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likoan [24]
3 years ago
14

Define the folowing terms. (a) Experimental unit (b) Treatment (c) Response variable (d) Factor (e) Placebo ( Confounding (a) De

fine experimental unit. Choose the correct answer below. O A. A person, object, or some other well-defined item upon which a treatment is applied ? B. The quantitative or qualitative variable or which the experimenter wishes to determine how, its value is affected by the explanatory vanable O C. Any combination of the values of the factors (explanatory variables) O D. An innocuous medication, such as a sugar tablet, that looks, tastes, and smells like the experimental medication (b) Define treatment. Choose the correct answer below. O A. The number of individuals in the experiment O B. The quantitative or qualitative variable for which the experimenter wishes to determine how its value is affected by the explanatory variable O C. Any combination of the values of the factors (explanatory variables) ? D. A controlled study to determine the effect varying one or more explanatory variables or factors has on a response variable (c) Define response variable. Choose the correct answer below. O A. The quantitative or qualitative variable for which the experimenter wishes to determine how its value is affected by the explanatory variable O B. The effect of two factors (explanatory variables on the response variable) cannot be distinguished O C. An innocuous medication, such as a sugar tablet, that looks, tastes, and smells like the experimental medication O D. The variable whose effect on the response variable is to be assessed by the experimenter
Mathematics
1 answer:
Rudik [331]3 years ago
4 0

Answer:A) A

B) B

C) A

Step-by-step explanation:Experimental unit is a unit of statistical analysis.

It is also a member in a set entities.

b) Treatment is a combination of factor levels. It is an independent variable and can be manipulated by the one doing the experiment.

c) Response variables is an independent variable in which changes can be made to effect a change in the result.

Factor is a circumstance that engenders to a result.

e) Placebo is an inert substance or treatment which delivers a therapeutic value.

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masha68 [24]

Answer:

107520x6+89344x5−241808x4−55444x3+121478x2+6030x−12600

Step-by-step explanation:

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7 0
2 years ago
Please help me with this​
Vladimir [108]
I also suggest using an app called photomath. It takes you through the steps of solving the equations and gives you the answer as well.

8 0
3 years ago
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I WILL GIVE 20 POINTS TO THOSE WHO ANSWER THIS QUESTION RIGHT NOOOO SCAMS PLEASE AND PLEASE EXPLAIN WHY THAT IS THE ANSWER
nekit [7.7K]

Answer:

1 - 60

2 - 120

3 - 60

4 - 120

5 - 60

6 - 120

7 - 60

8 - 120

Step-by-step explanation:

Via supplementary angles, you can conclude that angle 5 is 60. Because of vertical angles, angle 8 is 120 and angle 7 is 60. Because of alternate exterior angles, angle 1 is congruent to angle 7 and angle 2 is congruent to angle 8, meaning angle 1 is 60 and angle 2 is 120. Because of vertical angles, angle 3 is 60 and angle 4 is 120.

7 0
2 years ago
At 2 PM, a thermometer reading 80°F is taken outside where the air temperature is 20°F. At 2:03 PM, the temperature reading yiel
Alexeev081 [22]
Use Law of Cooling:
T(t) = (T_0 - T_A)e^{-kt} +T_A
T0 = initial temperature, TA = ambient or final temperature

First solve for k using given info, T(3) = 42
42 = (80-20)e^{-3k} +20 \\  \\ e^{-3k} = \frac{22}{60}=\frac{11}{30} \\  \\ k = -\frac{1}{3} ln (\frac{11}{30})

Substituting k back into cooling equation gives:
T(t) = 60(\frac{11}{30})^{t/3} + 20

At some time "t", it is brought back inside at temperature "x".
We know that temperature goes back up to 71 at 2:10 so the time it is inside is 10-t, where t is time that it had been outside.
The new cooling equation for when its back inside is:
T(t) = (x-80)(\frac{11}{30})^{t/3} + 80 \\  \\ 71 = (x-80)(\frac{11}{30})^{\frac{10-t}{3}} + 80
Solve for x:
x = -9(\frac{11}{30})^{\frac{t-10}{3}} + 80
Sub back into original cooling equation, x = T(t)
-9(\frac{11}{30})^{\frac{t-10}{3}} + 80 = 60 (\frac{11}{30})^{t/3} +20
Solve for t:
60 (\frac{11}{30})^{t/3} +9(\frac{11}{30})^{\frac{-10}{3}}(\frac{11}{30})^{t/3}  = 60 \\  \\  (\frac{11}{30})^{t/3} = \frac{60}{60+9(\frac{11}{30})^{\frac{-10}{3}}} \\  \\  \\ t = 3(\frac{ln(\frac{60}{60+9(\frac{11}{30})^{\frac{-10}{3}}})}{ln (\frac{11}{30})}) \\  \\ \\  t = 4.959

This means the exact time it was brought indoors was about 2.5 seconds before 2:05 PM
8 0
3 years ago
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Rama09 [41]

Answer:

66 2/3 %

Step-by-step explanation:

First find the students not in the 8th grade

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16 students are not in the 8th grade

Take the fraction of the students not in the 8th grade over the total

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Change to a decimal

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Multiply by 100 to change to a percent

66.666666%

66 2/3 %

6 0
3 years ago
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