cdot0.5+0.5^2}_{(*)}-0.5^2+0.5=0\\\\(x+0.5)^2-0.25+0.5=0\\\\(x+0.5)^2+0.25=0\ \ \ \ |-0.25\\\\(x+0.5)^2=-0.25\to D)$

Answer: D) (x + 0.5)2 = -0.25

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Two different simple random samples are drawn from two different populations. The first sample consists of 30 people with 16 hav

Furkat [3]

Answer:

There is no significant evidence that p1 is different than p2 at 0.01 significance level.

99% confidence interval for p1-p2 is -0.171 ±0.237 that is (−0.408, 0.066)

Step-by-step explanation:

Let p1 be the proportion of the common attribute in population1

And p2 be the proportion of the same common attribute in population2

$H_{0}$ : p1-p2=0

$H_{a}$ : p1-p2≠0

Test statistic can be found using the equation:

$z=\frac{p2-p1}{\sqrt{{p*(1-p)*(\frac{1}{n1} +\frac{1}{n2}) }}}$ where

p1 is the sample proportion of the common attribute in population1 ( $\frac{16}{30} =0.533$ )

p2 is the sample proportion of the common attribute in population2 ( $\frac{1337}{1900} =0.704$ )

p is the pool proportion of p1 and p2 ( $\frac{16+1337}{30+1900}=0.701$ )

n1 is the sample size of the people from population1 (30)

n2 is the sample size of the people from population2 (1900)

Then $z=\frac{0.704-0.533}{\sqrt{{0.701*0.299*(\frac{1}{30} +\frac{1}{1900}) }}}$ ≈ 2.03

p-value of the test statistic is 0.042>0.01, therefore we fail to reject the null hypothesis. There is no significant evidence that p1 is different than p2.

99% confidence interval estimate for p1-p2 can be calculated using the equation

p1-p2± $z*\sqrt{\frac{p1*(1-p1)}{n1}+\frac{p2*(1-p2)}{n2}}$ where