Answer:
The range is the difference between the highest and lowest number
Hello,
sum=-8+5=-3
prod=-8*5=-40
==>x²+3x-40
Answer A
Answer:
8.5
I hope this helps.
Step-by-step explanation:
7^2 + x^2 = 11^2
Answer:
a. 49.5 and 54.5
Step-by-step explanation:
Class interval is a range of a value that is used to group data into equal size for easy analysis and representation of the data. It is applicable in the divisions of a histogram or bar chart into classes. Examples of class interval are 50-54, 55-59, 60-64, 65-69, 70-74 etc.
Class limit is the minimum and maximum value the class interval may contain. The minimum value is called the lower class limit and the maximum value is called the upper class limit. For class interval 50-54, the lower class limit is 50 and the upper class limit is 54.
Class boundaries are the numbers used to separate classes. It is the real limits of a class. For non-overlapping classes, the lower class boundary of each class is calculated by subtracting 0.5 from the lower class limit. The upper class boundary of each class is calculated by adding 0.5 to the upper class limit. Example: For class interval 50-54, the lower class boundary is 49.5 and the upper class class boundary is 54.5
Considering the question given, to get the real limits of the interval 50-54, 0.5 is subtracted from the lower class limit to give 49.5. Also, 0.5 is added to the upper class limit to give 54.5.
Here's the general formula for bacteria growth/decay problems
Af = Ai (e^kt)
where:
Af = Final amount
Ai = Initial amount
k = growth rate constant
<span>t = time
But there's another formula for a doubling problem.
</span>kt = ln(2)
So, Colby (1)
k1A = ln(2) / t
k1A = ln(2) / 2 = 0.34657 per hour.
So, Jaquan (2)
k2A = ln(2) / t
<span>k2A = ln(2) /3 = 0.23105 per hour.
</span>
We need to use the rate of Colby and Jaquan in order to get the final amount in 1 day or 24 hours.
Af1 = 50(e^0.34657(24))
Af1 = 204,800
Af2 = 204,800 = Ai2(e^0.23105(24))
<span>Af2 = 800</span>
