For example, think about 8.4 - 2.9.
You have to regroup because you can't subract 4 - 9.
4 turns into a 14 since you borrow from the left number, 8.
8 turns into a 7 after that.
Now, you have to subtract 14 - 9 and 7 - 2. You're answer is 5.5.
You can check your work by doing inverse opperations:
By adding:
5.5 + 2.9 = 8.4
Or by subracting 8.4 - 5.5 = 2.9
Hope this helped you! If you need a better explaination or another example, I will be happy to help you! :)
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:
1496.79 mm squared
Step-by-step explanation:
Have a good day :)
Answer:
<h3>The given polynomial of degree 4 has atleast one imaginary root</h3>
Step-by-step explanation:
Given that " Polynomial of degree 4 has 1 positive real root that is bouncer and 1 negative real root that is a bouncer:
<h3>To find how many imaginary roots does the polynomial have :</h3>
- Since the degree of given polynomial is 4
- Therefore it must have four roots.
- Already given that the given polynomial has 1 positive real root and 1 negative real root .
- Every polynomial with degree greater than 1 has atleast one imaginary root.
<h3>Hence the given polynomial of degree 4 has atleast one imaginary root</h3><h3> </h3>