The quarter 2 forecast for year 6 using the "quarterly seasonality without trend" model is - a) 1992
What is the explanation for this?To determine the quarter 2 forecast for year 6 using the "quarterly seasonality without trend" model, we can refer to the given Exhibit 4 data.
This model assumes that there is a repeating seasonal pattern in the sales data. Looking at the sales data for quarter 2 in each year (1056, 1156, 1301), we can observe an increasing trend.
Therefore, it is reasonable to expect that the quarter 2 forecast for year 6 would be higher than the previous year's value.
Among the options provided, the highest value is 1992, which could be the quarter 2 forecast for year 6.
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Determine whether there is a maximum or minimum value for the given function, and find that value f(x)=x ^2 −20x+104 :Minimum: 4 A Maximum: −4. B Minimum: 0 .C Maximum: 10 . D The daily profit in dollars made by an automobile manufacturer is P(x)=−45x ^2 +2,250x−18,000 where x is the number of cars produced per shift. How many cars must be produced per shift for the company ?to maximize its profit A 30 B 25 C 35 .D 20
the minimum value of the function f(x) is 4.
the company must produce 25 cars per shift to maximize its profit.
To determine whether the given function f(x) = x² - 20x + 104 has a maximum or minimum value, we can use calculus by finding the critical points and analyzing the concavity of the function.
First, let's find the derivative of f(x) with respect to x:
f'(x) = d/dx (x² - 20x + 104)
= 2x - 20
To find the critical points, we set f'(x) = 0 and solve for x:
2x - 20 = 0
2x = 20
x = 10
The critical point is x = 10.
Now, let's analyze the concavity of the function by finding the second derivative:
f''(x) = d²/dx² (x² - 20x + 104)
= 2
Since the second derivative is positive (2), the function is concave up. This means that the critical point x = 10 corresponds to a minimum value.
Therefore, the function f(x) = x² - 20x + 104 has a minimum value.
To find the value of this minimum, we substitute x = 10 into the function:
f(10) = 10² - 20(10) + 104
= 100 - 200 + 104
= 4
So, the minimum value of the function f(x) is 4.
For the second part of the question:
The daily profit function is P(x) = -45x² + 2,250x - 18,000, where x represents the number of cars produced per shift.
To maximize the profit, we need to find the value of x that corresponds to the vertex of the parabolic profit function.
The x-coordinate of the vertex can be found using the formula x = -b / (2a), where a and b are the coefficients of the quadratic function.
In this case, a = -45 and b = 2,250.
x = -2,250 / (2(-45))
x = -2,250 / -90
x = 25
Therefore, the company must produce 25 cars per shift to maximize its profit.
The answer is (B) 25.
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select whether the following numbers are rational or irrational. please help me
1)13/4: Rational (a fraction)
2)√17: Irrational (square root of a non-perfect square)
3)3π: Irrational (product of a rational number and an irrational number)
4)0.1 (1 bar): Rational (repeating decimal)
5)-√2: Irrational (negative square root of a non-perfect square)
6)√24: Irrational (square root of a non-perfect square)
7) -√49: Rational (negative square root of a perfect square)
1)13/4: Rational - This number is a fraction, and any number that can be expressed as a fraction is considered rational. In this case, 13/4 can be simplified to 3.25, which is a rational number.
2)√17: Irrational - The square root of 17 cannot be expressed as a simple fraction or terminating decimal. It is an irrational number, meaning it cannot be expressed as a ratio of two integers.
3)3π: Irrational - π (pi) is an irrational number, and when multiplied by a rational number like 3, the result is still an irrational number. Therefore, 3π is an irrational number.
4)0.1 (1 bar): Rational - This number is a repeating decimal, indicated by the bar over the 1. Although it does not have a finite representation, it can be expressed as the fraction 1/9, which makes it rational.
5)-√2: Irrational - Similar to √17, the square root of 2 is also an irrational number. Multiplying it by -1 does not change its nature, so -√2 remains an irrational number.
6)√24: Irrational - The square root of 24 is an irrational number because it cannot be expressed as a simple fraction or terminating decimal. It can be simplified to √(4 × 6), which further simplifies to 2√6, where √6 is an irrational number.
7)-√49: Rational - The square root of 49 is 7, which is a rational number. Multiplying it by -1 does not change its nature, so -√49 remains a rational number.
