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!
9/3x = 5/8
Using the reciprocal, multiply each side by 3/9.
5/8 * 3/9 = 15/72
Simplify it : 5/24
x = 5/24
Hope this helps :)
Answer:
160
Step-by-step explanation:
The top right quadrant is 90 and the given is 70, adding them equal 160. It is the only option that is greater than 90 degrees anyway.