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3 years ago

Which of the following is not a possible r-value? A.0.3 B.-0.3 C.-1.6 D.0.6

Mathematics

2 answers:

sveta [45]3 years ago

8 0

|r| ≤ 1
selection C cannot be a value of r.

aliya0001 [1]3 years ago

5 0

Answer:

option C $-1.6$

Step-by-step explanation:

we know that

The <u>correlation coefficient</u> r is used in statistics to measure how strong a relationship is between two variables. The value of r is always between $+1$ and $-1$

The value of $-1.6$ is not a possible r-value

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The graph of a function f(x) is shown. What is the value of x where f(x) = -5? And f(0)?

max2010maxim [7]

Answer:

x = -3 when f(x) = -5

f(0) = 3

Step-by-step explanation:

When f(x) = -5, that means they have told you the y value. When y = -5, x = -3.

If the number is in the parentheses of the function, they are giving you the x value. When x = 0, y = 3

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2 years ago

Prove that:

Lorico [155]

A.)

$\csc^2(x) \tan^2 (x)- 1 = \tan^2(x)$

Use the identities $\csc x = 1 / \sin x$ and $\tan x = \sin x / \cos x$ on the left-hand side

$\begin{aligned}
\text{LHS} &= \csc^2(x) \tan^2 (x)- 1 \\
&= \frac{1}{\sin^2 (x)} \cdot \frac{\sin^2 (x)}{\cos^2 (x)} - 1 \\
&= \frac{1}{\cos^2 (x)} - 1
\end{aligned}$

Make 1 have a common denominator to allow for fraction subtraction
Multiply the numerator and denominator of 1 by cos^2 x

$\begin{aligned} \text{LHS} &= \frac{1}{\cos^2 (x)} - 1 \cdot \tfrac{\cos^2 (x)}{\cos^2 (x)} \\
&= \frac{1}{\cos^2 (x)} - \frac{\cos^2 (x)}{\cos^2 (x)} \\
&= \frac{1 - \cos^2 x}{\cos^2 (x)}
\end{aligned}$

Use Pythagorean identity for the numerator.

If $\sin^2 (x) + \cos^2(x) = 1$ then subtracting both sides by $\cos^2 (x)$ yields $\sin^2(x) = 1 - \cos^2(x)$ . We can substitute that into the numerator

$\begin{aligned} \text{LHS} &= \frac{1 - \cos^2 (x)}{\cos^2 (x)} \\
&= \frac{\sin^2 (x)}{\cos^2 (x)} \\
&= \tan^2 (x) && \text{Since } \tan x = \tfrac{\sin x }{\cos x} \\
&= \text{RHS}
\end{aligned}$

======

b.)

$\dfrac{\sec(x)}{\cos(x)} - \dfrac{\tan(x)}{\cot(x)} = 1$

For the left-hand side:
By definition, $\sec(x) = 1/\cos(x)$ and $\tan (x) = 1/\cot (x)$

$\begin{aligned}
\text{LHS} &= \dfrac{\sec(x)}{\cos(x)} - \dfrac{\tan(x)}{\cot(x)} \\
&= \dfrac{ \frac{1}{\cos(x)} }{\cos(x)} - \dfrac{\frac{1}{\cot(x)}}{\cot(x)} \\
&= \frac{1}{\cos^2 (x)} - \frac{1}{\cot^2(x)} 
\end{aligned}$

Since $\cot (x) = \cos (x) / \sin (x)$

$\begin{aligned} \text{LHS} &= \frac{1}{\cos^2 (x)} - \frac{1}{\frac{\cos^2(x)}{\sin^2(x)} } \\ &= \frac{1}{\cos^2 (x)} -\frac{\sin^2(x)}{\cos^2(x)} \\ &= \frac{1 - \sin^2(x)}{\cos^2 (x)} \end{aligned}$

Using Pythagorean identity, $\cos^2(x) = 1 - \sin^2(x)$ so

$\begin{aligned} \text{LHS} &= \frac{\cos^2(x)}{\cos^2 (x)} \\
&= 1 \\
&= \text{RHS}
\end{aligned}$

6 0

3 years ago

If f(1) = 160 and f(n + 1) = –2f(n), what is f(4)?

sashaice [31]

Hello,

f(1)=160
f(2)=160*(-2)
f(3)=160*(-2)²
f(4)=160*(-2)^3=-1280

f(n)=160*(-2)^(n-1)

8 0

3 years ago

Read 2 more answers

Sena determines the remainder of 4x32−14x16+2x3−8x−1, using the remainder theorem. How does she proceed to the correct answer? D

Flauer [41]

Answer: x = 1 and -16

Explanation:

(4x^32 - 14x^16 + 2x^3 - 8)/x - 1

x - 1 = 0
x = 1

f(x) = 4x^32 - 14x^16 + 2x^3 - 8
f(1) = 4(1)^32 - 14(1)^16 + 2(1)^3 - 8
f(1) = 4 - 14 + 2 - 8
f(1) = -10 - 6
f(1) = -16

Therefore, the remainder is -16

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3 years ago

What would x equal for this equation

zaharov [31]

Answer:

20 ft

Step-by-step explanation:

48÷12=4

5×4=20

so 20 is your answer

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3 years ago