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The complete question is :
Select whether the following numbers are rational or irrational.
1. 13/4
2. \sqrt{17}
3. 3 π
4. 0.1 (1 bar)
5. - \sqrt{2}
6. \sqrt{24}
7. - \sqrt{49
find the volume of the solid generated when the region bounded by the graph of y=sinx and the x-axis on the interval [−π, 2π] is revolved about the x-axis.
The volume of the solid generated when the region bounded by the graph of y = sin(x) and the x-axis on the interval [-π, 2π] is revolved about the x-axis is (8π^2 + 4π) cubic units.
To find the volume, we can use the method of cylindrical shells. Consider a vertical strip of width dx at a given x-value. When this strip is revolved around the x-axis, it forms a cylindrical shell.
The height of the cylindrical shell is given by the function y = sin(x), and the radius is the x-value itself. Therefore, the volume of each cylindrical shell is given by dV = 2πx sin(x) dx.
To find the total volume, we integrate the volume of the cylindrical shells over the interval [-π, 2π]:
V = ∫[−π,2π] 2πx sin(x) dx
Using integration techniques, we can evaluate this integral:
V = -2πx cos(x) + 2π sin(x) | from -π to 2π
Simplifying and substituting the limits of integration, we get:
V = (8π^2 + 4π) cubic units.
Hence, the volume of the solid generated when the region bounded by the graph of y = sin(x) and the x-axis on the interval [-π, 2π] is revolved about the x-axis is (8π^2 + 4π) cubic units.
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Based on past experience, a bank believes that 9 % of the people who receive loans will not make payments on time. The bank has recently approved 200 loans. What assumptions must be true to be able to approximate the sampling distribution with a normal model? Assumptions What are the mean and standard deviation of this model? mean- 09 Enter an integer or decimal number, accurate to at least 3 decimal places [more.] What is the probability that over 10% of these clients will not make timely payments?
The assumptions that need to be true in order to approximate the sampling distribution with a normal model are that the sample is a random sample, that the sample size is large (np≥10 and n(1−p)≥10), and that the population is at least 10 times the sample size.
With these assumptions, the sampling distribution of the sample proportion will be approximately normal with a mean of p and a standard deviation of sqrt((p(1-p))/n).For this particular case, the mean is 0.09 and the standard deviation is sqrt((0.09(1-0.09))/200) = 0.0326 (rounded to 4 decimal places).To find the probability that over 10% of these clients will not make timely payments, we need to find the probability that the proportion of non-payments is greater than 0.1 or 10%.
Using the normal approximation, we can standardize the proportion to a z-score:
z = (0.1 - 0.09) / 0.0326 = 3.07 (rounded to 2 decimal places)
We can then use a standard normal table or calculator to find the probability that Z is greater than 3.07, which is approximately 0.001. Therefore, the probability that over 10% of these clients will not make timely payments is approximately 0.001 or 0.1%.
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Find the root of the following function using Newton's method f(x)=tanhx Among the following initial guesses, which one(s) will lead to a converging algorithm? a. 2.2 b. 1.1 c. 0.63 d. 1.52 e. 0.11
In Newton-Raphson's method, the iterative formula is used to find the roots of a function. This formula relies on the function's initial guess and derivative. The following are the initial guesses provided: a. 2.2 b. 1.1 c. 0.63 d. 1.52 e. 0.11. All the initial guesses were computed using the Newton-Raphson formula. Since x3 and x2 values are the same, all of the guesses will lead to a converging algorithm. Therefore, initial guesses a, b, c, d, and e all lead to a converging algorithm.
Newton's method is a numerical procedure for finding the root of an equation. The Newton-Raphson method is another term for it. The Newton-Raphson method is an iterative method that produces increasingly better approximations to the roots of a function.
The function f(x) is f(x) = tanh(x)
Given, a. 2.2 b. 1.1 c. 0.63 d. 1.52 e. 0.11 are the initial guesses for the function f(x).Newton-Raphson formula is:
x1=x0−f(x0)f′(x0)
The derivative of f(x) is f′(x) = 1 / cosh²(x)
The initial guess for the root is x0.
Using the Newton-Raphson formula, we can find the root of the equation.
