Which of the following scatterplots do not show a clear relationship and would not have a trend line?

Mathematics

1 answer:

docker41 [41]2 years ago

5 0

Answer:

the second one

Step-by-step explanation:

it is not going in any general direction

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PLEASE HELP!!!!!!<br> Counts for most of my grade

MrMuchimi

Answer:

the answer is 9/ 15

Step-by-step explanation:

Sin y = p / h

= 9/ 15 (from y reference )

cos x =b/h = 9 / 15

as from x reference b= 9

I hope this will be helpful to you

if yes give 5 star

5 0

3 years ago

Read 2 more answers

Solve 2x^2 + x - 4 = 0 <br> X2 +

damaskus [11]

Answer:

$\large \boxed{\sf \ \ x = -\dfrac{\sqrt{33}+1}{4} \ \ or \ \ x = \dfrac{\sqrt{33}-1}{4} \ \ }$

Step-by-step explanation:

Hello, please find below my work.

$2x^2+x-4=0\\\\\text{*** divide by 2 both sides ***}\\\\x^2+\dfrac{1}{2}x-2=0\\\\\text{*** complete the square ***}\\\\x^2+\dfrac{1}{2}x-2=(x+\dfrac{1}{4})^2-\dfrac{1^2}{4^2}-2=0\\\\\text{*** simplify ***}\\\\(x+\dfrac{1}{4})^2-\dfrac{1+16*2}{16}=(x+\dfrac{1}{4})^2-\dfrac{33}{16}=0$

$\text{*** add } \dfrac{33}{16} \text{ to both sides ***}\\\\(x+\dfrac{1}{4})^2=\dfrac{33}{16}\\\\\text{**** take the root ***}\\\\x+\dfrac{1}{4}=\pm \dfrac{\sqrt{33}}{4}\\\\\text{*** subtract } \dfrac{1}{4} \text{ from both sides ***}\\\\x = -\dfrac{1}{4} -\dfrac{\sqrt{33}}{4} \ \ or \ \ x = -\dfrac{1}{4} +\dfrac{\sqrt{33}}{4}$