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kondor19780726 [428]
2 years ago
11

The concentration of dichlorodiphenyltrichloroethane (DDT - an infamous pesticide) in Lake Michigan has been declining exponenti

ally over many years. In 1971, the DDT concentration was 13.00 ppm (parts per million). In 1984, the DDT concentration was 2.22 ppm.
Using these data points, determine the exponential decay function to model the DDT concentration decline in terms of years, tt, where t=0t=0 represents the year 1971.


A) Write your function as y = Ce^{rt}y=Ce^rt. Round your value of rr to four decimal places.


B) Another way to write the function down is known as the half-life formula, written as y = C(0.5)^{t/H}y=C(0.5)^t/H, where HH is the time required for the concentration to get cut in half. Using the initial two data points, find the value of HH, rounded to 2 decimal places. Show all Algebraic work.


C) The goal was to reduce the DDT concentration to 0.98 ppm by 1995. Was this possible according to your model?
Mathematics
1 answer:
mojhsa [17]2 years ago
8 0

Answer: A) y=13e^{-0.1359t}

              B) H = 5.10

              C) Yes

Step-by-step explanation: <u>Exponential</u> <u>Decay</u> <u>function</u> is a model that describes the reducing of an amount by a constant rate over time. Generally, it is written in the form: y(t)=Ce^{rt}

A) C is initial quantity, in this case, the initial concentration of DDT. To determine r, using the data given:

y(t)=Ce^{rt}

2.22=13e^{13r}

e^{13r}=0.1708

Using a natural logarithm property called <em>power rule:</em>

13r=ln(0.1708)

r=\frac{ln(0.1708)}{13}

r=-0.1359

The decay function for concentration of DDT through the years is y(t)=13e^{-0.1359t}

B) The value of H is calculated by y=C(0.5)^{\frac{t}{H} }

2.22=13(0.5)^{\frac{13}{H} }

(0.5)^{\frac{13}{H} }=0.1708

Again, using power rule for logarithm:

\frac{13}{H} log(0.5)=log(0.1708)

\frac{13}{H} =\frac{log(0.1708)}{log(0.5)}

\frac{13}{H} =2.55

H = 5.10

Constant H in the half-life formula is H=5.10

C) Using model y(t)=13e^{-0.1359t} to determine concentration of DDT in 1995:

y(24)=13e^{-0.1359.24}

y(24) = 0.5

By 1995, the concentration of DDT is 0.5 ppm, so using this model is possible to reduce such amount and more of DDT.

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If ƒ(x) = x + 2 and g(x) = 3x2 + 7x, find ....
Tamiku [17]

Answer:

The correct answer is option C

(f o g)(x) = 3x² + 7x + 2

Step-by-step explanation:

<u>Points to remember</u>

<u>Composite functions</u>

Let f(x) and g(x) be the two functions then (f o g)(x) can be written as

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<u>To find the value of (f o g)(x)</u>

Here f(x) =x + 2  and g(x) = 3x² + 7x

(f o g)(x) = f(g(x))

 = f(3x² + 7x)

 = 3x² + 7x + 2

Therefore the correct answer is option C

(f o g)(x) = 3x² + 7x + 2

7 0
3 years ago
Joseph has 10 wooden boards, each is 2 meters in length. he needs 8 yards of wood to repair his fence and 25 feet to make shelve
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the answer is A he can complete both projects.

Step-by-step explanation:

3 0
3 years ago
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2 years ago
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(a) Find the size of each of two samples (assume that they are of equal size) needed to estimate the difference between the prop
zalisa [80]

Answer:

(a) The sample sizes are 6787.

(b) The sample sizes are 6666.

Step-by-step explanation:

(a)

The information provided is:

Confidence level = 98%

MOE = 0.02

n₁ = n₂ = n

\hat p_{1} = \hat p_{2} = \hat p = 0.50\ (\text{Assume})

Compute the sample sizes as follows:

MOE=z_{\alpha/2}\times\sqrt{\frac{2\times\hat p(1-\hat p)}{n}

       n=\frac{2\times\hat p(1-\hat p)\times (z_{\alpha/2})^{2}}{MOE^{2}}

          =\frac{2\times0.50(1-0.50)\times (2.33)^{2}}{0.02^{2}}\\\\=6786.125\\\\\approx 6787

Thus, the sample sizes are 6787.

(b)

Now it is provided that:

\hat p_{1}=0.45\\\hat p_{2}=0.58

Compute the sample size as follows:

MOE=z_{\alpha/2}\times\sqrt{\frac{\hat p_{1}(1-\hat p_{1})+\hat p_{2}(1-\hat p_{2})}{n}

       n=\frac{(z_{\alpha/2})^{2}\times [\hat p_{1}(1-\hat p_{1})+\hat p_{2}(1-\hat p_{2})]}{MOE^{2}}

          =\frac{2.33^{2}\times [0.45(1-0.45)+0.58(1-0.58)]}{0.02^{2}}\\\\=6665.331975\\\\\approx 6666

Thus, the sample sizes are 6666.

7 0
2 years ago
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