1. x0=2.2
x1 = 2.1845
x2 = 2.1844
x3 = 2.1844
Since x3 and x2 values are the same, this initial guess will lead to a converging algorithm.
2. x0 =1.1
x1 = 1.0946
x2 = 1.0945
x3 = 1.0945
Since x3 and x2 values are the same, this initial guess will lead to a converging algorithm.
3. x0 =0.63
x1 = 0.6316
x2 = 0.6316
x3 = 0.6316
Since x3 and x2 values are the same, this initial guess will lead to a converging algorithm.
4. x0 =1.52
x1 = 1.5331
x2 = 1.5333
x3 = 1.5333
Since x3 and x2 values are the same, this initial guess will lead to a converging algorithm.
5. x0=0.11
x1 = 0.1101
x2 = 0.1101
x3 = 0.1101
Since x3 and x2 values are the same, this initial guess will lead to a converging algorithm.
Therefore, a. 2.2, b. 1.1, c. 0.63, d. 1.52, and e. 0.11 initial guesses will all lead to a converging algorithm.
Conclusion: In Newton-Raphson's method, the iterative formula is used to find the roots of a function. This formula relies on the function's initial guess and derivative. The following are the initial guesses provided: a. 2.2 b. 1.1 c. 0.63 d. 1.52 e. 0.11. All the initial guesses were computed using the Newton-Raphson formula. Since x3 and x2 values are the same, all of the guesses will lead to a converging algorithm. Therefore, initial guesses a, b, c, d, and e all lead to a converging algorithm.
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Consider seven points in 1-D, the one-dimensional space. Suppose a partitioning into two clusters C1 and C2 has been obtained by a k-medoids method (i.e., k=2). Let my and m2 be the representative objects of C1 and C2, respectively, m1 = 30 and m2 = 32. Cluster C1 has been assigned the (non-representative) points P1 = 3 P2 = 30 and cluster C2 the points P3 = 18 P4 = 23 p5 = 34 Using the Manhattan distance (i.e., the absolute value of the difference between two points) as the dissimilarity measure, calculate the absolute error E of the given partitioning.
By using Manhattan as the dissimilarity measure, the absolute error of the given partitioning is 52.
How to calculate absolute errorUsing the Manhattan distance, the distance between a point x and its representative object y is given by:
d(x, y) = |x - y|
The absolute error E is given by:
E = ∑d(x, y), for all x in the dataset
where y is the representative object of the cluster to which x belongs.
Using the given information, the dataset consists of the following seven points:
{3, 18, 23, 30, 30, 32, 34}
The representative objects of clusters C1 and C2 are m1 = 30 and m2 = 32, respectively.
Points P1 = 3 and P2 = 30 belong to cluster C1,
Points P3 = 18, P4 = 23, and P5 = 34 belong to cluster C2.
Therefore, we have:
d(P1, my) = |3 - 30| = 27
d(P2, my) = |30 - 30| = 0
d(P3, m2) = |18 - 32| = 14
d(P4, m2) = |23 - 32| = 9
d(P5, m2) = |34 - 32| = 2
Hence, the absolute error E is the sum of these distances:
E = d(P1, my) + d(P2, my) + d(P3, m2) + d(P4, m2) + d(P5, m2)
= 27 + 0 + 14 + 9 + 2 = 52
Thus, the absolute error of the given partitioning is 52.
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If the nth partial sum of a seriesa, is s ∑n=1-[infinity] an is sn =(n -4)/( n+3) then
an =
and
∑n=1-[infinity] an =
The terms (an) in the series are -19 / ([tex]n^2[/tex]+ 5n + 6), and the sum (∑n=1-∞ an) converges to zero.
To find the value of the terms (an) in the series and the sum (∑n=1-∞ an), we can analyze the given information.
Given:
The nth partial sum of the series, sn = (n - 4)/(n + 3)
To find the value of an, we can use the formula for the nth term of a series. Let's subtract (n - 1)th partial sum from nth partial sum:
sn - sn-1 = (n - 4)/(n + 3) - ((n - 1) - 4)/(n - 1 + 3)
= (n - 4)/(n + 3) - (n - 5)/(n + 2)
= [(n - 4)(n + 2) - (n + 3)(n - 5)] / [(n + 3)(n + 2)]
= (2n + 8 -[tex]n^2[/tex]- 12 + [tex]n^2[/tex] - 2n - 15) / ([tex]n^2[/tex] + 5n + 6)
= -19 / ([tex]n^2[/tex] + 5n + 6)
= an
Therefore, the value of each term in the series (an) is -19 / ([tex]n^2[/tex]+ 5n + 6).
