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aleksandrvk [35]

3 years ago

15

PLEASE HELP ME WITH THIS WORKSHEET!!!!!

1 answer:

Debora [2.8K]3 years ago

8 0

Answer:

Sorry, but I can't see the worksheet.

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Please help! 40 points! 2 answer choice questions.<br> will give: Brainliest, Rattings, Thanks, etc.

ivann1987 [24]

Answer:

7. Positive association

8. No association

Step-by-step explanation:

7. In this scatterplot, the points form a rough line that goes higher as the x values increase. In other words, it's tilting up, and its slope is positive. This signifies a positive association or correlation.

8. In this scatterplot, there seems to be no particular pattern; the points are random. Notice how there is no rough line or shape formed by these points. Since there is no relationship between "Salary" and "Height," there is no association or correlation.

7 0

3 years ago

Which is the slope of the line that passes through the points (-2,4 and (5,-1)?

12345 [234]

Use the slope formula which is y2-y1/x2-x1. -1-4/5-(-2)= -5/7

5 0

4 years ago

Read 2 more answers

Find a value for x that makes -x greater than 2x

IrinaVladis [17]

Answer:

-3x

Step-by-step explanation:

When you multiply two negatives, they become positive. The answer can be -3, -4, -5 ..... any negative number more than 2

3 0

4 years ago

Which equation best shows that 55 is a multiple of 11? Choose 1 answer: (Choice A) A 55 = 44 + 1155=44+1155, equals, 44, plus, 1

Charra [1.4K]

Answer: 11*5= 55

Step-by-step explanation:

3 0

3 years ago

Without solving, determine the number of real solutions for each quadratic equation.

masha68 [24]

Answer:

1. x^2 − 4x + 3 = 0

$b^2 -4ac = (-4)^2 -4(1)*(3)= 4 >0$ So we have two real solutions

2. 2n^2 + 7 = −4n + 5

$b^2 -4ac = (4)^2 -4(2)*(2)= 0$ So we just one real solution

3. x − 3x^2 = 5 + 2x − x^2

$b^2 -4ac = (1)^2 -4(2)*(5)= -19$ No real solutions

4. 4x + 7 = x^2 − 5x + 1

$b^2 -4ac = (-9)^2 -4(1)*(-6)= 105 >0$ So we have two real solutions

Step-by-step explanation:

1. x^2 − 4x + 3 = 0

We need to compare this function with the general equation for a quadratic formula given by:

$f(x) = ax^2 + bx + c$

On this case we see that a=1, b = -4 and c =3

We can find the discriminat with the following formula:

$\sqrt{b^2 -4ac}$

$b^2 -4ac = (-4)^2 -4(1)*(3)= 4 >0$ So we have two real solutions

2. 2n^2 + 7 = −4n + 5

We can rewrite the expression like this:

$2n^2 +4n +2$

On this case we see that a=2, b = 4 and c =2

We can find the discriminat with the following formula:

$\sqrt{b^2 -4ac}$

$b^2 -4ac = (4)^2 -4(2)*(2)= 0$ So we just one real solution

3. x − 3x^2 = 5 + 2x − x^2

We can rewrite the expression like this:

$2x^2 +x +5$

On this case we see that a=2, b = 1 and c =5

We can find the discriminat with the following formula:

$\sqrt{b^2 -4ac}$

$b^2 -4ac = (1)^2 -4(2)*(5)= -19$ No real solutions

4. 4x + 7 = x^2 − 5x + 1

We can rewrite the expression like this:

$x^2 -9x -6$

On this case we see that a=1, b = -9 and c =-6

We can find the discriminat with the following formula:

$\sqrt{b^2 -4ac}$

$b^2 -4ac = (-9)^2 -4(1)*(-6)= 105 >0$ So we have two real solutions

8 0

4 years ago