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murzikaleks [220]

2 years ago

11

The _________ measures the strength and direction of the linear relationship between the dependent and the independent variable.

1 answer:

Molodets [167]2 years ago

6 0

Answer:

Correlation Coefficient

Step-by-step explanation:

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NEED HELP WORTH 100 POINTS SHOW YOUR WORK PLEASE

tia_tia [17]

Answer:

Q1:

$-5\sqrt{27c^2}$

$-5\sqrt{27}\sqrt{c^2}$

$-5\times\sqrt{9} \sqrt{3} \sqrt{c^2}$

$-5\times3\sqrt{3}\sqrt{c^2}$

$-5\times3\sqrt{3}c$

$-15\sqrt{3}c$

Q2:

$5\sqrt{72}-3\sqrt{32}$

$5\sqrt{36} \sqrt{2} -3\sqrt{16} \sqrt{2}$

$5\times6\sqrt{2}-3\times4\sqrt{2}$

$30\sqrt{2}-12\sqrt{2}$

$18\sqrt{2}$

Q3:

$\sqrt{108yz}+3\sqrt{98yz}+2\sqrt{75yz}$

$\sqrt{36} \sqrt{3} \sqrt{yz} +3\sqrt{49} \sqrt{2} \sqrt{yz} +2\sqrt{25} \sqrt{3} \sqrt{yz}$

$6\sqrt{3}\sqrt{yz}+21\sqrt{2}\sqrt{yz}+10\sqrt{3}\sqrt{yz}$

$16\sqrt{3}\sqrt{yz}+21\sqrt{2}\sqrt{yz}$

Q4:

$\sqrt{15n^2}\sqrt{10n^3}$

$\sqrt{15}n\sqrt{10n^3}$

$\sqrt{15}n\sqrt{10}\sqrt{n^2}\sqrt{n}$

$\sqrt{15}\sqrt{10}n^2\sqrt{n}$

$\sqrt{5}\sqrt{3}\sqrt{5}\sqrt{2}n^2\sqrt{n}$

$5\sqrt{3}\sqrt{2}n^2\sqrt{n}$

$5\sqrt{6}n^2\sqrt{n}$

Q5:

$\frac{\sqrt{3x^2y^3}}{4\sqrt{5xy^3}}$

$\frac{\sqrt{3}xy\sqrt{y}}{4\sqrt{5}y\sqrt{x}\sqrt{y}}$

Cancel off $\sqrt{x}$ :

$\frac{\sqrt{3}y\sqrt{x}\sqrt{y}}{4\sqrt{5}y\sqrt{y}}$

Cancel off $y:$

$\frac{\sqrt{3}\sqrt{x}\sqrt{y}}{4\sqrt{5}\sqrt{y}}$

Cancel off $\sqrt{y} :$

$\frac{\sqrt{3}\sqrt{x}}{4\sqrt{5}}$

Multiply $\sqrt{5}$ on the numerator and denominator:

$\frac{\sqrt{3}\sqrt{x}\sqrt{5}}{4\sqrt{5}\sqrt{5}}$

$\frac{\sqrt{15}\sqrt{x}}{20}$

5 0

3 years ago

2.5y; when y = -5.2

attashe74 [19]

The answer is -13. You figure it out by multiplying 2.5×-5.2=13

7 0

2 years ago

Pls helppppppppppp nowwww

nika2105 [10]

Answer:

D is the answer. The graph was reflected across the y-axis, and shifted 1 unit downwards.

Step-by-step explanation:

Count the number of spaces to D, then to D'. They are the same (besides one being negative). A'B'C'D' was then shifted down 1 unit, which is why it is lower than shape ABCD.

3 0

2 years ago

Read 2 more answers

Describe the end behavior of the function f(x)=-2x^4- x^3 +3.

Hoochie [10]

Answer:

B

Step-by-step explanation:

For a polynomial , the behaviour as x gets large is determined by the term of greatest degree,

f(x) = - 2 $x^{4}$ - x³ + 3

Is of even degree with a negative leading coefficient , then

f(x) → - ∞ as x → + ∞ and f(x) → - ∞ as x → - ∞

6 0

1 year ago

If cos(θ)=2853 with θin Q IV, what is sin(θ)?

marysya [2.9K]

Answer: $\sin \theta=\frac{-45}{53}$

Step-by-step explanation:

Since we have given that

$\cos\theta=\frac{28}{53}$

And we know that θ is in the Fourth Quadrant.

So, Except cosθ and sec θ, all trigonometric ratios will be negative.

As we know the "Trigonometric Identity":

$\cos^2\theta+\sin^2\theta=1\\\\\sin \theta=\sqrt{1-\cos^2\theta}\\\\\sin \theta=\sqrt{1-(\frac{28}{53})^2}=\sqrt{\frac{53^2-28^2}{53^2}}\\\\\sin \theta=\sqrt{\frac{2025}{53^2}}\\\\\sin \theta=\frac{45}{53}$

It must be negative due to its presence in Fourth quadrant.

Hence, $\sin \theta=\frac{-45}{53}$

7 0

3 years ago