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lyudmila [28]
3 years ago
9

A graduate student wants to test whether a reading curriculum is better than a computerized reading lab application, by comparin

g the reading test scores between a group who only gets the lab and another group who only gets the curriculum. Students are assigned to one or the other reading instructional group based on their schedules: lab is in the morning, the curriculum is offered in the afternoon. What type of experimental design is this? Comparison groups Time series analysis Randomized control and experimental groups Randomized block design
Mathematics
1 answer:
Soloha48 [4]3 years ago
4 0

Answer:

Comparison Group

Step-by-step explanation:

Answered this question for another person on Brainly :)

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Is the quotient of two integers always a rational number? Explain.
Troyanec [42]

Answer:

It is true that the quotient of two integers is always a rational number. This is because a rational number is any number that can be expressed as the quotient or fraction p/q of two integers, a numerator p and a non-zero denominator q.

4 0
3 years ago
Explain the algebra 5a - 9a - (-2a) =
sleet_krkn [62]

Hello!

5a - 9a - (-2a) =

= 5a - 9a + 2a =

= -4a + 2a =

= -2a

Good luck! :)

7 0
3 years ago
An equation is shown below.<br> 3.5p = 28
grin007 [14]

Answer:

ncbdnifvdvdsovsv

Step-by-step explanation:

4 0
3 years ago
4)
erastovalidia [21]

Answer:

D.) 4x + y

Step-by-step explanation:

3x + x + y

Add  3 x  and  x .

4 x  +  y

7 0
3 years ago
A ship leaves port at 10 miles per hour, with a heading of N 35° W. There is a warning buoy located 5 miles directly north of th
Leno4ka [110]

The value of the angle subtended by the distance of the buoy from the

port is given by sine and cosine rule.

  • The bearing of the buoy from the is approximately <u>307.35°</u>

Reasons:

Location from which the ship sails = Port

The speed of the ship = 10 mph

Direction of the ship = N35°W

Location of the warning buoy = 5 miles north of the port

Required: The bearing of the warning buoy from the ship after 7.5 hours.

Solution:

The distance travelled by the ship = 7.5 hours × 10 mph = 75 miles

By cosine rule, we have;

a² = b² + c² - 2·b·c·cos(A)

Where;

a = The distance between the ship and the buoy

b = The distance between the ship and the port = 75 miles

c = The distance between the buoy and the port = 5 miles

Angle ∠A = The angle between the ship and the buoy = The bearing of the ship = 35°

Which gives;

a = √(75² + 5² - 2 × 75 × 5 × cos(35°))

By sine rule, we have;

\displaystyle \frac{a}{sin(A)} = \mathbf{ \frac{b}{sin(B)}}

Therefore;

\displaystyle sin(B)= \frac{b \cdot sin(A)}{a}

Which gives;

\displaystyle sin(B) = \mathbf{\frac{75 \cdot sin(35^{\circ})}{\sqrt{75^2 + 5^2 - 2 \times 75\times5\times cos(35^{\circ}) } }}

\displaystyle B = arcsin\left( \frac{75 \cdot sin(35^{\circ})}{\sqrt{75^2 + 5^2 - 2 \times 75\times5\times cos(35^{\circ}) } }\right) \approx 37.32^{\circ}

Similarly, we can get;

\displaystyle B = arcsin\left( \frac{75 \cdot sin(35^{\circ})}{\sqrt{75^2 + 5^2 - 2 \times 75\times5\times cos(35^{\circ}) } }\right) \approx \mathbf{ 142.68^{\circ}}

The angle subtended by the distance of the buoy from the port, <em>C</em> is therefore;

C ≈ 180° - 142.68° - 35° ≈ 2.32°

By alternate interior angles, we have;

The bearing of the warning buoy as seen from the ship is therefore;

Bearing of buoy ≈ 270° + 35° + 2.32° ≈ <u>307.35°</u>

Learn more about bearing in mathematics here:

brainly.com/question/23427938

5 0
2 years ago
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