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dimulka [17.4K]
2 years ago
13

. Given f(x) = e 2x e 2x + 3e x + 2 : (a) Make the substitution u = e x to convert Z f(x) dx into an integral in u (HINT: The ea

siest way to do this in Python is to substitute into f(x) u 0 (x) ) (b) Your result from part (a) should be a rational function in u. Find the partial fraction decomposition (directly in Python) and use it to integrate the function. (Remember to substitute x back in when done!) (c) Evaluate Z f(x) dx directly.
Mathematics
1 answer:
MrRa [10]2 years ago
8 0

Answer:

Step-by-step explanation:

Given;

f(x)=\frac{e^{2x}}{e^{2x}+3e^x+2}

a)

substitute u=e^x\\du=e^x dx\\\\\int\frac{e^{2x}}{e^{2x}+3e^x+2}dx=\int\frac{e^x\dot e^x}{e^x^{2x}+3e^x+2}dx\\\\=\int\frac{udu}{u^2+3u+2}

b)

Apply partial fraction in (a), we get;

\frac{u}{u^2+3u+2}=\frac{2}{u+2}-\frac{1}{u+1}\\\\\therefore u^2+3u+2\\=u^2+2u+u+2\\=u(u+2)+1(u+1)\\=(u+2)(u+1)\\\\Now\,\int\frac{u}{u^2+3u+2}\,du=\int\frac{2}{u+2}du-\int\frac{1}{u+1}du\\\\=2ln|u+2|-ln|u+1|+c\\=2ln|e^x+2|-lm|e^x+1|+c

where C is an arbitrary constant

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Prove the following with Euler's formula to write e^itheta and e^-itheta in terms of sine and cosine. Then subtract them.
kirill115 [55]
Euler's formula tells us that

e^{i\theta}=\cos\theta+i\sin\theta
e^{-i\theta}=\cos\theta-i\sin\theta

Suppose we subtract the two. This eliminates the cosine terms.

e^{i\theta}-e^{-i\theta}=i\sin\theta-(-i\sin\theta)=2i\sin\theta

Divide both sides by 2i and you're done.
8 0
2 years ago
If a cow has a mass of 9×102 kilograms, and a blue whale has a mass of 1.8×105 kilograms, which of these statements is true?
lana66690 [7]

Answer:

The mass of the Blue whale is 200 times the mass of the cow

Step-by-step explanation:

Given

Mass of Cow = 9 * 10² kg

Mass of Blue Whale = 1.8 * 10⁵ kg

Required

Determine the relationship between both weights

Represent the mass of the cow with C and the mass of the whale with B

C = 9 * 10^2kg

C = 9 *100kg

C = 900kg

B = 1.8 * 10^5kg

B = 1.8 * 100000kg

B = 180000kg

Divide the bigger weight by the smaller weight

\frac{B}{C} = \frac{180000kg}{900kg}

\frac{B}{C} = \frac{180000}{900}

\frac{B}{C} = {200}{}

Multiply both sides by C

C * \frac{B}{C} = {200}{} * C

B = {200}{} * C

B = 200}C

<em>From the expression above, it can be concluded that the mass of the Blue whale is 200 times the mass of the cow</em>

7 0
2 years ago
The school playground is in the shape of a pentagon. There is a drinking fountain at each of the 5 corners of the playground. Th
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The answer is 21 save
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2 years ago
In a study by Peter D. Hart Research Associates for the Nasdaq Stock Market, it was determined that 20% of all stock investors a
solong [7]

Answer:

The answer to the questions are;

a. The probability that exactly six are retired people is 0.1633459.

b. The probability that 9 or more are retired people is 0.04677.

c. The number of expected retired people in a random sample of 25 stock investors is 0.179705.

d. In a random sample of 20 U.S. adults the probability that exactly eight adults invested in mutual funds is 0.179705.

e. The probability that fewer than five adults invested in mutual funds out of a random sample of 20 U.S. adults is 5.095×10⁻².

f. The probability that exactly one adult invested in mutual funds out of a random sample of 20 U.S. adults is 4.87×10⁻⁴.

g. The probability that 13 or more adults out of a random sample of 20 U.S. adults invested in mutual funds is 2.103×10⁻².

h. 4, 1, 13. They tend to converge to the probability of the expected value.

Step-by-step explanation:

To solve the question, we note that the binomial distribution probability mass function is given by

f(n,p,x) = \left(\begin{array}{c}n&x&\end{array}\right) × pˣ × (1-p)ⁿ⁻ˣ = ₙCₓ × pˣ × (1-p)ⁿ⁻ˣ

Also the mean of the Binomial distribution is given by

Mean = μ = n·p = 25 × 0.2 = 5

Variance = σ² = n·p·(1-p) = 25 × 0.2 × (1-0.2) = 4

Standard Deviation = σ = \sqrt{n*p*(1-p)}

Since the variance < 5 the normal distribution approximation is not appropriate to sole the question

We proceed as follows

a. The probability that exactly six are retired people is given by

f(25, 0.2, 6) = ₂₅C₆ × 0.2⁶ × (1-0.2)¹⁹ = 0.1633459.

