Resources

Resources

Bytelearn - cat image with glasses

AI tutor

Welcome to Bytelearn!

Let’s check out your problem:

After 24 hours,
30% of a 25 milligram dose of a new antibiotic remains in the body. Which of the following functions,
M, models the amount of the antibiotic (in milligrams) that remains in the body after
h hours?
Choose 1 answer:
(A)
M(h)=30*(0.3)^((h)/( 24))
(B)
M(h)=25*(0.3)^((h)/( 24))
(c)
M(h)=25*(0.3)^(h)
(D)
M(h)=25*(0.7)^((h)/( 24))

After $24$ hours, $30 \%$ of a $25$ milligram dose of a new antibiotic remains in the body. Which of the following functions, $M$ , models the amount of the antibiotic (in milligrams) that remains in the body after $h$ hours? $\newline$ Choose $1$ answer: $\newline$ (A) $M(h)=30 \cdot(0.3)^{\frac{h}{24}}$ $\newline$ (B) $M(h)=25 \cdot(0.3)^{\frac{h}{24}}$ $\newline$ (C) $M(h)=25 \cdot(0.3)^{h}$ $\newline$ (D) $M(h)=25 \cdot(0.7)^{\frac{h}{24}}$

View step-by-step help

View step-by-step help

Compare linear and exponential growth

Full solution

Q. After

24

hours,

30 \%

of a

25

milligram dose of a new antibiotic remains in the body. Which of the following functions,

M

, models the amount of the antibiotic (in milligrams) that remains in the body after

h

hours?

\newline

Choose

1

answer:

\newline

(A)

M(h)=30 \cdot(0.3)^{\frac{h}{24}}

\newline

(B)

M(h)=25 \cdot(0.3)^{\frac{h}{24}}

\newline

(C)

M(h)=25 \cdot(0.3)^{h}

\newline

(D)

M(h)=25 \cdot(0.7)^{\frac{h}{24}}

Identify Decay Model: We need to find a function that models the decay of the antibiotic in the body over time. We know that after $24$ hours, $30\%$ of the initial dose remains. This means that the amount of antibiotic decreases to $30\%$ of its initial amount every $24$ hours. We can use an exponential decay function to model this.
Exponential Decay Function: The general form of an exponential decay function is $M(h) = M_0 \times (\text{decay\_rate})^{(h / \text{time\_period})}$ , where $M_0$ is the initial amount, $\text{decay\_rate}$ is the percentage that remains after each time period, and $\text{time\_period}$ is the length of the time period after which the $\text{decay\_rate}$ is applied.
Substitute Values: In this case, $M_0$ is the initial dose, which is $25$ milligrams. The $\text{decay\_rate}$ is $30\%$ , or $0.3$ , since $30\%$ of the antibiotic remains after $24$ hours. The $\text{time\_period}$ is $24$ hours, as this is the period after which the $\text{decay\_rate}$ is applied.
Final Exponential Decay Function: Substituting these values into the general form of the exponential decay function, we get $M(h) = 25 \times (0.3)^{\frac{h}{24}}$ . This matches option (B) from the given choices.

More problems from Compare linear and exponential growth

Question

Jill bought a used motorcycle from a seller online for $1,200. The seller will charge her $5 a day to store the motorcycle at his house until she is able to pick it up. You can use a function to describe the total amount of money Jill will owe the seller if she waits

x

days to pick up the motorcycle.

\newline

Write an equation for the function. If it is linear, write it in the form

g(x) = mx + b

. If it is exponential, write it in the form

g(x) = a(b)^x

\newline

g(x) = \underline{\hspace{3em}}

Posted 5 months ago

Question

f(t)=-2t-7

change over the interval from

t=-7

t=-6

\newline

\newline

\left[\text{f(t) decreases by } 2\right]

\newline

\left[\text{f(t) increases by } 2\right]

\newline

\left[\text{f(t) increases by } 200\%\right]

\newline

\left[\text{f(t) decreases by a factor of } 2\right]

Posted 5 months ago

Question

g(t)=9^t

change over the interval from

t=1

t=3

\newline

\newline

\left[\text{g(t) decreases by a factor of } 81\right]

\newline

\left[\text{g(t) increases by a factor of } 81\right]

\newline

\left[\text{g(t) increases by } 18\%\right]

\newline

\left[\text{g(t) decreases by } 9\%\right]

Posted 5 months ago

Question

Both of these functions grow as

x

gets larger and larger. Which function eventually exceeds the other?

\newline

\newline

(A)f(x) = 8x + 3.3

\newline

(B)g(x) = 3.3^x - 5

Posted 6 months ago

Question

Both of these functions grow as

x

gets larger and larger. Which function eventually exceeds the other?

\newline

\newline

(A) \ f(x) = 2x + 9

\newline

(B)\ g(x) = 2^x

Posted 6 months ago

Question

f(x)

g(x)

are differentiable.

\newline

h(x)

h(x)=\frac{g(x)}{f(x)}

\newline

f(8)=1

f'(8)=-6

g(8)=8

g'(8)=-10

h'(8)

\newline

Simplify any fractions.

\newline

h'(8)=

Posted 6 months ago

Question

p(x)

q(x)

are differentiable.

\newline

r(x)

r(x)= \frac{p(x)}{q(x)}

\newline

p(3)= 2

p'(3)= 4

q(3)= 6

q'(3)= 9

r'(3)

\newline

Simplify any fractions.

\newline

r'(3)=

Posted 7 months ago

Question

u(x)

v(x)

are differentiable.

\newline

w(x)

w(x)= \frac{u(x)}{v(x)}

\newline

u(5)= 3

u'(5)= -2

v(5)= 7

v'(5)= 1

w'(5)

\newline

Simplify any fractions.

\newline

w'(5)=

Posted 7 months ago

Question

a(x)

b(x)

are differentiable.

\newline

c(x)

c(x)= \frac{a(x)}{b(x)}

\newline

a(2)= 4

a'(2)= 5

b(2)= 1

b'(2)= -3

c'(2)

\newline

Simplify any fractions.

\newline

c'(2)=

Posted 7 months ago

Question

m(x)

n(x)

are differentiable.

\newline

o(x)

o(x)= \frac{m(x)}{n(x)}

\newline

m(7)= 2

m'(7)= -1

n(7)= 4

n'(7)= 8

o'(7)

\newline

Simplify any fractions.

\newline

o'(7)=

Posted 7 months ago

Related topics

Algebra - Order of Operations Algebra - Distributive Property `X` and `Y` Axes Geometry - Scalene Triangle Common Multiple Geometry - Quadrant