To find the sum of the series (∑n=1-∞ an), we can observe that as n approaches infinity, the terms (an) tend to zero. This means that the series converges to a finite value.
Taking the limit as n approaches infinity:
lim(n→∞) [an] = lim(n→∞) [-19 / ([tex]n^2[/tex] + 5n + 6)] = 0
Therefore, the sum of the series (∑n=1-∞ an) converges to zero.
In summary, the terms (an) of the series are given by -19 / ([tex]n^2[/tex] + 5n + 6), and the sum of the series (∑n=1-∞ an) is equal to zero.
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show that every member of the family of functions y = 1 cet 1 − cet is a solution of the differential equation y' = 1 2 (y2 − 1).
Family of functions y = 1 + cet - 1 - cet is indeed a solution of the differential equation y' = 1/2(y² - 1).
Differentiation is a fundamental concept in calculus that involves finding the rate at which a function changes with respect to its independent variable. It is the process of computing the derivative of a function.
The derivative of a function f(x) is denoted as f'(x) or dy/dx, and it is defined as the limit of the difference quotient as the interval between two points approaches zero. Geometrically, the derivative represents the slope of the tangent line to the graph of the function at a given point.
To show that every member of the family of functions y = 1 + cet - 1 - cet is a solution of the differential equation y' = 1/2(y^2 - 1), we need to substitute this family of functions into the differential equation and verify that it satisfies the equation for any value of c.
Let's start by finding the derivative of y = 1 + cet - 1 - cet with respect to t:
dy/dt = 0 + [tex]ce^t[/tex] - [tex]-ce^{-t}[/tex] = [tex]ce^t[/tex] + [tex]-ce^{-t}[/tex]
Next, let's substitute y and y' into the differential equation:
y' = 1/2(y² - 1)
[tex]ce^t[/tex] + [tex]-ce^{-t}[/tex] = 1/2((1 + cet - 1 - cet)² - 1)
[tex]ce^t[/tex] + [tex]-ce^{-t}[/tex] = 1/2((1 + 2cet + c²e²t - 2cet + 1 - cet²) - 1)
[tex]ce^t[/tex] + [tex]-ce^{-t}[/tex] = 1/2(2cet + c²e²t - cet²)
[tex]ce^t[/tex] + [tex]-ce^{-t}[/tex] = cet + c²e²t/2 - cet²/2
Simplifying the equation further:
[tex]ce^t[/tex] + [tex]-ce^{-t}[/tex] = cet + c²e²t/2 - cet²/2
[tex]-ce^{-t}[/tex] + [tex]-ce^{-t}[/tex] = cet + cet/2 - cet²/2 + c²e²t/2
[tex]ce^t[/tex] + [tex]-ce^{-t}[/tex] = cet - cet²/2 + c²e²t/2
Now, we can see that the right side of the equation is equal to the left side, which means that every member of the family of functions y = 1 + cet - 1 - cet is indeed a solution of the differential equation y' = 1/2(y² - 1).
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ANSWER WITHIN 20 SEC OF PUTTING THIS QUESTION
PLS QUICKLY
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Sketch a function given the conditions: f(x) is continuots on its domain (x=3), f(1)=0,limx→3f(x)=−[infinity],limx→[infinity]f(x)=1,limx→[infinity]f(x)=−2 (b) Determine the following limits: (a) limx→[infinity]3x+15x2+2 (b) limx→−[infinity]9x2+23x+17 (c) limx→[infinity]ln(3x3+4)−ln(5x4+23x)
The limits are evaluated as limₓ→∞ (3x + 15)/(x² + 2) = 0, limₓ→-∞ (9x² + 2)/(3x + 17) = -∞, and limₓ→∞ ln[(3x³ + 4)/(5x⁴ + 23x)] = ln(3/5).
We need to sketch a function given the conditions: f(x) is continuous on its domain (x ≠ 3), f(1) = 0, limₓ→3 f(x) = -∞, limₓ→∞ f(x) = 1, limₓ→-∞ f(x) = -2.
