Please help me i really need this grade !
1 answer:
You might be interested in
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:
see explanation
Step-by-step explanation:
In an arithmetic sequence the common difference d is
d = a₂ - a₁ = 10 - 8 = 2
To obtain the next term in the sequence add d to the previous term, that is
a₅ = 14 + 2 = 16
a₆ = 16 + 2 = 18
a₇ = 18 + 2 = 20
The next 3 terms in the sequence are 16, 18, 20
The n th term equation for an arithmetic sequence is
= a₁ + (n - 1)d
where a₁ is the first term and d the common difference
Here a₁ = 8 and d = 2, thus
= 8 + 2(n - 1) = 8 + 2n - 2 = 2n + 6