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Marizza181 [45]
3 years ago
15

When there is a positive trend (or positive correlation), do the y-values of the points on the scatter plot tend to increase or

decrease as the x-values increase?
Mathematics
2 answers:
mestny [16]3 years ago
7 0
As the x-values increase, y-values should decrease as the rule for positive correlations is that while one variable increases the other one decreases. In this problem's case, the x values are increasing therefore, the y values must decrease or it would not be a positive correlation. 
Assoli18 [71]3 years ago
4 0
As the y-axes decreases the x-axes increases 

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NEED HELP WORTH 100 POINTS SHOW YOUR WORK PLEASE
tia_tia [17]

Answer:

Q1:

-5\sqrt{27c^2}

-5\sqrt{27}\sqrt{c^2}

-5\times\sqrt{9} \sqrt{3} \sqrt{c^2}

-5\times3\sqrt{3}\sqrt{c^2}

-5\times3\sqrt{3}c

-15\sqrt{3}c

Q2:

5\sqrt{72}-3\sqrt{32}

5\sqrt{36} \sqrt{2} -3\sqrt{16} \sqrt{2}

5\times6\sqrt{2}-3\times4\sqrt{2}

30\sqrt{2}-12\sqrt{2}

18\sqrt{2}

Q3:

\sqrt{108yz}+3\sqrt{98yz}+2\sqrt{75yz}

\sqrt{36} \sqrt{3} \sqrt{yz} +3\sqrt{49} \sqrt{2} \sqrt{yz} +2\sqrt{25} \sqrt{3} \sqrt{yz}

6\sqrt{3}\sqrt{yz}+21\sqrt{2}\sqrt{yz}+10\sqrt{3}\sqrt{yz}

16\sqrt{3}\sqrt{yz}+21\sqrt{2}\sqrt{yz}

Q4:

\sqrt{15n^2}\sqrt{10n^3}

\sqrt{15}n\sqrt{10n^3}

\sqrt{15}n\sqrt{10}\sqrt{n^2}\sqrt{n}

\sqrt{15}\sqrt{10}n^2\sqrt{n}

\sqrt{5}\sqrt{3}\sqrt{5}\sqrt{2}n^2\sqrt{n}

5\sqrt{3}\sqrt{2}n^2\sqrt{n}

5\sqrt{6}n^2\sqrt{n}

Q5:

\frac{\sqrt{3x^2y^3}}{4\sqrt{5xy^3}}

\frac{\sqrt{3}xy\sqrt{y}}{4\sqrt{5}y\sqrt{x}\sqrt{y}}

Cancel off \sqrt{x}:

\frac{\sqrt{3}y\sqrt{x}\sqrt{y}}{4\sqrt{5}y\sqrt{y}}

Cancel off y:

\frac{\sqrt{3}\sqrt{x}\sqrt{y}}{4\sqrt{5}\sqrt{y}}

Cancel off \sqrt{y} :

\frac{\sqrt{3}\sqrt{x}}{4\sqrt{5}}

Multiply \sqrt{5} on the numerator and denominator:

\frac{\sqrt{3}\sqrt{x}\sqrt{5}}{4\sqrt{5}\sqrt{5}}

\frac{\sqrt{15}\sqrt{x}}{20}

5 0
3 years ago
a boy is standing on a pole of height 14.7m throws a stone upwards. it moves in a vertical line slightly away from the pole and
fomenos

Answer:

The time taken for the upward motion is 1 second. The same time is taken for the downward motion

It reaches a maximum height of 4.9 meters.

Step-by-step explanation:

The equation of motion is:

x(t) = -4.9t^{2} + 9.8t

Since the term which multiplies t squared is negative, the graph is concave down, that is, x increases until the vertex, where it reaches it's maximum height, then it decreases.

Vertex of a quadratic equation:

Quadratic equation in the format x(t) = at^{2} + bt + c

The vertex is the point (t_{v}, x(t_{v})), in which

t_{v} = -\frac{b}{2a}

In this question:

x(t) = -4.9t^{2} + 9.8t

So a = -4.9, b = 9.8

Vertex:

t_{v} = -\frac{9.8}{2*(-4.9)} = 1

The time taken for the upward motion is 1 second.

x(t_{v}) = x(1) = 9.8*1 - 4.9*(1)^{2} = 4.9

It reaches a maximum height of 4.9 meters.

Downward motion:

From the vertex to the ground.

The ground is t when x = 0. So

-4.9t^{2} + 9.8t = 0

4.9t^{2} - 9.8t = 0

4.9t(t - 2) = 0

4.9t = 0

t = 0

Or

t - 2 = 0

t = 2

It reaches the ground when t = 2 seconds.

The downward motion started at the vertex, when t = 1.

So the duration of the downward motion is 2 - 1 = 1 second.

5 0
2 years ago
Choose the pair(s) of integers that can be substituted into the equation ax2 - 6x + c = 0 so that it has two real solutions.
Maksim231197 [3]

The options are not provided, but method is stated below

Answer:

Quadratic equation ax2 - 6x + c = 0

options would be given for a and c

  1. substitute a and c
  2. check for Discriminant
  3.  b^2 - 4ac \geq  0
  4. 36 -4ac \geq 0

These conditions will fetch us the result required among the options.

Note : the \geq sign will give us the result for Two real unequal solutions and two real equal solutions.  If we only need Real unequal solutions we only use > sign instead of \geq

4 0
2 years ago
An organism measures 2.9 x 10-5 centimeters in diameter. What is this number in standard notation?
stepladder [879]

Answer: The answer is A. Hope this helps :)

Step-by-step explanation:

3 0
3 years ago
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These the answers ❤️.
irina [24]
No ❤️ hahaha i can’t see anything
4 0
2 years ago
Read 2 more answers
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