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julia-pushkina [17]

3 years ago

13

Charles wants to determine which car is the fastest. The actual speed of four cars is shown below: Car A B C D Speed (in km/h) 6

0.34 60.32 60.33 60.35 If the speedometer can only measure to the nearest tenth of a second, which car will have a different measurement?

1 answer:

marysya [2.9K]3 years ago

5 0

it would be car d because of what the time was on the chart (i did the test and it was correct)

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2 years ago

Pregnant women metabolize some drugs at a slower rate than the rest of the population. The half-life of caffeine is about 4 hour

UkoKoshka [18]

Answer:

Husband:

The husband will have 16.35 mg of caffeine in his body at 7 pm.

Woman:

The pregnant woman will have 51.33 mg of caffeine in her body at 7 pm.

Step-by-step explanation:

The amount of caffeine in the body can be modeled by the following equation:

$C(t) = C(0)e^{rt}$

In which C(t) is the amount of caffeine t hours after 8 am, C(0) is how much coffee they took and r is the rate the the amount of caffeine decreases in their bodies.

110 mg of caffeine at 8 am,

So $C(0) = 110$

Husband

Half life of 4 hours. So

$C(4) = 0.5C(0) = 0.5*110 = 55$

$C(t) = C(0)e^{rt}$

$55 = 110e^{4r}$

$e^{4r} = 0.5$

Applying ln to both sides

$\ln{e^{4r}} = \ln{0.5}$

$4r = \ln{0.5}$

$r = \frac{\ln{0.5}}{4}$

$r = -0.1733$

So for the husband

$C(t) = 110e^{-0.1733t}$

At 7 pm

7 pm is 11 hours after 8 am, so this is C(11)

$C(t) = 110e^{-0.1733t}$

$C(11) = 110e^{-0.1733*11} = 16.35$

The husband will have 16.35 mg of caffeine in his body at 7 pm.

Pregnant woman

Half life of 10 hours. So

$C(10) = 0.5C(0) = 0.5*110 = 55$

$C(t) = C(0)e^{rt}$

$55 = 110e^{10}$

$e^{10r} = 0.5$

Applying ln to both sides

$\ln{e^{10r}} = \ln{0.5}$

$10r = \ln{0.5}$

$r = \frac{\ln{0.5}}{10}$

$r = -0.0693$

At 7 pm

7 pm is 11 hours after 8 am, so this is C(11)

$C(t) = 110e^{-0.0693t}$

$C(11) = 110e^{-0.0693*11} = 51.33$

The pregnant woman will have 51.33 mg of caffeine in her body at 7 pm.

7 0

3 years ago