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3 years ago

The slope of the trend line is 15. What does that mean in regard to the data of the scatterplot? Check all that apply.

Mathematics

2 answers:

rodikova [14]3 years ago

5 0

Answer: A, C, E

Step-by-step explanation:

i just answered it and got it correct

lys-0071 [83]3 years ago

5 0

Answer:

A, C, E.

Step-by-step explanation:

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132/4 = 33, x = 33 boxes

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3 years ago

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100 decreased by a number k equals 44

Arte-miy333 [17]

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Step-by-step explanation:

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2 years ago

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a bowl of candies contains the following makeup of colors: 4 red, 2 yellow, 5 green, and 1 purple. taina is selecting a candy at

Ivanshal [37]

Candies

4 red

2 yellow

5 green

1 purple

- Taina doesn't like red candies.

4/4+2+5+1 = 4/12

4/12 is the probability Taina will get a red candy, but we're looking for the opposite.

12/12 - 4/12 = 8/12 = 2/3

The probability Taina will not select a red candy is 2/3.

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2 years ago

6 The area of a square is 32.5 square inches. Which best represents the length of one side of the square? F 32.5 -16 in. 16 . 5

laiz [17]

Answer:

The answer will be 5.701

Step-by-step explanation:

Find the square root of the area of the square

7 0

2 years ago

Prove that if x is an positive real number such that x + x^-1 is an integer, then x^3 + x^-3 is an integer as well.

Shkiper50 [21]

Answer:

By closure property of multiplication and addition of integers,

If $x + \dfrac{1}{x}$ is an integer

∴ $\left ( x + \dfrac{1}{x} \right) ^3 = x^3 + \dfrac{1}{x^3} +3\cdot \left (x + \dfrac{1}{x} \right )$ is an integer

From which we have;

$x^3 + \dfrac{1}{x^3}$ is an integer

Step-by-step explanation:

The given expression for the positive integer is x + x⁻¹

The given expression can be written as follows;

$x + \dfrac{1}{x}$

By finding the given expression raised to the power 3, sing Wolfram Alpha online, we we have;

$\left ( x + \dfrac{1}{x} \right) ^3 = x^3 + \dfrac{1}{x^3} +3\cdot x + \dfrac{3}{x}$

By simplification of the cube of the given integer expressions, we have;

$\left ( x + \dfrac{1}{x} \right) ^3 = x^3 + \dfrac{1}{x^3} +3\cdot \left (x + \dfrac{1}{x} \right )$

Therefore, we have;

$\left ( x + \dfrac{1}{x} \right) ^3 - 3\cdot \left (x + \dfrac{1}{x} \right )= x^3 + \dfrac{1}{x^3}$

By rearranging, we get;

$x^3 + \dfrac{1}{x^3} = \left ( x + \dfrac{1}{x} \right) ^3 - 3\cdot \left (x + \dfrac{1}{x} \right )$

Given that $x + \dfrac{1}{x}$ is an integer, from the closure property, the product of two integers is always an integer, we have;

$\left ( x + \dfrac{1}{x} \right) ^3$ is an integer and $3\cdot \left (x + \dfrac{1}{x} \right )$ is also an integer

Similarly the sum of two integers is always an integer, we have;

$\left ( x + \dfrac{1}{x} \right) ^3 + \left(- 3\cdot \left (x + \dfrac{1}{x} \right ) \right )$ is an integer

$\therefore x^3 + \dfrac{1}{x^3} = \left ( x + \dfrac{1}{x} \right) ^3 - 3\cdot \left (x + \dfrac{1}{x} \right )= \left ( x + \dfrac{1}{x} \right) ^3 + \left(- 3\cdot \left (x + \dfrac{1}{x} \right ) \right )$ is an integer

From which we have;

$x^3 + \dfrac{1}{x^3}$ is an integer.

4 0

3 years ago