HELP ME PLZ ASAP!!!!!
2 answers:
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Answer:
B. No, this distribution does not appear to be normal
Step-by-step explanation:
Hello!
To observe what shape the data takes, it is best to make a graph. For me, the best type of graph is a histogram.
The first step to take is to calculate the classmark`for each of the given temperature intervals. Each class mark will be the midpoint of each bar.
As you can see in the graphic (2nd attachment) there are no values of frequency for the interval [40-44] and the rest of the data show asymmetry skewed to the left. Just because one of the intervals doesn't have an observed frequency is enough to say that these values do not meet the requirements to have a normal distribution.
The answer is B.
I hope it helps!
Answer:
On a coordinate plane, a line goes through (0, 3) and (2, 4) and another line goes through (0, 3) and (0.75, 0).
This answer almost coincide with option C. I suppose there was a mistype.
Step-by-step explanation:
The system of equations is formed by:
–x + 2y = 6
4x + y = 3
In the picture attached, the solution set is shown.
The first equation goes through (0, 3) and (2, 4), as can be checked by:
–(0) + 2(3) = 6
–(2) + 2(4) = 6
The second goes through (0, 3) and (0.75, 0), as can be checked by:
4(0) + (3) = 3
4(0.75) + (0) = 3