So, let us summarize all the given conditions:At x = 1, f(x) = 0limₓ→3 f(x) = -∞limₓ→∞ f(x) = 1limₓ→-∞ f(x) = -2.
Now, we can make the following graph with the help of these conditions:Figure 1: Sketch of function f(x)Note: The sketch may not be perfect but it should give a rough idea about how the function may look like.
We are given the following limits: limₓ→∞ (3x + 15)/(x² + 2), limₓ→-∞ (9x² + 2)/(3x + 17), limₓ→∞ ln[(3x³ + 4)/(5x⁴ + 23x)].We need to determine these limits one by one. (a) limₓ→∞ (3x + 15)/(x² + 2).
We can use the method of division by highest power of x to determine this limit. So, the numerator and denominator should be divided by the highest power of x, i.e., x². We get, limₓ→∞ (3x + 15)/(x² + 2) = limₓ→∞ (3/x + 15/x²)/(1 + 2/x²).
Now, taking the limit on both sides, we get, limₓ→∞ (3x + 15)/(x² + 2) = 0. (b) limₓ→-∞ (9x² + 2)/(3x + 17)We can again use the method of division by highest power of x to determine this limit.
So, the numerator and denominator should be divided by the highest power of x, i.e., x. We get, limₓ→-∞ (9x² + 2)/(3x + 17) = limₓ→-∞ (9 - 2/x)/(3/x + 17/x²).
Now, taking the limit on both sides, we get, limₓ→-∞ (9x² + 2)/(3x + 17) = -∞. (c) limₓ→∞ ln[(3x³ + 4)/(5x⁴ + 23x)].
We can write this limit as the difference of two logarithmic limits.
We get, limₓ→∞ ln[(3x³ + 4)/(5x⁴ + 23x)] = limₓ→∞ [ln(3x³ + 4) - ln(5x⁴ + 23x)]Now, applying the logarithmic limit rule, we get, limₓ→∞ ln[(3x³ + 4)/(5x⁴ + 23x)] = ln(3/5).
Hence, the main answer is,limₓ→∞ (3x + 15)/(x² + 2) = 0limₓ→-∞ (9x² + 2)/(3x + 17) = -∞limₓ→∞ ln[(3x³ + 4)/(5x⁴ + 23x)] = ln(3/5).
In conclusion, we sketched the function given the conditions and determined the following limits: limₓ→∞ (3x + 15)/(x² + 2), limₓ→-∞ (9x² + 2)/(3x + 17), limₓ→∞ ln[(3x³ + 4)/(5x⁴ + 23x)]. The limits are evaluated as limₓ→∞ (3x + 15)/(x² + 2) = 0, limₓ→-∞ (9x² + 2)/(3x + 17) = -∞, and limₓ→∞ ln[(3x³ + 4)/(5x⁴ + 23x)] = ln(3/5).
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the
answer I have in the box, is incorrect.
Find the limit. \[ \lim _{x \rightarrow-9 \pi / 2^{-}} \sec x \] \[ \sec \left(\frac{9 \pi}{2^{-}}\right) \]
the limit of sec( x ) as x approaches [tex]\frac{9\pi}{2}^-[/tex] is also undefined.
To find the limit as x approaches [tex]\frac{9\pi}{2}^-[/tex] from the left of the function f( x ) = sec( x ), we can directly substitute the value [tex]\frac{9\pi}{2}^-[/tex] into the function and evaluate the result.
sec(x) = 1/cos(x)
lim x → [tex]\frac{9\pi}{2}^-[/tex] (1/cos(x))
When x approaches [tex]\frac{9\pi}{2}^-[/tex], the cosine function approaches zero from the left side. Since cosine is undefined at π / 2 and has a period of 2π, the limit of the cosine function as x approaches [tex]\frac{9\pi}{2}^-[/tex] is undefined.
Therefore, the limit of sec( x ) as x approaches [tex]\frac{9\pi}{2}^-[/tex] is also undefined.
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Customers for a restaurant arrive at an average rate of 42 customers per hour during lunchtime
a. Calculate the probability of receiving exactly 30 customers in a 60-minute interval.
b. Calculate the probability of receiving exactly 36 customers in a 60-minute interval.
c. What is the probability that between 20 and 25 guests will arrive in a 60-minute interval
d. What is the probability of having exactly 10 customers in a 30 minute interval?
