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Whitepunk [10]
3 years ago
11

Three high school participated in a study to evaluate the effectiveness of a new computer-based mathematics curriculum. Four 24-

student sections of freshman algebra were available for the study. The two types of instruction (standard curriculum, computer-based curriculum) were randomly assigned to the four sections in each of the three schools. At the end of the term, a standard mathematics achievement test was given to each of the 24 students in each section.
a. Is this study experimental, observational, or mixed experimental and observational? Why?
b. Identify important factors, factor levels, and/or factor-level combinations.
c. Classify these factors as within- or between-subjects
d. what’s the dependent measures?
e. What is the basic experiment unit of study?
Mathematics
1 answer:
snow_tiger [21]3 years ago
8 0

Answer:

A) experimental

B)  The important factors are : Type of instruction, scores and  High schools,

   The factor levels are :

   Level -1 which comprises of  3 highschool,

   Level -2 which comprises of  highschool and the highschool has  4   sections   and each section has 24 students

   Factor level combinations:

  Level 1 : computer and standard based combination

C) Within the subject : 24 students

   Between subject : High school

D) score of the test

E) student

Step-by-step explanation:

A) The study is experimental because of the random assignment of two different types of instruction to the students which will control/affect the technique of teaching without just observing the performance of the students from the side

B) The important factors are : Type of instruction, scores and  High schools,

   The factor levels are :

   Level -1 which comprises of  3 highschool,

   Level -2 which comprises of  highschool and the highschool has  4   sections   and each section has 24 students

   Factor level combinations:

  Level 1 : computer and standard based combination

C) Within the subject : 24 students

   Between subject : High school

D) The dependent measure  is the score of the test

E) The basic experiment unit of the study is : student

 

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Duffy was analyzing the trajectory made after throwing a football. his coach was able to measure that his throw reached a maximu
gregori [183]

Answer:

4.35 seconds.

Step-by-step explanation:

Let x represent the time the independent variable and y be dependent variable (height of the ball).

We are told that Duffy figured that the ball left his hand at a height of 5 feet. This means at x equals 0, y was 5 or initial height of ball is 5 feet.

We have been given that Duffy's coach measured that his throw reached a maximum height of 18 feet after 2 seconds. This means that at x equals 2 y was 18.

As point (2,18) represents maximum height of ball, so it will be vertex of parabola.  

Since initial height of ball is less than maximum height, so our parabola will be downward opening and leading coefficient will be negative.

We know that vertex form of a downward opening parabola is in form: y=-a(x-h)^2+k

y=-a(x-2)^2+18

Let us find value of a using point (0,5).

5=-a(0-2)^2+18

5=-a*4+18

5-18=-a*4+18-18

-13=-4a

\frac{-13}{-4}=\frac{-4a}{-4}

\frac{13}{4}=a

Therefore, the equation y=-(\frac{13}{4})(x-2)^2+18 can be used to find the height of ball after x seconds.

To find the time it will take the ball to hit the ground, we will substitute y equals 0 in our equation.

0=-(\frac{13}{4})(x-2)^2+18

0-18=-(\frac{13}{4})(x-2)^2+18-18

0-18=-(\frac{13}{4})(x-2)^2

-18*(\frac{4}{13})=-(x-2)^2

\frac{72}{13}=(x-2)^2

Taking square root of both sides of our equation we will get,

\sqrt{\frac{72}{13}}=x-2

x-2\pm 2.35339

x-2=2.35339\text{ (or) } x-2=2.35339

x-2+2=2.35339+2\text{ (or) } x-2+2=-2.35339+2

x=4.35339\text{ (or) } x=-0.35339

x\approx 4.35\text{ (or) } x\approx -0.35

Since time can not be negative, therefore, the ball will hit the ground approximately after 4.35 seconds.  

7 0
2 years ago
Simplify 3 sqrt 5x * 3 sqrt 25x^ 2 completely.​
tia_tia [17]

Answer:

=45x√5x

Step-by-step explanation:

The provided expression is: 3√5x×3√25x²

We can simplify 3√25x² because √25x²=5x

3×5x=15x

Combine with the rest of the expression through multiplication.

3√5x×15x

=45x√5x

The expression 3√5x×3√25x² is equivalent to 45x√5x.

The expected simplified answer is 45x√5x

8 1
3 years ago
Read 2 more answers
An infant is 32.625 inches long write this as a common fraction
elena-14-01-66 [18.8K]
32.625 is the same as 32 and 5/8 .
8 0
2 years ago
Read 2 more answers
PLEASE HELP! THANKS! The measures, in degrees, of the 3 angles of a triangle are given by 2x +1, 3x - 3, and 9x. What is the mea
drek231 [11]
So all of the internal angles of a triangle will add up to 180 degrees.  So if we add all of the expressions given and set them equal to 180, we can solve for x, then solve for each angle...

(2x + 1) + (3x - 3) + 9x = 180
2x + 3x + 9x + 1 - 3 = 180
14x - 2 = 180
14x = 182
x = 13

Now we can substitute 13 into each expression to find the value of each angle...

2x + 1 =
2(13) + 1 =
26 + 1 = 27 degrees

3x - 3 =
3(13) - 3 =
39 - 3 = 36 degrees

9x =
9(13) = 117 degrees

So the 3 angles are 27 degrees, 36 degrees, and 117 degrees.

The smallest is 27 degrees

** Note-- I don't see 27 degrees in the answer choices you gave, so I double checked my math (I didn't find any mistakes).  Please check to see that the question is written correctly (pay close attention to plus and minus signs).
7 0
3 years ago
From experience, it is known that on average 10% of welds performed by a particular welder are defective. if this welder is requ
bulgar [2K]
Binomial distribution can be used because the situation satisfies all the following conditions:1. Number of trials is known and remains constant (n)2. Each trial is Bernoulli (i.e. exactly two possible outcomes) (success/failure)3. Probability is known and remains constant throughout the trials (p)4. All trials are random and independent of the othersThe number of successes, x, is then given byP(x)=C(n,x)p^x(1-p)^{n-x}whereC(n,x)=\frac{n!}{x!(n-x)!}
Here we're given
p=0.10  [ success = defective ]
n=3

(a) x=0
P(x)=C(n,x)p^x(1-p)^{n-x}
=C(3,0)0.1^0(1-0.1)^{3-0}
=1(1)(0.729)
=0.729

(b) x=2
P(x)=C(n,x)p^x(1-p)^{n-x}
=C(3,2)0.1^2(1-0.1)^{3-2}
=3(0.01)(0.9)
=0.027

(c) x ≥ 2
P(x)=\sum_{x=2}^3C(n,x)p^x(1-p)^{n-x}
=P(2)+P(3)
=C(n,2)p^2(1-p)^{n-2}+C(n,3)p^3(1-p)^{n-3}
=C(3,2)0.1^2(1-0.1)^{3-2}+C(3,3)0.1^3(1-0.1)^{3-3}
=3(0.01)(0.9)+1(0.001)1
=0.027+0.001
=0.028


8 0
3 years ago
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