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3 years ago

Factor the polynomial completely.

Mathematics

2 answers:

sveta [45]3 years ago

3 0

Answer:

Option D is correct

Step-by-step explanation:

The original polynomial is:

–2x2 + 2 + 5x3 – 5x

Arranging in decreasing power of x:

$5x^3 - 2x^2 -5x+2$

Factoring the given polynomial by grouping:

$5x^3 - 2x^2 -5x+2\\=5x^3-5x- 2x^2+2\\=5x(x^2-1)-2(x^2-1)\\=(5x-2)(x^2-1)$

Now, (x^2-1) can be further solved using formula:

(a^2-b^2)=(a-b)(a+b)

Solving:

$=(5x-2)(x^2-1)\\=(5x-2)(x-1)(x+1)$

So, $(5x-2)(x-1)(x+1)$ represents the factors of $-2x2 + 2 + 5x3-5x$

Hence Option D is correct.

AnnZ [28]3 years ago

3 0

Answer: the answer is D

Step-by-step explanation:

the screen has proof it is correct... also please STOP deleting my answers its all ways the same person and its getting really annoying when im giving you the correct answer, im just trying to help :(

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The table showing the stock price changes for a sample of 12 companies on a day is contained in the Excel file below.

AfilCa [17]

Answer:

(a) The sample variance for the daily price change is 0.2501.

(b) The sample standard deviation for the daily price change is 0.5001.

Step-by-step explanation:

Let the random variable X denote the stock price changes for a sample of 12 companies on a day.

The data provided is:

X = {0.82 , 1.44 , -0.07 , 0.41 , 0.21 , 1.33 , 0.97 , 0.30 , 0.14 , 0.12 , 0.42 , 0.15}

(a)

The formula to compute the sample variance for the daily price change is:

$s^{2}=\frac{1}{n-1}\sum\limits^{12}_{i=1}{(X_{i}-\bar X)^{2}}$

The sample mean is computed using the formula:

$\bar X=\frac{1}{n}\sum\limits^{12}_{i=1}{X_{i}}$

Consider the Excel output attached below.

In Excel the formula to compute the sample mean and sample variance are:

$\bar X$ =AVERAGE(A2:A13)

$s^{2}$ =VAR.S(A2:A13)

Thus, the sample variance for the daily price change is 0.2501.

(b)

The formula to compute the sample standard deviation for the daily price change is:

$s=\sqrt{\frac{1}{n-1}\sum\limits^{12}_{i=1}{(X_{i}-\bar X)^{2}}}$

Consider the Excel output attached below.

In Excel the formula to compute the sample standard deviation is:

$s$ =STDEV.S(A2:A13)

Thus, the sample standard deviation for the daily price change is 0.5001.

(c)

The (1 - α)% confidence interval for population variance is:

$CI=[\frac{(n-1)s^{2}}{\chi^{2}_{\alpha/2} } \leq \sigma^{2}\leq \frac{(n-1)s^{2}}{\chi^{2}_{1-\alpha/2} } ]$

Compute the critical value of Chi-square for α = 0.05 and (n - 1) = (12 - 1) = 11 degrees of freedom as follows:

$\chi^{2}_{\alpha/2, (n-1)}=\chi^{2}_{0.05/2,11}=21.920$

$\chi^{2}_{1-\alpha/2, (n-1)}=\chi^{2}_{(1-0.05/2),11}=\chi^{2}_{0.975,11}=3.816$

*Use a Chi-square table.

Compute the 95% confidence interval estimates of the population variance as follows:

$CI=[\frac{(n-1)s^{2}}{\chi^{2}_{\alpha/2} } \leq \sigma^{2}\leq \frac{(n-1)s^{2}}{\chi^{2}_{1-\alpha/2} } ]$

$=[\frac{(12-1)\times 0.2501}{21.920 } \leq \sigma^{2}\leq \frac{(12-1)\times 0.2501}{3.816} ]$

$=[0.125506\leq \sigma^{2}\leq 0.720938]\\\approx [0.1255, 0.7210]$

Thus, the 95% confidence interval estimates of the population variance is (0.1255, 0.7210).

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4 years ago

Alexa is buying lace to sew around the edge of a round tablecloth. The radius of the tablecloth is 2 3/4 feet. If lace is sold b

sergey [27]

To find the circumference of a circle the formula is pi times diameter. The radius is half the diameter, so you have to find the diameter by multiplying 2.75 by 2. you get 5.5. Pi is around 3.14, so 5.5 times 3.14 is 17.27. Because you have to convert feet to yards, divide 17.27 by 3 (there are 3 feet in a yard). You get around 5.76. Because if Alexa bought only five meters of lace, she will run short, she needs to buy 6 meters of lace.

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Evaluate 9/m +4 when m =3

iVinArrow [24]

Answer:

the answer is 7

Step-by-step explanation:

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Read 2 more answers

Are the two expressions |x-5| and |x|-5 equivalents

Mama L [17]

Answer:

Step-by-step explanation:

They are not equivalent because in |x-5|, the answer is always positive.

In |x|-5, the answer can be negative if x is less than 5.

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