Can you please add the picture with the problem, then I will answer your question ?
Answer:
4.45 ccs will remain after 5 hours.
Step-by-step explanation:
The anti-biotic in his body decreases at a rate proportional to the amount, Q(t), present at time t.
(dQ/dt) = - kQ ((Minus sign because it's a rate of reduction)
(dQ/dt) = -kQ
(dQ/Q) = -kdt
∫ (dQ/Q) = -k ∫ dt
Solving the two sides as definite integrals by integrating the left hand side from Q₀ to Q and the Right hand side from 0 to t.
We obtain
In (Q/Q₀) = -kt
(Q/Q₀) = e⁻ᵏᵗ
Q(t) = Q₀ e⁻ᵏᵗ
the initial injection was 8 ccs and 5 ccs remain after 4 hours
Q₀ = 8 ccs,
At t = 4 hours, Q = 5 ccs
5 = 8 e⁻ᵏᵗ
e⁻ᵏᵗ = 0.625
-kt = In (0.625) = -0.47
-4k = 0.47
k = 0.1175 /hour
Q(t) = Q₀ e⁻⁰•¹¹⁷⁵ᵗ
At t = 5 hours, Q = ?
Q = 8 e⁻⁰•¹¹⁷⁵ᵗ
0.1175 × 5 = 0.5875
Q = 8 e(^-0.5875)
Q = 4.45 ccs
Hope this Helps!!!
Answer:
(1/6)2
Step-by-step explanation:
(1/6)2 has a value of 1/36
The mean of the ages of the men that are given will be 22.1.
<h3>How to calculate the mean?</h3>
It should be noted that mean simply means average. The mean of the men will be:
= (27 + 20 + 20 + 19 + 19 + 25 + 24 + 21 + 22 + 24) / 10
= 221/10
= 22.1
Therefore, the mean of the ages of the men that are given will be 22.1.
Learn more about mean on:
brainly.com/question/1136789
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