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elena-s [515]
4 years ago
6

(Distribute , then simplify the remaining expression. Final answer must be in standard form) Please help !!

Mathematics
1 answer:
Sergeu [11.5K]4 years ago
8 0
Problem 22
4x^2(2x^2+x-5) - x(x^3+5x^2-3) + 17
4x^2(2x^2)+4x^2(x)+4x^2(-5) - x(x^3)-x(5x^2)-x(-3) + 17
8x^4+4x^3-20x^2 - x^4-5x^3+3x + 17
7x^4-x^3-20x^2+3x + 17

So the final simplified answer to problem 22 is
7x^4-x^3-20x^2+3x + 17

--------------------------------

Problem 23
-2x(x^3-6x^2+6)+4x^3-(5x^4+10x)
-2x(x^3)-2x(-6x^2)-2x(6)+4x^3-(5x^4)-(10x)
-2x^4+12x^3-12x+4x^3-5x^4-10x
-7x^4+16x^3-22x

So problem 23 simplifies to
-7x^4+16x^3-22x

---------------------------------------------------------

Problem 24
4b[2a^2-5(3ab-2b^2)]+29ab^2
4b[2a^2-5(3ab)-5(-2b^2)]+29ab^2
4b[2a^2-15ab+10b^2]+29ab^2
4b[2a^2]+4b[-15ab]+4b[10b^2]+29ab^2
8a^2b-60ab^2+40b^3+29ab^2
8a^2b-31ab^2+40b^3

So problem 24 simplifies to 
8a^2b-31ab^2+40b^3

-------------------------------------------------

Problem 25
2y(6x-4x^2y)+13x^2y^2-6
2y(6x)+2y(-4x^2y)+13x^2y^2-6
12xy-8x^2y^2+13x^2y^2-6
12xy+5x^2y^2-6

The final answer for problem 25 is
12xy+5x^2y^2-6
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Step-by-step explanation:

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A doctor is measuring the average height of male students at a large college. The doctor measures the heights, in inches, of a s
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Answer:

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Step-by-step explanation:

The (1 - <em>α</em>)% confidence interval for population mean is:

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The (1 - α)% confidence interval for population parameter implies that there is a (1 - α) probability that the true value of the parameter is included in the interval.

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The 95% confidence interval for the average height of male students at a large college is, (63.5 inches, 74.4 inches).

The 95% confidence interval for the average height of male students (63.5, 74.4) implies that, there is a 0.95 probability that the true mean height of all male student at the large college is between the interval (63.5, 74.4).

Or, there is a 95% confidence that the true mean height of all male student at the large college is between the interval (63.5, 74.4).

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