The average of the five consecutive numbers ending with b in discuss when expressed in terms of a is; Choice D; a+3.
<h3>What is the average of five consecutive integers ending with b?</h3>
First, since it was given in the task content that the average of six positive consecutive odd integers starting with a is equal to b, it therefore follows that;
(a+a+2+a+4+a+6+a+8+a+10)/6 = b
6b=6a+30
b=a+5
Also, let the average of the consecutive intergers ending with b be denoted by; x.
(b+b-1+b-2+b-3+b-4)/5 = x
=(5b-10)/5
=b–2
The average, x=b – 2 (where b = a-5)
Ultimately, the value of the required average is; = a+5-2 = a+3.
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Given :
The percent of concentration of a certain drug in the bloodstream x hours after the drug is administered is given by 
.
To Find :
Find the time at which the concentration is a maximum. b. Find the maximum concentration.
Solution :
For maximum value of x, K'(x) = 0.
Since, time cannot be negative, so ignoring x = -3 .
Putting value of x = 3, we get, K(3) = 15/( 9 + 9) = 5/6
Therefore, maximum value drug in bloodstream is 5/6 at time x = 3 units.
Hence, this is the required solution.
Answer:
2, 5, 6
Step-by-step explanation:
from top to bottom, it's the second, fifth, and sixth
4.16=416/100
416/100=4•104/4•25
=104/25
4.16=104/25
The answer is 66% because you divide 5.28 divided by 8
