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:
57
Step-by-step explanation:
It is given in the question that length CDA = 57.
Since the shape is a parallelogram, then we know that length AD=BC and AB=CD.
CDA = CD + AD
BCD = BC + CD
Since BC=AD and CD=CD
BCD = BC + CD is the same as CD + AD = CDA
Therefore BCD is the same length as CDA = 57
In other words, CDA is made up of a long side and a short side = 57
BCD is also made up of a long side and a short side, and since the longs sides are equal to each other and the short sides are also equal to each other in a parallelogram, BCD is the same length as CDA = 57.
Hope this helped!
Because the shadow doubled, the tree is actually only 11 FEET TALL.
Y=-2x+0 reason: rise over run as you can see in the graph -2 rise (drop) and -1 run and b=0 y intercept
