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Marrrta [24]
3 years ago
10

James had $ 6500 in a bank account that paid 4% interest at the end of each year. How much money did Mr. James have in his accou

nt at the end of 1 year ?
$6700

$ 6760

$ 6750

$6775​
Mathematics
2 answers:
Kazeer [188]3 years ago
8 0
He would have $6,760 in his account after year one.
Do 6,500*0.04=260
6,500+260=6,760
Elena L [17]3 years ago
3 0

Answer:

6760

Step-by-step explanation:

4% of 6500

= (6500 x 4) ÷ 100

= 26000 ÷ 100

= 260 $

6500 + 260 = 6760$

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\sqrt{-6 + \frac{y}{3}} = x

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3 years ago
After an antibiotic tablet is taken, the concentration of the antibiotic in the bloodstream is modelled by the function C(t)=8(e
Alexxx [7]

Answer:

the maximum concentration of the antibiotic during the first 12 hours is 1.185 \mu g/mL at t= 2 hours.

Step-by-step explanation:

We are given the following information:

After an antibiotic tablet is taken, the concentration of the antibiotic in the bloodstream is modeled by the function where the time t is measured in hours and C is measured in \mu g/mL

C(t) = 8(e^{(-0.4t)}-e^{(-0.6t)})

Thus, we are given the time interval [0,12] for t.

  • We can apply the first derivative test, to know the absolute maximum value because we have a closed interval for t.
  • The first derivative test focusing on a particular point. If the function switches or changes from increasing to decreasing at the point, then the function will achieve a highest value at that point.

First, we differentiate C(t) with respect to t, to get,

\frac{d(C(t))}{dt} = 8(-0.4e^{(-0.4t)}+ 0.6e^{(-0.6t)})

Equating the first derivative to zero, we get,

\frac{d(C(t))}{dt} = 0\\\\8(-0.4e^{(-0.4t)}+ 0.6e^{(-0.6t)}) = 0

Solving, we get,

8(-0.4e^{(-0.4t)}+ 0.6e^{(-0.6t)}) = 0\\\displaystyle\frac{e^{-0.4}}{e^{-0.6}} = \frac{0.6}{0.4}\\\\e^{0.2t} = 1.5\\\\t = \frac{ln(1.5)}{0.2}\\\\t \approx 2

At t = 0

C(0) = 8(e^{(0)}-e^{(0)}) = 0

At t = 2

C(2) = 8(e^{(-0.8)}-e^{(-1.2)}) = 1.185

At t = 12

C(12) = 8(e^{(-4.8)}-e^{(-7.2)}) = 0.059

Thus, the maximum concentration of the antibiotic during the first 12 hours is 1.185 \mu g/mL at t= 2 hours.

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