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natima [27]
3 years ago
5

Bai is conducting a study on learning. When she manipulates an independent variable, it is possible that some other factor, such

as noise in the hall, can affect learning in one of the groups but not in the other. This possibility reflects the presence of:
a) random assignment
b) a confound
c) selection bias
d) a dependent variable
Mathematics
1 answer:
nlexa [21]3 years ago
3 0

Answer:

The answer is : b) a confound

Step-by-step explanation:

While manipulating, is possible that some factors like noise in the hall, can affect learning in one of the groups but not in the other.

This possibility reflects the presence of a confound.

We can define a confounding variable as an external influencing factor which results in bringing changes in the effects of a dependent and independent variable.

This variable changes the outcome of an experiment and produces useless results.

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8 divided by 2/3 in a word problem
sattari [20]

Answer:

12

Step-by-step explanation:

8/1 ÷ 2/3

Keep the first fraction and flip the second one: 8×3/ 1×2

Multiply top and bottom: 24÷2

Simplified to be 12

7 0
3 years ago
An adult has a total of about 22.5 square feet (ft2) of skin. Use the fact that 1 m is approximately equal to 3.281 feet to conv
zheka24 [161]

Answer:

There are about 3.281 * 3.281 = 10.764 square feet in one square meter. Therefore, 22.5 square feet is 22.5 / 10.764 = 2.09 square meters.

3 0
4 years ago
6.7z = 5.2z + 12.3 z=​
vladimir1956 [14]

Answer:

z = 8.2

Step-by-step explanation:

6.7z = 5.2z + 12.3

combine like terms by subtracting 5.2z from both sides

1.5z = 12.3

divide by 1.5

z = 8.2

7 0
4 years ago
Read 2 more answers
Honors Geometry QUiz- Please Help
viktelen [127]

Answer:

QUESTION 2:

The area of the polygon is 21.5 units²

QUESTION 3:

The area of the combined figure is 12 + 18.25·π units²

QUESTION 4:

The area of the figure is 40 units²

QUESTION 5

The area of the compost shape is 40 in.²

Step-by-step explanation:

QUESTION 2.

The area of the polygon is given as follows;

The area of two triangles + The area of trapezoid

1/2×6×3 + 1/2×3×4 + 1/2×(6 + 7) = 21.5 units²

QUESTION 3

The diameter of the semicircle, 'd', is given as follows;

d = √(8² + 3²) = √73

The area of the semicircle = π·d²/4 = π·73/4 = 18.25·π unit²

The area of the right triangle beside the semicircle is given as follows;

1/2 × 8 × 3 = 12 unit²

The area of the combined figure = The area of the semicircle + The area of the right triangle

∴ The area of the combined figure = 12 + 18.25·π units²

QUESTION 4:

The dimensions of the parallelogram are;

Width, W = √(4² + 1²) = √(17)

Length, L = √(8² + 2²) = √(68)

The area of the parallelogram, A_R = W × L = √(17) × √(68) = √(17 × 68) = 34

The area of the parallelogram, A_R = 34 units²

The length of a parallel side, 'A', of the parallelogram is A = 6

The height of the parallelogram, is h = 1

The area of the parallelogram is A = B·h = 6 × 1 = 6 units²

The area of the figure is the sum of the area of the rectangle and the area of the parallelogram

Therefore, we have;

The area of the figure = 34 units² + 6 units² = 40 units²

QUESTION 5

The composite shape consists of a triangle and a rectangle

Therefore, we have;

The area of the rectangle = 5 in. × 7 in. = 35 in.²

The area of the triangle = 1/2 × 2 in. × 5 in. = 5 in.²

The area of the compost shape = The area of the rectangle + The area of the triangle

∴ The area of the compost shape = 35 in.² + 5 in.² = 40 in.²

The area of the compost shape = 40 in.².

5 0
3 years ago
The time a basketball player spends in the air when shooting a basket is called the​ "hang time." the vertical leap l measured i
Marina86 [1]
Dénote l the vertical leap, then the given equation can be written like this:
l(t)=4t^2\\\text{Since one inch equate 0.0833333 feet then 44 inches equate  }\\44\times 0.0833333=3.6feet\\\text{Add to 33 feet like this:}\\33+3.6=36.6feet\\\text{Now we have to solve the equation:}\\4t^2=36.6\\\text{Divide over 4 we get:}\\t^2= \frac{36.6}{4}=9.15\\\text{Take the square root we get:}\\t=\sqrt{9.15}=3

The hang time equates then 3 seconds  
7 0
3 years ago
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