Answer:
(a)
- Independent Variable- Dosage of Omega-3 Fatty Acids
- Dependent Variable - Number of news item remembered
(b)Discrete
(c)Ratio Scale and Discrete Variable
(d) Experimental Method
Step-by-step explanation:
The psychologist wants to evaluate the claim that omega-3 fatty acids can help improve memory in normal adult humans.
(a)In the study, the participants in the two groups were given fish extracts containing Omega-3 (500 mg) and no Omega-3 (0 mg).
The memory test involves measuring the number of items each participant remembers from the past three weeks of news.
Therefore:
- Independent Variable- Dosage of Omega-3
- Dependent Variable - Number of news item remembered
(b) The dependent variable is discrete since the number of news items remembered can only be whole numbers.
(c)The independent variable is in milligrams of Omega-3 where the placebo is 0 mg. This is a ratio scale since it has an absolute zero.
Since the dosage is given in multiples of 50mg, it is a discrete variable.
(d)Since the psychologist seeks to manipulate the conditions of the study by introducing Omega-3 to some of the participants and placebo to other participants, it is an experimental distribution.
For the smaller triangle ;
15^2 + b^2 = 17^2
b = 8
To check the work
15^2 + 8^2 = 289
17^2 = 289
(This is if the bottom number on the smaller triangle says 15, its a bit hard to read..)
Step-by-step explanation:m okay so the answer is 56 x= 56
Answer:
Step-by-step explanation:
Given: 76,45,64,80,92
Required: Determine the standard deviation
We start by calculating the mean
Where x-> 76,45,64,80,92 and n = 5
Subtract Mean (71.4) from each of the given data
Determine the absolute value of the above result
Square Individual Result
Calculate the mean of the above result to give the variance
Hence, Variance = 255.298
Standard Deviation is calculated by 
