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dlinn [17]
3 years ago
8

Maureen is taking an antibiotic. The table below shows the amount of antibiotic f(t), in mg, that was present in her body after

time t:
t (hours) 1 2 3 4 5
f(t) (mg) 150 90 54 32.4 19.4


Ken was administered 200 mg of the same antibiotic. The amount of antibiotic f(t) in his body after time t is shown by the equation below:

f(t) = 200(0.976)t

Which statement best describes the rate at which Maureen's and Ken's bodies eliminated the antibiotic?
Maureen's body eliminated the antibiotic faster than Ken's body.
Maureen's body eliminated the antibiotic at the same rate as Ken's body.
Maureen's body eliminated the antibiotic at half of the rate at which Ken's body eliminated the antibiotic.
Maureen's body eliminated the antibiotic at one-fourth of the rate at which Ken's body eliminated the antibiotic.
Mathematics
1 answer:
MariettaO [177]3 years ago
3 0
One way to test this is by using the equation for Ken on Maureen. Let's say in the first hour,

f(t) = 200(0.976)^1 = 195.2 mg

In the second hour,

f(t) = 200(0.976)^2 = 190.52 mg

In the third hour,

f(t) = 200(0.976)^3 = 186 mg

If you compare this with Maureen's data which is 150, 90 and 54 for the first, second and third hour, respectively, you will see that Maureen's rate is much faster. However, you cannot tell by what factor because the function is exponential, not a multiple. There is no constant difference between their rates. Therefore, we only know that Maureen's rate is much faster.

The answer is <span>Maureen's body eliminated the antibiotic faster than Ken's body.</span>
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Past experience has indicated that the breaking strength of yarn used in manufacturing drapery material is normally distributed
vlabodo [156]

Answer:

95% two-sided confidence interval on the true mean breaking strength is (94.8cm, 99.2cm)

Step-by-step explanation:

Our sample size is 11.

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Then, we need to subtract one by the confidence level \alpha and divide by 2. So:

\frac{1-0.95}{2} = \frac{0.05}{2} = 0.025

Now, we need our answers from both steps above to find a value T in the t-distribution table. So, with 10 and 0.025 in the two-sided t-distribution table, we have T = 1.812

Now, we find the standard deviation of the sample. This is the division of the standard deviation by the square root of the sample size. So

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Now, we multiply T and s

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For the upper end of the interval, we add the sample mean and M. So the upper end of the interval here is

L = 97 + 2.19 = 99.19 = 99.2cm

So

95% two-sided confidence interval on the true mean breaking strength is (94.8cm, 99.2cm).

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Simplify 4+(-3)-2 times (-6)
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3 years ago
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