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2 years ago

Central limit theorem: Sample proportion

Mathematics

1 answer:

marysya [2.9K]2 years ago

8 0

Answers wiebsuwbsyjwvs he dustbdysfais he eibssavabsbsusv god blesses you

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Complete each statement in the steps to solve x2 – 6x – 7 = 0 using the process of completing the square. Isolate the constant b

Tomtit [17]

"Isolate the constant by adding 7 to both sides of the equation."
This step separates the non-squareable 7 and the squareable $x^2 - 6x$ .

"Add 9 to both sides of $x^2 - 6x = 7$ to form a perfect square trinomial while keeping the equation balanced."
After separating the non-squareable, add the number which makes the first or left side a perfect square trinomial. The formula to find the number is: $c = {\frac{b}{2} }^{2}$ .
When we plug the values: $c = { \frac{ - 6}{2} }^{2}$
Simplify: $c = {( - 3)}^{2} = 9$

"Write the trinomial $x^2 - 6x + 9$ as $(x - 3)$ squared."
When you factor $x^2 - 6x + 9$ , you will get $(x - 3) \times (x - 3)$ .

"Use the square root property of equality to get $x - 3 = ± \sqrt{16}$ ."
The 16 is coming from the part when we add 9. We needed 9 on the left side for a perfect square, but to protect the balance of the equality, we need to add 9 to the right side too. When we add 7 and 9, we got 16, and that is where it came from.

"Isolate the variable x to get solutions of -1 and 7."
To isolate x we branched the plus-minus sign:
$x - 3 = 4 \: \: \: x = 4 + 3 = 7 \\ \\ x - 3 = - 4 \: \: \: x = - 4 + 3 = - 1$

8 0

2 years ago

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Carson buys eggs and lemons at the store. he pays a total of $39.62. he pays $6.74 for the eggs. he buys 8 bags of lemons that e

Nikolay [14]

The algebraic equation is used to determine the costs of each bag of lemons is 8x = $39.62 - $6.74

According to the question,

Cost of 1 bag of lemons = $x

Cost of 8 bags of lemons = 8*x = 8x

Cost of 8 bags of lemons = Total cost - cost of the egg

8x = $39.62 - $6.74

Hence, the algebraic equation is used to determine the costs of each bag of lemons is 8x = $39.62 - $6.74.

Learn more about algebraic equation here

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8 0

1 year ago

\int (x+1)\sqrt(2x-1)dx

Nezavi [6.7K]

Answer:

$\int (x+ 1) \sqrt{2x-1} dx = \frac{1}{3}(x+1) (2x - 1)^{\frac{3}{2} } - \ \frac{1}{15}(2x-1)^{\frac{5}{2}} + C$

Step-by-step explanation:

$\int (x+1)\sqrt {(2x-1)} dx\\Integrate \ using \ integration \ by\ parts \\\\u = x + 1, v'= \sqrt{2x - 1}\\\\v'= \sqrt{2x - 1}\\\\integrate \ both \ sides \\\\\int v'= \int \sqrt{2x- 1}dx\\\\v = \int ( 2x - 1)^{\frac{1}{2} } \ dx\\\\v = \frac{(2x - 1)^{\frac{1}{2} + 1}}{\frac{1}{2} + 1}} \times \frac{1}{2}\\\\v= \frac{(2x - 1)^{\frac{3}{2}}}{\frac{3}{2}} \times \frac{1}{2}\\\\v = \frac{2 \times (2x - 1)^{\frac{3}{2}}}{3} \times \frac{1}{2}\\\\v = \frac{(2x - 1)^{\frac{3}{2}}}{3}$

$\int (x+1)\sqrt(2x-1)dx\\\\ = uv - \int v du$

$= (x +1 ) \cdot \frac{(2x - 1)^{\frac{3}{2}}}{3} - \int \frac{(2x - 1)^{\frac{3}{2}}}{3} dx \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ [ \ u = x + 1 => du = dx \ ]$

$= \frac{1}{3}(x+1) (2x - 1)^{\frac{3}{2} } - \ \frac{1}{3} \int (2x - 1)^{\frac{3}{2}}} dx\\\\= \frac{1}{3}(x+1) (2x - 1)^{\frac{3}{2} } - \ \frac{1}{3} \times ( \frac{(2x-1)^{\frac{3}{2} + 1}}{\frac{3}{2} + 1}) \times \frac{1}{2}\\\\= \frac{1}{3}(x+1) (2x - 1)^{\frac{3}{2} } - \ \frac{1}{3} \times ( \frac{(2x-1)^{\frac{5}{2}}}{\frac{5}{2} }) \times \frac{1}{2}\\\\= \frac{1}{3}(x+1) (2x - 1)^{\frac{3}{2} } - \ \frac{1}{15} \times (2x-1)^{\frac{5}{2}} + C\\\\$

6 0

2 years ago

I was working on this question and and a bit confused, somebody plz help!! :)

ladessa [460]

The answer should be 6s^3 because you add the two powers together since the number (6) is the same

5 0

3 years ago

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A table tennis ball has a diameter of 40 mm.What is the approximate volume of the table tennis ball in cubic millimeters?

Alexxandr [17]

33510.32mm to the power of 3

7 0

2 years ago