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3 years ago

A scatter plot is made with the data shown.

Mathematics

2 answers:

crimeas [40]3 years ago

6 0

This is a negative linear association because when ever the x (top line) is increasing and the y (bottom line) is decreasing, that shows that you are going farther to the end of the x axis and lower down the y axis.

ss7ja [257]3 years ago

3 0

Negative linear association

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What is 1/8 as a percent and how did you get it. Also, what is 2/3 as a fraction and how did you get it.

Vesnalui [34]

This is my work. Also 2/3 is a repeating decimal so it’s actually 0.6...

4 0

4 years ago

(x-6)(x+ 3) = A. X^2 - 18 B. X^2 - 3x - 18 C. X^2 + 3x -18 D. X2 + 9x - 18

nlexa [21]

Answer:

x^2 - 3x - 18

Step-by-step explanation:

Just follow in terms of FOIL

First, outer, inner, last. First means multiply the terms which occur first in each binomial, Outer means multiply the outermost terms in the product, Inner means multiply the innermost terms, and Last means multiply the terms which occur last in each binomial.

4 0

3 years ago

Read 2 more answers

Six different second-year medical students at Bellevue Hospital measured the blood pressure of the same person. The systolic re

Sauron [17]

Answer:

What? Hey no no no yes what no no yes what yes yes

7 0

3 years ago

A pair of events are observed to have coordinates (0s,0m) and (50.0s,9.00×109m) in a frame s. what is the proper time interval δ

Ludmilka [50]

To find out time interval δt we need to substract initial time from the final time. In this question first number in the coordinates representes time:

δt=50 - 0
δt= 50s

TIme interval is 50s.

5 0

3 years ago

What are the solutions to the equation

frosja888 [35]

Answer:

$x_1=\frac{1}{4}+(\frac{\sqrt{7}}{4})i$ and $x_2=\frac{1}{4}-(\frac{\sqrt{7} }{4})i$

Step-by-step explanation:

You have the quadratic function $2x^2-x+1=0$ to find the solutions for this equation we are going to use Bhaskara's Formula.

For the quadratic functions $ax^2+bx+c=0$ with $a\neq 0$ the Bhaskara's Formula is:

$x_1=\frac{-b+\sqrt{b^2-4.a.c} }{2.a}$

$x_2=\frac{-b-\sqrt{b^2-4.a.c} }{2.a}$

It usually has two solutions.

Then we have $2x^2-x+1=0$ where a=2, b=-1 and c=1. Applying the formula:

$x_1=\frac{-b+\sqrt{b^2-4.a.c} }{2.a}\\\\x_1=\frac{-(-1)+\sqrt{(-1)^2-4.2.1} }{2.2}\\\\x_1=\frac{1+\sqrt{1-8} }{4}\\\\x_1=\frac{1+\sqrt{-7} }{4}\\\\x_1=\frac{1+\sqrt{(-1).7} }{4}\\x_1=\frac{1+\sqrt{-1}.\sqrt{7}}{4}$

Observation: $\sqrt{-1}=i$

$x_1=\frac{1+\sqrt{-1}.\sqrt{7}}{4}\\\\x_1=\frac{1+i.\sqrt{7}}{4}\\\\x_1=\frac{1}{4}+(\frac{\sqrt{7}}{4})i$

And,

$x_2=\frac{-b-\sqrt{b^2-4.a.c} }{2.a}\\\\x_2=\frac{-(-1)-\sqrt{(-1)^2-4.2.1} }{2.2}\\\\x_2=\frac{1-i.\sqrt{7} }{4}\\\\x_2=\frac{1}{4}-(\frac{\sqrt{7}}{4})i$

Then the correct answer is option C.

$x_1=\frac{1}{4}+(\frac{\sqrt{7}}{4})i$ and $x_2=\frac{1}{4}-(\frac{\sqrt{7} }{4})i$

3 0

3 years ago