A bag contains 10 pounds of dog food.
A family feeds their dogs 0.85 pound of dog food a day so cin seven day the dog eats (0.85 x 7) = 5.95 pounds
so, <span>dog food is left in the bag after 7 days is </span>(10-5.95) = 4.05 pounds
Answer:
See below for answers and explanations
Step-by-step explanation:
<u>Part A</u>
The average rate of change of a function over the interval ![[a,b]](https://tex.z-dn.net/?f=%5Ba%2Cb%5D)
is equal to 
, hence:
Therefore, the average rate of change of 
over the interval ![[5,9]](https://tex.z-dn.net/?f=%5B5%2C9%5D)
is 
.
<u>Part B</u>
Do the same thing as in Part A:
Therefore, the average rate of change of 
over the interval ![[0.25,1]](https://tex.z-dn.net/?f=%5B0.25%2C1%5D)
is 
.
<u>Part C</u>
To interpret our answer from Part B in terms of the real world it represents, we say that between 0.25 seconds and 1 second, the ball falls at a rate of 4 feet per second (since our average rate of change is negative).
Step-by-step explanation:
First:
(Error/Total )x 100%
(0.49/23) x100%= 2.13%
Second:
(Absolute uncertainty/mean) x100%
(3/44)x100%= 6.81%
S(8)=3500(1+(.047/4))^32
S(8)=$5086.40 in the account after 8 years.
a)The relative growth rate is .25, or 25%
b)at t=0, the population is 955e^.25(0)=955
c)at t=5; the population is 955*e^.25(5)=955*3.49=3333.28 bacterium.
<h3>Answer: 7366.96 dollars</h3>
========================================================
Use the compound interest formula:
A = P(1+r/n)^(n*t)
where in this case,
A = 12000 = amount after t years
P = unknown = deposited amount we want to solve for
r = 0.05 = the decimal form of 5% interest
n = 1 = refers to the compounding frequency (annual)
t = 10 = number of years
-------
Plug all these values into the equation, then solve for P
A = P(1+r/n)^(n*t)
12000 = P(1+0.05/1)^(1*10)
12000 = P(1.05)^(10)
12000 = P(1.62889462677744)
12000 = 1.62889462677744P
1.62889462677744P = 12000
P = 12000/1.62889462677744
P = 7366.95904248911
P = 7366.96