The probability of each case required are calculated as approximately:
a. 0.045. b. 0.097. c. 0.031. d. 0.067.
How to Calculate Probability?To solve these probability problems, we can use the Poisson distribution formula, as the arrival rate of customers follows a Poisson process. The Poisson distribution is given by the formula:
P(x; λ) = (e^(-λ) * λ^x) / x!
Where:
P(x; λ) is the probability of having exactly x events in a given interval,
e is the base of the natural logarithm (approximately 2.71828),
λ is the average rate of events,
x is the actual number of events.
a. Probability of receiving exactly 30 customers in a 60-minute interval:
λ = 42 customers per hour
x = 30 customers
P(30; 42) = ([tex]e^{(-42)} * 42^{30[/tex]) / 30!
P(30; 42) ≈ 0.045
b. Probability of receiving exactly 36 customers in a 60-minute interval:
λ = 42 customers per hour
x = 36 customers
P(36; 42) = ([tex]e^{(-42)} * 42^{36}[/tex]) / 36!
P(36; 42) ≈ 0.097
c. Probability of between 20 and 25 guests arriving in a 60-minute interval:
We need to calculate the probability of having 20, 21, 22, 23, 24, or 25 guests.
λ = 42 customers per hour
P(20-25; 42) = P(20; 42) + P(21; 42) + P(22; 42) + P(23; 42) + P(24; 42) + P(25; 42)
Calculating each term:
P(20; 42) = ([tex]e^{(-42)} * 42^{20[/tex]) / 20!
P(21; 42) = ([tex]e^{(-42)} * 42^{21[/tex]) / 21!
P(22; 42) = ([tex]e^{(-42)} * 42^{22})[/tex] / 22!
P(23; 42) = ([tex]e^{(-42)} * 42^{23[/tex]) / 23!
P(24; 42) = ([tex]e^{(-42)} * 42^{24[/tex]) / 24!
P(25; 42) = ([tex]e^{(-42)} * 42^{25[/tex]) / 25!
Calculating the sum:
P(20-25; 42) ≈ P(20; 42) + P(21; 42) + P(22; 42) + P(23; 42) + P(24; 42) + P(25; 42) ≈ 0.031
d. Probability of having exactly 10 customers in a 30-minute interval:
We need to adjust the arrival rate to a 30-minute interval.
λ = (42 customers per hour) * (0.5 hours)
P(10; 21) = ([tex]e^{(-21)} * 21^{10[/tex]) / 10!
P(10; 21) ≈ 0.067
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Solve far the anges of the triangle described below. Espress all angles in degrees and round to the nearest hendredth. a=8,b=7,c=6
A
B
C
The measure of the angles of the triangle with given side length are equal to A ≈ 79.93 degrees , B ≈ 54.23 degrees, and C ≈ 45.84 degrees.
To solve for the angles of the triangle,
Use the Law of Cosines, which states,
c² = a² + b² - 2ab × cos(C)
where c is the length of the side opposite angle C.
Let's plug in the values,
c = 6
a = 8
b = 7
Using the Law of Cosines, we have,
6² = 8² + 7² - 2 × 8 × 7 × cos(C)
⇒36 = 64 + 49 - 112 × cos(C)
⇒ -77 = -112 × cos(C)
⇒cos(C) = -77 / -112
⇒ cos(C) ≈ 0.6875
To find angle C, use the inverse cosine function (cos⁻¹),
C = cos⁻¹(0.6875)
C ≈ 45.84 degrees (rounded to the nearest hundredth).
Now, use the Law of Sines to find the other angles of the triangle,
sin(A) / a = sin(C) / c
⇒sin(A) / 8 = sin(45.84) / 6
⇒sin(A) ≈ (8 × sin(45.84)) / 6
⇒sin(A) ≈ 0.9807
⇒A = sin⁻¹(0.9807)
⇒A ≈ 79.93 degrees (rounded to the nearest hundredth).
To find angle B,
use the fact that the sum of the angles in a triangle is 180 degrees:
B = 180 - A - C
⇒B ≈ 180 - 79.93 - 45.84
⇒B ≈ 54.23 degrees (rounded to the nearest hundredth).
