Recursive algorithm for computing n²: 1. If n equals 0, return 0 (base case), 2. Otherwise, recursively compute (n-1)², 3. Compute n² by using the formula: n² = (n-1)² + 2n - 1.
What is recursive algorithm?
A recursive algorithm is a problem-solving approach where a function calls itself to solve smaller instances of the same problem until a base case is reached.
To prove the correctness of the recursive algorithm for computing n², we will use mathematical induction.
Base case: For n = 0, the algorithm returns 0, which is correct since 0² equals 0.
Inductive step: Assume that the algorithm correctly computes n² for a given nonnegative integer k, i.e., n² = k². Now, we will show that it also computes (k+1)² correctly.
According to the algorithm, (k+1)² is computed by first calculating k² using the recursive step, and then applying the formula n² = (n-1)² + 2n - 1.
Using the assumption, we have k² = (k-1)² + 2k - 1.
Expanding (k-1)², we get k² - 2k + 1 + 2k - 1 = k².
Therefore, (k+1)² = k² + 2(k+1) - 1, which simplifies to (k+1)² = k² + 2k + 1.
This matches the definition of (k+1)², so the algorithm correctly computes (k+1)².
By induction, we have proven that the algorithm correctly computes n² for any nonnegative integer n.
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Find the area A of the region that is bounded between the curve f(x) = 4-2-1 and the line g(20) = 2x +1 over the interval -2,Find the area A of the region that is bounded between the curve f(x)=4−ex−1 and the line g(x)=2x+1 over the interval [−2,2]. Enter an exact answer.
The exact numerical value of this expression depends on the precision of the decimal approximations of e.
To find the area A of the region bounded between the curve f(x) = 4 - e²x - 1 and the line g(x) = 2x + 1 over the interval [-2, 2], we need to calculate the definite integral of the absolute difference between the two functions over that interval.
Let's denote the absolute difference between f(x) and g(x) as h(x):
h(x) = |f(x) - g(x)| = |(4 - e²x - 1) - (2x + 1)| = |3 - e²x - 2x|
To find the area A, we integrate h(x) over the interval [-2, 2]:
A = ∫[-2,2] |3 - e²x - 2x| dx
Since the integrand is not continuous on the interval, we need to break it down into two separate integrals based on the intervals where the function changes sign. In this case, it happens at x = -1.
Therefore, the area A is given by:
A = ∫[-2,-1] (e²x + 2x - 3) dx + ∫[-1,2] (-e²x - 2x + 3) dx
Evaluating the integrals, we get:
A = [e²x + x²2 - 3x] [-2,-1] + [-e²x - x²2 + 3x] [-1,2]
Substituting the limits of integration, we have:
A = [(e²-1 + 1 - 3) - (e²-2 + 4 - 6)] + [(-e²2 - 4 + 6) - (-e²-1 - 1 + 3)]
Simplifying, we get:
A = [-e²-2 + e²-1 - 1] + [-e²2 + e²-1 + 3]
Therefore, the exact area A of the region bounded between the given curve and line over the interval [-2, 2] is:
A = -e²-2 + e²-1 - 1 - e²2 + e²-1 + 3
Note: The exact numerical value of this expression depends on the precision of the decimal approximations of e.
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Q1 (Q 5.5, Page 188 of the text book): Let X₁, X2, ..., Xn be a random sample From N(u, o²). Р ns² a. Prove that Yo², where Yn = and S² = 1(X - X)²/n. n-1 b. Prove that S²0², where S² = ₁
Prove that Yo², where Yn = (X1 + X2 + ... + Xn) / n and
S² = ∑(Xi - X)² / (n - 1)
For a random sample X1, X2, ..., Xn from a normally distributed population N(u, σ²),
the sample mean Yn is also normally distributed with mean u and variance σ²/n.
Hence, the standardized random variable Z = (Yn - u) / (σ / sqrt(n)) is distributed as N(0, 1).
Now consider the random variable:
(n - 1) S² / σ² = ∑(Xi - Yn)² / σ²,
The sum of squares of deviations of the individual observations from the sample mean. It can be shown that this is a chi-square random variable with (n - 1) degrees of freedom.
This is because the squared deviation of each observation from the mean is distributed as N(0, σ²) and the sum of squares of n independent standard normal random variables is distributed as chi-square with n degrees of freedom.
Hence, we have:(n - 1) S² / σ² ~ chi-square(n - 1)From this, we get the following result:
E(S²) = σ²,
Var(S²) = 2σ⁴ / (n - 1).
