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3 years ago

Positive linear relationships are represented by values of the correlation, r, that are _____.

Mathematics

1 answer:

Mandarinka [93]3 years ago

8 0

The answer is greater than zero. hope this helped

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2 years ago

Read 2 more answers

: Xem xét 2 nhà đầu tư A và B với đường cầu về cổ phiếu như sau: Nhà đầu tư A: 2P = 300 – 2Q Nhà đầu tư B: P = 150 – 3Q a. Tại m

ycow [4]

Answer:

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2 years ago

Can someone help me find the answer to these 3 problems if so thank you .

alexandr1967 [171]

Answer:

1. r+s²=49

2. pq²=24

3. -4xy-6x²+2x²y-6y= -54

Step-by-step explanation:

1. r+s²=0+7²=49

2. pq²=(6)(2)²=6(4)=24

3. -4xy-6x²+2x²y-6y= -4(3)(2)-6(3)²+2(3)²(2)-6(2)

= -24-6(9)+2(9)(2)-12

= -24-54+36-12

= -78+36-12

= -42-12

= -54

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2 years ago

From first principles, find the indicated derivatives

LenaWriter [7]

By definition of the derivative,

$\displaystyle\frac{dr}{ds} = \lim_{h\to0} \frac{\left(\frac{(s + h)^3}2 + 1\right) - \left(\frac{s^3}2 + 1\right)}{h}$

$\displaystyle\frac{dr}{ds} = \lim_{h\to0} \frac{\left(\frac{s^3+3s^2h+3sh^2+h^3}2 + 1\right) - \left(\frac{s^3}2 + 1\right)}{h}$

$\displaystyle\frac{dr}{ds} = \lim_{h\to0} \frac{\frac{3s^2h+3sh^2+h^3}2}{h}$

$\displaystyle\frac{dr}{ds} = \lim_{h\to0} \frac12 \frac{3s^2h+3sh^2+h^3}{h}$

$\displaystyle\frac{dr}{ds} = \lim_{h\to0} \frac12 (3s^2+3sh+h^2)$

$\displaystyle\frac{dr}{ds} = \frac{3s^2}2$

6 0

2 years ago

Solve for x<br> x^9=12^2x12^7

jekas [21]

Answer:

Step-by-step explanation:

Hi!

Here is your answer,

I will explain in step by step,

${x}^{9} = {12}^{2} \times {12}^{7}$

Here we can simplify this further using this law,

${a}^{m} \times {a}^{n} = {a}^{m + n}$

So we will get,

${x}^{9} = {12}^{7 + 2} \\ {x}^{9} = {12}^{9}$

Since the powers are same we can cancel it out and we will get,

$x = 12$

(Hope this answer helped :))

4 0

3 years ago