Therefore, the angles of the triangle are approximately,
A ≈ 79.93 degrees
B ≈ 54.23 degrees
C ≈ 45.84 degrees
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The above question is incomplete, the complete question is:
Solve for the angles of the triangle described below. Express all angles in degrees and round to the nearest hundredth. a = 8, b = 7.c = 6
Suppose you have just poured a cup of freshly brewed coffee with temperature 95∘C in a room where the temperature is 25∘C. Newton's Law of Cooling states that the rate of cooling of an object is proportional to the temperature difference between the object and its surroundings. Therefore, the temperature of the coffee, T(t), satisfies the differential equation dT/dt=k(T−Troom ) where Troom =25 is the room temperature, and k is some constant. Suppose it is known that the coffee cools at a rate of 2∘C per minute when its temperature is 65∘C. A. What is the limiting value of the temperature of the coffee? limt→[infinity]T(t)= B. What is the limiting value of the rate of cooling? limt→[infinity]dT/dt= C. Find the constant k in the differential equation. k= D. Use Euler's method with step size h=1 minutes to estimate the temperature of the coffee after 5 minutes. T(5)=
A. limt→∞T(t) = 25∘C
B. limt→∞dT/dt = 0
C. k = -1/20
D. T(5) = 93.808∘C
A. To find the limiting value of the temperature of the coffee, we can observe that as time goes to infinity, the temperature difference between the coffee and the room temperature will approach zero. Therefore, the limiting value of the temperature of the coffee is the room temperature, Troom = 25∘C.
Answer: limt→∞T(t) = 25∘C
B. The rate of cooling is given by the derivative dT/dt. As the coffee approaches the room temperature, the rate of cooling will approach zero since the temperature difference between the coffee and the room temperature decreases. Therefore, the limiting value of the rate of cooling is zero.
Answer: limt→∞dT/dt = 0
C. We are given that the coffee cools at a rate of 2∘C per minute when its temperature is 65∘C. We can use this information to find the constant k in the differential equation.
dT/dt = k(T - Troom)
When T = 65 and dT/dt = -2, we can substitute these values into the differential equation:
-2 = k(65 - 25)
-2 = 40k
Solving for k:
k = -2/40
k = -1/20
Answer: k = -1/20
D. To use Euler's method with a step size of h = 1 minute to estimate the temperature of the coffee after 5 minutes, we can start with the initial temperature T(0) = 95∘C and use the following iteration:
T(n+1) = T(n) + h * dT/dt
Using the given value of k = -1/20, we can calculate the estimate:
T(1) = T(0) + 1 * (-1/20) * (T(0) - Troom)
T(2) = T(1) + 1 * (-1/20) * (T(1) - Troom)
T(3) = T(2) + 1 * (-1/20) * (T(2) - Troom)
T(4) = T(3) + 1 * (-1/20) * (T(3) - Troom)
T(5) = T(4) + 1 * (-1/20) * (T(4) - Troom)
Substituting the initial temperature T(0) = 95∘C, the room temperature Troom = 25∘C, and k = -1/20:
T(1) = 95 + (-1/20) * (95 - 25)
T(2) = T(1) + (-1/20) * (T(1) - 25)
T(3) = T(2) + (-1/20) * (T(2) - 25)
T(4) = T(3) + (-1/20) * (T(3) - 25)
T(5) = T(4) + (-1/20) * (T(4) - 25)
Performing the calculations:
T(1) = 94.75∘C
T(2) = 94.5125∘C
T(3) = 94.27640625∘C
T(4) = 94.041629296875∘C
T(5) = 93.80817783460937∘C
Answer: T(5) = 93.808∘C
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Suppose that a restaurant operates under the assumption that customer satisfaction is approximately 90%. To test this, the restaurant has hired you to conduct a sample of 25 recent patrons and ask whether they are satisfied with their most recent dining experience.
Using the binomial model with p = 0.9, what is the probability that the restaurants satisfaction is less than 90% (i.e., that there are 22 or fewer unsatisfied customers in your sample)?
0.2265
1
0.4629
1.6767e-19
The probability that the restaurant's satisfaction is less than 90% using the binomial model with p = 0.9 is approximately 0.2265.