This can be proved using the moment-generating function of the chi-square distribution.Part b:Prove that S²0²,
where S² = ₁∑(Xi - X)² / (n - 1) and σ² is the population variance.
Now consider the random variable:S² / σ² = ∑(Xi - X)² / (σ²(n - 1))It can be shown that this is a chi-square random variable with (n - 1) degrees of freedom.
This is because the squared deviation of each observation from the population mean is distributed as N(0, σ²) and the sum of squares of n independent standard normal random variables is distributed as chi-square with n degrees of freedom.
Hence, we have:S² / σ² ~ chi-square(n - 1)From this, we get the following result:
E(S² / σ²) = 1,
Var(S² / σ²) = 2 / (n - 1).
This can be proved using the moment-generating function of the chi-square distribution.Now we have:
S² / σ² - 1 ~ chi-square(n - 1)
Therefore:
S² / σ² - Var(S² / σ²) = chi-square(n - 1) - 2 / (n - 1)
Now we use the fact that if U ~ chi-square(n) and V ~ chi-square(m) are independent, then U / n and V / m are independent and have an F-distribution with (n, m) degrees of freedom. Hence, we have:
S² / σ² / (n - 1) Var(S² / σ²) / (n - 2) = F(n - 1, n - 2)
Now let Fα(n - 1, n - 2) be the (1 - α)-th percentile of the F-distribution with (n - 1, n - 2) degrees of freedom. Then we have:
P[S² / σ² / (n - 1) < Fα(n - 1, n - 2)]
= α/2P[S² / σ² / (n - 1) > F1-α(n - 1, n - 2)]
= α/2
Therefore:
P[S² / σ² > F1-α(n - 1, n - 2) (n - 1) / n] = α
Now we know that σ² is a scale parameter, which means that if Y ~ N(0, 1), then σY ~ N(0, σ²). Therefore, we have:
S² / (σ²n) ~ chi-square(n - 1) / n
Now let χ²α(n - 1) be the (1 - α)-th percentile of the chi-square distribution with (n - 1) degrees of freedom. Then we have:
P[S² / (σ²n) < χ²α(n - 1) / n]
= α/2P[S² / (σ²n) > χ²1-α(n - 1) / n]
= α/2
Therefore:
P[S² / (σ²n) > χ²1-α(n - 1) / n] - P[S² / (σ²n) < χ²α(n - 1) / n] = α
Now we use the fact that if X ~ N(u, σ²) and Y ~ N(v, τ²) are independent, then X - Y ~ N(u - v, σ² + τ²).
Therefore, we have:(n - 1) S² / σ² ~ chi-square(n - 1)(n - 1) S0² / σ² ~ chi-square(n - 1)
Now let χ²α(n - 1) be the (1 - α)-th percentile of the chi-square distribution with (n - 1) degrees of freedom. Then we have:
P[(n - 1) S0² / σ² < χ²α(n - 1)]
= α/2P[(n - 1) S0² / σ² > χ²1-α(n - 1)]
= α/2
Therefore:
P[(n - 1) S0² / σ² > χ²1-α(n - 1)] - P[(n - 1) S0² / σ² < χ²α(n - 1)] = α
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a veterinarian charges $25.00 for each vaccination administered. if the veterinarian administered four vaccines into a patient, how much should the veterinarian charge the patient’s owner?
The veterinarian should charge the patient's owner a total of $100.00 for administering four vaccines, as each vaccine is priced at $25.00.
The veterinarian charges a fixed price of $25.00 for each vaccination administered. In this case, since the veterinarian administered four vaccines to the patient, the total charge can be calculated by multiplying the cost of one vaccine ($25.00) by the number of vaccines administered (4). Therefore, the veterinarian should charge the patient's owner a total of $100.00 for the four vaccinations.
To break it down further, each vaccination carries an individual cost of $25.00. When the veterinarian administers the first vaccine, the owner is charged $25.00. Similarly, for the second, third, and fourth vaccines, the charges remain the same, totaling $75.00. Adding up all these charges, the total amount comes to $100.00. Hence, the veterinarian should charge the patient's owner $100.00 for administering four vaccines.
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Find the area of each triangle to the nearest tenth.
Answer:
The area is 52.6 square inches.
Step-by-step explanation:
If you don't know base or height but you have info on a triangle, such as two sides and the angle in between, and a little bit of trig, you can find the area.
There's an SAS Trig Triangle area formula:
A = 1/2×side×side×sin
this works for two sides and the angle in between, which is exactly the info you have in your question.
A = 1/2(14)(8)sin70°
A = 56sin70°
A = 52.6 sq inches
