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Aleks [24]
2 years ago
7

The scatter plot shows the number of absences in a week for classes of different sizes. Trevor concluded that there is a positiv

e correlation between class size and the number of absences.
image

Which of the following statements BEST describes why Trevor’s conclusion was incorrect?

A.
The data shows a negative relationship between class size and number of absences.

B.
The data shows no relationship between class size and the number of absences.

C.
The largest class does not have the most absences.

D.
The smallest class does not have the least number of absences.
Mathematics
1 answer:
QveST [7]2 years ago
8 0
This is right the answer is A.
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KonstantinChe [14]

Answer:

(2, -2)

Step-by-step explanation:

Given the equation

x - y = 4 ... 1

6x+y = 10 ...2

Add both expressions

x+6x = 4 + 10

7x = 14

x = 14/7

x = 2

Substitute x = 2 into equation 1;

From 1;

x - y = 4

2 - y = 4

y = 2-4

y = -2

Hence the solution to the system of equation is (2, -2)

4 0
3 years ago
What is 80 ÷ (-50)?
Harrizon [31]

Answer:

- 1.6

Step-by-step explanation:

If divide these to numbers without the negative, you'll get 1.6

Then you can add the negative sign to the 50 and 1.6 and you'll get the correct answer... btw I'm just lazy and that's my way of dividing negative numbers.

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2 years ago
Help!!! Please Answer
adell [148]

Answer:

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6 0
3 years ago
Pleasee help mee<br><br>12sin(x)-5cos(x)=6,5​
abruzzese [7]

We turn -5,12 into polar coordinates.  It's a Pythagorean Triple so

r = 13   Ф=arctan(-12/5) + 180°   ( in the second quadrant )

so -5 = 13 cos Ф, 12 = 13 sin Ф

12 sin x - 5 cos x = 6.5

13 sinФ sin x + 13 cos Ф cos x = 6.5  

13 cos(x - Ф) = 6.5

cos(x - Ф) = 1/2

cos(x - Ф) = cos 60°

x - Ф = ± 60° + 360° k     integer k

x = Ф ± 60° + 360° k  

x = 180° + arctan(-12/5) ± 60° + 360° k  

That's the exact answer;

x ≈  180° - 67.38° ± 60° + 360° k  

x ≈  122.62° ± 60° + 360° k  

x ≈  { 62.62°, 182.62°} + 360° k,  integer k

3 0
2 years ago
PLEASE HELP 100 POINTS!!!!!!
horrorfan [7]

Answer:

A)  See attached for graph.

B)  (-3, 0)  (0, 0)  (18, 0)

C)   (-3, 0) ∪ [3, 18)

Step-by-step explanation:

Piecewise functions have <u>multiple pieces</u> of curves/lines where each piece corresponds to its definition over an <u>interval</u>.

Given piecewise function:

g(x)=\begin{cases}x^3-9x \quad \quad \quad \quad \quad \textsf{if }x < 3\\-\log_4(x-2)+2 \quad  \textsf{if }x\geq 3\end{cases}

Therefore, the function has two definitions:

  • g(x)=x^3-9x \quad \textsf{when x is less than 3}
  • g(x)=-\log_4(x-2)+2 \quad \textsf{when x is more than or equal to 3}

<h3><u>Part A</u></h3>

When <u>graphing</u> piecewise functions:

  • Use an open circle where the value of x is <u>not included</u> in the interval.
  • Use a closed circle where the value of x is <u>included</u> in the interval.
  • Use an arrow to show that the function <u>continues indefinitely</u>.

<u>First piece of function</u>

Substitute the endpoint of the interval into the corresponding function:

\implies g(3)=(3)^3-9(3)=0 \implies (3,0)

Place an open circle at point (3, 0).

Graph the cubic curve, adding an arrow at the other endpoint to show it continues indefinitely as x → -∞.

<u>Second piece of function</u>

Substitute the endpoint of the interval into the corresponding function:

\implies g(3)=-\log_4(3-2)+2=2 \implies (3,2)

Place an closed circle at point (3, 2).

Graph the curve, adding an arrow at the other endpoint to show it continues indefinitely as x → ∞.

See attached for graph.

<h3><u>Part B</u></h3>

The x-intercepts are where the curve crosses the x-axis, so when y = 0.

Set the <u>first piece</u> of the function to zero and solve for x:

\begin{aligned}g(x) & = 0\\\implies x^3-9x & = 0\\x(x^2-9) & = 0\\\\\implies x^2-9 & = 0 \quad \quad \quad \implies x=0\\x^2 & = 9\\\ x & = \pm 3\end{aligned}

Therefore, as x < 3, the x-intercepts are (-3, 0) and (0, 0) for the first piece.

Set the <u>second piece</u> to zero and solve for x:

\begin{aligned}\implies g(x) & =0\\-\log_4(x-2)+2 & =0\\\log_4(x-2) & =2\end{aligned}

\textsf{Apply log law}: \quad \log_ab=c \iff a^c=b

\begin{aligned}\implies 4^2&=x-2\\x & = 16+2\\x & = 18 \end{aligned}

Therefore, the x-intercept for the second piece is (18, 0).

So the x-intercepts for the piecewise function are (-3, 0), (0, 0) and (18, 0).

<h3><u>Part C</u></h3>

From the graph from part A, and the calculated x-intercepts from part B, the function g(x) is positive between the intervals -3 < x < 0 and 3 ≤ x < 18.

Interval notation:  (-3, 0) ∪ [3, 18)

Learn more about piecewise functions here:

brainly.com/question/11562909

3 0
1 year ago
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