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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!
This is a way to factoring trinomials (there exist different equivalent methods).
Multiply the trinomial but the term accompanying . This is the second line. Then, you could take the square of the
, ant try to create a factor () () that will correspond to the expression in the second line. That is, we want
In ? we put the corresponding numbers that, if we multiply them we will obtain 42, and if we add them we will obtain 13. This numbers are 6 and 7. Then, we have
The last step is divide by the number that we multipy in the first step.