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3 years ago

15

3. Use the following table to answer questions (a – c).

1 answer:

sineoko [7]3 years ago

7 0

Answer:

See below

Step-by-step explanation:

B) The correlation coefficient is $r=0.75$ , which can be determined by plugging the data into a TI-84 calculator.

C) A correlation coefficient of $r=0.75$ indicates that the correlation between the independent and dependent variable (x and y in this case) is moderately strong with a positive correlation. The closer $r$ is to 1, the stronger the positive correlation. The closer $r$ is to -1, the stronger the negative correlation. If $r$ is closer to 0, then there's no correlation.

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I NEED THIS ASAP ILL GIVE YPU BRAINLIST!

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3 years ago

Find the values of the six trigonometric functions of an angle in standard position if the point with coordinates (84, 13) lies

tia_tia [17]

Answer:

$\sin \theta = 0.153$ , $\cos \theta = 0.988$ , $\tan \theta = 0.155$ , $\cot \theta = 6.462$ , $\sec \theta = 1.012$ , $\csc \theta = 6.538$

Step-by-step explanation:

Let be the point $(x,y)$ , the six trigonometric functions of the angle are represented by following formulas:

$\sin \theta = \frac{y}{\sqrt{x^{2}+y^{2}}}$ (1)

$\cos \theta = \frac{x}{\sqrt{x^{2}+y^{2}}}$ (2)

$\tan \theta = \frac{\sin \theta}{\cos \theta} = \frac{y}{x}$ (3)

$\cot \theta = \frac{1}{\tan \theta} = \frac{x}{y}$ (4)

$\sec \theta = \frac{1}{\cos \theta} = \frac{\sqrt{x^{2}+y^{2}}}{x}$ (5)

$\csc \theta = \frac{1}{\sin \theta} = \frac{\sqrt{x^{2}+y^{2}}}{y}$ (6)

If we know that $x = 84$ and $y = 13$ , then the values of the six trigonometric functions is:

$\sin \theta = \frac{y}{\sqrt{x^{2}+y^{2}}}$

$\sin \theta = 0.153$

$\cos \theta = \frac{x}{\sqrt{x^{2}+y^{2}}}$

$\cos \theta = 0.988$

$\tan \theta = \frac{y}{x}$

$\tan \theta = 0.155$

$\cot \theta = \frac{x}{y}$

$\cot \theta = 6.462$

$\sec \theta = \frac{\sqrt{x^{2}+y^{2}}}{x}$

$\sec \theta = 1.012$

$\csc \theta = \frac{\sqrt{x^{2}+y^{2}}}{y}$

$\csc \theta = 6.538$

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3 years ago

Which expression is equivalent to log w (x^2 -6)^4/ 3 sqrt x^2+8?

evablogger [386]

Answer:

C $4\log_w(x^2-6)-\dfrac{1}{3}\log_w(x^2+8)$

Step-by-step explanation:

First use the property of logarithms

$\log _ab-\log_ac=\log_a\dfrac{b}{c}.$

For the given expression you get

$\log_w\dfrac{(x^2-6)^4}{\sqrt[3]{x^2+8} }=\log_w(x^2-6)^4-\log_w\sqrt[3]{x^2+8}=\log_w(x^2-6)^4-\log_w(x^2+8)^{\frac{1}{3}}$

Now use property of logarithms

$\log_ab^k=k\log_ab.$

For your simplified expression, you get

$\log_w(x^2-6)^4-\log_w(x^2+8)^{\frac{1}{3}}=4\log_w(x^2-6)-\dfrac{1}{3}\log_w(x^2+8).$

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pychu [463]

Because there is no solution if any number divide by 0

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3 years ago