b. The probability that 9 or more are retired people is given by

P(x>9) = 1- P(x≤8) = 1- ∑f(25, 0.2, x where x = 0 →8)

Therefore we have

f(25, 0.2, 0) = ₂₅C₀ × 0.2⁰ × (1-0.2)²⁵ = 3.78×10⁻³

f(25, 0.2, 1) = ₂₅C₁ × 0.2¹ × (1-0.2)²⁴ = 2.36 ×10⁻²

f(25, 0.2, 2) = ₂₅C₂ × 0.2² × (1-0.2)²³ = 7.08×10⁻²

f(25, 0.2, 3) = ₂₅C₃ × 0.2³ × (1-0.2)²² = 0.135768

f(25, 0.2, 4) = ₂₅C₄ × 0.2⁴ × (1-0.2)²¹ = 0.1866811

f(25, 0.2, 5) = ₂₅C₅ × 0.2⁵ × (1-0.2)²⁰ = 0.1960151

f(25, 0.2, 6) = ₂₅C₆ × 0.2⁶ × (1-0.2)¹⁹ = 0.1633459

f(25, 0.2, 7) = ₂₅C₇ × 0.2⁷ × (1-0.2)¹⁸ = 0.11084187

f(25, 0.2, 8) = ₂₅C₈ × 0.2⁸ × (1-0.2)¹⁷ = 6.235×10⁻²

∑f(25, 0.2, x where x = 0 →8) = 0.953226

and P(x>9) = 1- P(x≤8)  = 1 - 0.953226 = 0.04677.

c. The number of expected retired people in a random sample of 25 stock investors is given by

Proportion of retired stock investors × Sample count

= 0.2 × 25 = 5.

d. In a random sample of 20 U.S. adults the probability that exactly eight adults invested in mutual funds is given by

Here we have p = 0.4 and n·p = 8 while n·p·q = 4.8 which is < 5 so we have

f(20, 0.4, 8) = ₂₀C₈ × 0.4⁸ × (1-0.4)¹² = 0.179705.

e. The probability that fewer than five adults invested in mutual funds out of a random sample of 20 U.S. adults is

P(x<5) = ∑f(20, 0.4, x, where x = 0 →4)

Which gives

f(20, 0.4, 0) = ₂₀C₀ × 0.4⁰ × (1-0.4)²⁰ = 3.66×10⁻⁵

f(20, 0.4, 1) = ₂₀C₁ × 0.4¹ × (1-0.4)¹⁹ = 4.87×10⁻⁴

f(20, 0.4, 2) = ₂₀C₂ × 0.4² × (1-0.4)¹⁸ = 3.09×10⁻³

f(20, 0.4, 3) = ₂₀C₃ × 0.4³ × (1-0.4)¹⁷ = 1.235×10⁻²

f(20, 0.4, 4) = ₂₀C₄ × 0.4⁴ × (1-0.4)¹⁶ = 3.499×10⁻²

Therefore P(x<5) = 5.095×10⁻².

f. The probability that exactly one adult invested in mutual funds out of a random sample of 20 U.S. adults is given by

f(20, 0.4, 1) = ₂₀C₁ × 0.2¹ × (1-0.2)¹⁹ = 4.87×10⁻⁴.

g. The probability that 13 or more adults out of a random sample of 20 U.S. adults invested in mutual funds is

P(x≥13) =  ∑f(20, 0.4, x where x = 13 →20) we have

f(20, 0.4, 13) = ₂₀C₁₃ × 0.4¹³ × (1-0.4)⁷ = 1.46×10⁻²

f(20, 0.4, 14) = ₂₀C₁₄ × 0.4¹⁴ × (1-0.4)⁶ = 4.85×10⁻³

f(20, 0.4, 15) = ₂₀C₁₅ × 0.4¹⁵ × (1-0.4)⁵ = 1.29×10⁻³

f(20, 0.4, 16) = ₂₀C₁₆ × 0.4¹⁶ × (1-0.4)⁴ = 2.697×10⁻⁴

f(20, 0.4, 17) = ₂₀C₁₇ × 0.4¹⁷ × (1-0.4)³ = 4.23×10⁻⁵

f(20, 0.4, 18) = ₂₀C₁₈ × 0.4¹⁸ × (1-0.4)² = 4.70×10⁻⁶

f(20, 0.4, 19) = ₂₀C₁₉ × 0.4¹⁹ × (1-0.4)⁴ = 3.299×10⁻⁷

f(20, 0.4, 20) = ₂₀C₂₀ × 0.4²⁰ × (1-0.4)⁰ = 1.0995×10⁻⁸

P(x≥13) = 2.103×10⁻².

h.  For part e we have exactly 4 with a probability of 3.499×10⁻²

For part f the  probability for the one adult is 4.87×10⁻⁴

For part g, we have exactly 13 with a probability of 1.46×10⁻²

The expected number is 8 towards which the exact numbers with the highest probabilities in parts e to g are converging.

5 0
2 years ago
M and N are two events P(M) = 0.60, P(N) = 0.20, and P (M and N) = 0.1.
saul85 [17]

Answer:

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Step-by-step explanation:

i just took the test :)

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