Respuesta :
The answer is A.)
f(x)=5000(0.98)^0.3x What is x=1 and x=5?
f(1)=5000(0.98)^0.3(1) <<Plug in 1
= 5000(0.98)^0.3 <<Multiply 1
=5000(0.993957517) <<mulitply 0.3 with 0.98
=4969.78759 <<mulitply the answer to ^^ with 5000
=4970 <<Round
f(5)=5000(0.98)^0.3(5) <<plug in 5
= 5000(0.98)^1.5 <<Mulitply 5
=5000(0.970150504) <<multiply 1.5 with 0.98
=4850.75252 <<muliply the answer to ^^ with 5000
= 4851 << round
Next, you do (y2 - y1)/(x2-x1)
(4970 - 4851)/(1 - 5) <<plug in numbers (btw you get the 1 and 5 from plugging them in in the beginning)
=(119)/(4) <<Subtract
= -29.75 << divide
I hope this helps!!
f(x)=5000(0.98)^0.3x What is x=1 and x=5?
f(1)=5000(0.98)^0.3(1) <<Plug in 1
= 5000(0.98)^0.3 <<Multiply 1
=5000(0.993957517) <<mulitply 0.3 with 0.98
=4969.78759 <<mulitply the answer to ^^ with 5000
=4970 <<Round
f(5)=5000(0.98)^0.3(5) <<plug in 5
= 5000(0.98)^1.5 <<Mulitply 5
=5000(0.970150504) <<multiply 1.5 with 0.98
=4850.75252 <<muliply the answer to ^^ with 5000
= 4851 << round
Next, you do (y2 - y1)/(x2-x1)
(4970 - 4851)/(1 - 5) <<plug in numbers (btw you get the 1 and 5 from plugging them in in the beginning)
=(119)/(4) <<Subtract
= -29.75 << divide
I hope this helps!!
The average decrease per day in white-blood cells per cubic millimeter between days 1 and 5 is -29.75 cubic millimeters per day. (Correct choice: A)
The average decrease per day in white-blood cells per cubic millimeter is determined by means of the secant line equation, that is to say:
[tex]\dot f = \frac{f(5)-f(1)}{5-1}[/tex]
If we know that [tex]f(x) = 5000\cdot (0.98)^{0.3\cdot x}[/tex], then the average decrease is:
[tex]f(1) = 5000\cdot (0.98)^{0.3\cdot 1}[/tex]
[tex]f(1) \approx 4969.788[/tex]
[tex]f(5) = 5000\cdot (0.98)^{0.3\cdot 5}[/tex]
[tex]f(5) \approx 4850.752[/tex]
[tex]\dot f = \frac{4850.752-4969.788}{5-1}[/tex]
[tex]\dot f = -29.759\,\frac{mm^{3}}{day}[/tex]
The average decrease per day in white-blood cells per cubic millimeter between days 1 and 5 is -29.75 cubic millimeters per day. (Correct choice: A)
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