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3 years ago

15

Based on the graph, is there an association between

1 answer:

zhuklara [117]3 years ago

3 0

Answer: There is an association because the distribution of correct guesses is the same among the mint groups.

Step-by-step explanation:

We would have to see the graph but I know it's either B or C because the other answer choices don't make sense. I chose B because it seems like the most logical choice of the two.

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trasher [3.6K]

nuuuu the picture is blocked im on my school computer sorry :(

8 0

3 years ago

What does the linear inequality y ≥ −x − 3 look like?

kupik [55]

It is a line that divides the space into two.. The line is y=-x-3

8 0

3 years ago

Find the value of csc theta, if cos =-3/5; 90 degrees < theta < 180 degrees

Sloan [31]

Answer: $\frac{5}{4}$

Step-by-step explanation:

Given

$\cos \theta =\frac{-3}{5}$

and $\theta$ lies between

$90^{\circ}$

and for this $\theta$ , $\sin$ and $\text{cosec}$ is Positive as they lie in 2 nd Quadrant

$\sin ^2\theta +\cos ^2\theta =1$

$\sin ^2\theta =1-(\frac{-3}{5})^2$

$\sin ^2\theta =1-\frac{9}{25}$

$\sin ^2\theta =\frac{16}{25}$

$\sin \theta =\frac{4}{5}$

$\therefore \text{cosec}\ \theta =\frac{5}{4}$

6 0

3 years ago

The midpoint of AB is M(3, -1). If the coordinates of A are (5,1), what are the

Vera_Pavlovna [14]

Given parameters:

Midpoint of AB = M(3, -1)

Coordinates of A = (5,1)

Unknown:

Coordinates of B = ?

Solution:

To find the mid point of any line, we use the expression below;

$x_{m} = \frac{x_{1} + x_{2} }{2}$ and $y_{m} = \frac{y_{1} + y_{2} }{2}$

where $x_{m}$ and $y_{m}$ = coordinates of the mid points = 3 and -1

x₁ = 5 and y₁ = 1

x₂ = ? and y₂ = ?

Now let us input the variables and solve,

3 = $\frac{5 + x_{2} }{2}$ and -1 = $\frac{1 + y_{2} }{2}$

5 + x₂ = 6 -2 = 1 + y₂

x₂ = 1 y₂ = -2 -1 = -3

The coordinates of B = 1, -3

6 0

3 years ago

Bx^(2)+2b^(2)-b^(3)-2x^(2)

8090 [49]

Answer:

(b+x)(−b+x)(b−2)

Step-by-step explanation:

Factor bx2 + 2b2 − b3 − 2x2 − b3 + bx2 + 2b2 − 2x2 = (b+x)(−b+x)(b−2)

hope this helps! :)

8 0

3 years ago