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Nataly_w [17]
3 years ago
12

Janet researcher compares the math placement scores of college applicants who have, or have not taken a special college preparat

ion program.
Independent Variable (IV):

How will you describe the IV?:


Dependent variables (DV): ]

How will you measure the DV?:


Hypothesis:
Mathematics
1 answer:
riadik2000 [5.3K]3 years ago
8 0

Answer:

Step-by-step explanation:

Independent Variable (IV):  Special college preparation program

How will you describe the IV: Independent variable or known as manipulated variable is a variable where the researcher purposely manipulate the variable to see how it affect the results of the experiment.

Dependent variables (DV): Math placement scores of college applicants

How will you measure the DV: DV can be measured by recording the math placement scores of each applicants who have or have not taken the special college preparation program.

Explanation: In this case, the researcher wants to see how taking special college preparation program (IV) can affect the math placement scores of the applicants (DV). Explanation: In this case, the researcher wants to see how taking special college preparation program (IV) can affect the math placement scores of the applicants (DV).

Hypothesis:

If the applicants take the special college preparation program, the applicants will have higher math placement scores compared to the one who have do not take the program.

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Anuta_ua [19.1K]

9514 1404 393

Answer:

  (c)  dZ/dt = 11/√544, so moving away at about 0.47 m/s

Step-by-step explanation:

The (x, y) coordinates of the car are related by the fact that they are on a circle centered at (12, 7) with a radius of 5. Then their rates of change are related by the derivative of the circle equation with respect to time.

  (x -12)^2 + (y -7)^2 = 25

  2(x -12)x' +2(y -7)y' = 0

  y' = -(x -12)/(y -7)x'

At the time and point of interest, we have ...

  y' = -(15 -12)/(11 -7)(1) = -3/4

__

The distance (z) to the observer is given by the Pythagorean theorem:

  z^2 = (x -0)^2 +(y -2)^2

and its rate of change with time is ...

  2z·z' = 2(x-0)x' +2(y -2)y'

The distance d at the point of interest is ...

  z = √((15 -0)^2 +(11 -2)^2) = √(225 +81) = √306 = 3√34

So, the rate of change of distance to the observer at the time and point of interest is ...

  z' = (x(x') +(y -2)y')/z

  z' = ((15)(1) +(11-2)(-3/4))/(3√34) = (33/4)/(3√34)

  z' = 11/√544 ≈ 0.47 . . . m/s

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3 years ago
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3 years ago
The point (-7,4) is reflected over the line x=-3. Then the resulting point is reflected over the line y=x. Where is the point lo
Ne4ueva [31]

Answer:

  (4, 1)

Step-by-step explanation:

Since the first reflection is over the vertical line x=-3, the y-coordinate remains the same. The x-coordinate of A' will make the point (-3, 4) on the line of reflection be the midpoint between A and A':

  (-3, 4) = (A +A')/2

  2(-3, 4) -A = A' = (-6-(-7), 8 -4) = (1, 4)

The reflection over the line y=x simply interchanges the two coordinate values:

  A'' = (4, 1)

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3 years ago
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Naily [24]

Answer:

A. 2,310 sq ft

Step-by-step explanation:

<u>R</u><u>e</u><u>c</u><u>t</u><u>e</u><u>n</u><u>g</u><u>l</u><u>e</u><u> </u><u>b</u><u>e</u><u>l</u><u>o</u><u>w</u><u>:</u>

  • A = l × b
  • A = 74 × 15
  • A = 1,110ft

<u>Rectangle Above</u><u>:</u>

  • A = l × b
  • A = 50 × 24
  • A = 1,200ft

<u>A</u><u>r</u><u>e</u><u>a</u><u> </u><u>o</u><u>f</u><u> </u><u>t</u><u>h</u><u>e</u><u> </u><u>P</u><u>l</u><u>a</u><u>y</u><u>g</u><u>r</u><u>o</u><u>u</u><u>n</u><u>d</u><u>:</u>

  • 1,110ft + 1,200ft
  • 1,310 sq ft (A)
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