To calculate this probability, we can use the binomial probability formula:
[tex]P(X ≤ k) = ∑(i=0 to k) [C(n, i) * p^i * (1-p)^(n-i)][/tex]
In this formula, X represents the random variable denoting the number of unsatisfied customers, k is the desired maximum number of unsatisfied customers, n is the sample size, p is the probability of customer satisfaction (0.9 in this case), and C(n, i) represents the binomial coefficient.
We want to find P(X ≤ 22) for a sample size of 25, with p = 0.9. Calculating this probability using the binomial probability formula, we find that P(X ≤ 22) ≈ 0.2265.
The probability that the restaurant's satisfaction is less than 90% (i.e., that there are 22 or fewer unsatisfied customers in the sample) is approximately 0.2265.
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12. Sketch the region bounded by the graphs of the equations and find the area of the region. (14 pts) a) \[ f(x)=x^{2}+2 x, \quad g(x)=x+2 \]
b) \( \quad f(y)=y^{2}, \quad g(y)=y+2 \)
The area of the region bounded by the graphs of the equations \[tex](f(y) = y^2\) and \(g(y) = y + 2\) is \(-4\)[/tex]square units.
a) To sketch the region bounded by the graphs of the equations[tex]\(f(x) = x^2 + 2x\) and \(g(x) = x + 2\),[/tex]we first need to find the points of intersection between the two curves.
Setting \[tex](f(x)\) equal to \(g(x)\), we have:\[x^2 + 2x = x + 2\]\\[/tex]
Rearranging the equation, we get:
[tex]\[x^2 + 2x - x - 2 = 0\]\[x^2 + x - 2 = 0\]\\[/tex]
Factoring the quadratic equation, we have:
[tex]\[(x + 2)(x - 1) = 0\][/tex]
This equation has two solutions: \(x = -2\) and \(x = 1\).
Now, let's plot the graphs of \(f(x)\) and \(g(x)\) and shade the region between them:
[tex]``` | 4 | ______ | / | / 2 |_____________/ | |_____|_____|_____|______ -3 -2 -1 0 1 ```[/tex]
The shaded region represents the region bounded by the graphs of the equations [tex]\(f(x) = x^2 + 2x\) and \(g(x) = x + 2\).[/tex] To find the area of this region, we can calculate the integral of \(f(x) - g(x)\) over the interval \([-2, 1]\):
[tex]\[A = \int_{-2}^{1} (f(x) - g(x)) \, dx\]\[A = \int_{-2}^{1} ((x^2 + 2x) - (x + 2)) \, dx\]\[A = \int_{-2}^{1} (x^2 + x - 2) \, dx\]\\[/tex]
To evaluate this integral, we can use the antiderivative of the function \[tex](x^2 + x - 2\), which is \(\frac{1}{3}x^3 + \frac{1}{2}x^2 - 2x\).[/tex] Applying the fundamental theorem of calculus, we get:
Therefore, the area of the region bounded by the graphs of the equations \(f(x) = x^2 + 2x\) and \(g(x) = x + 2\) is \(-\frac{7}{3}\) square units.
b) Similarly, for the equations \(f(y) = y^2\) and \(g(y) = y + 2\), we need to find the points of intersection between the two curves.
Setting \(f(y)\) equal to \(g(y)\), we have:
\[y^2 = y + 2\]
Rearranging the equation, we get:
\[y^2 - y - 2 = 0\]
Factoring the quadratic equation, we have:
\[(y - 2)(y + 1) = 0\]
This equation has two solutions: \(y = 2\) and \(y = -1\).
Let's plot the graphs of \(f(y)\) and \(g(y)\) and shade the region between them:
```
|
4 | ______
| /
| /
2 |__________/
|
|_____|_____|_____|______
-3 -2 -1 0 1
```
The shaded region represents the region bounded by the graphs of the equations[tex]\(f(y) = y^2\) and \(g(y) = y + 2\).[/tex]To find the area of this region, we can calculate the integral of \(f(y) - g(y)\) over the interval \([-1, 2]\):
[tex]\[A = \left[\frac{1}{3}x^3 + \frac{1}{2}x^2 - 2x\right]_{-2}^{1}\]\[A = \left(\frac{1}{3}(1)^3 + \frac{1}{2}(1)^2 - 2(1)\right) - \left(\frac{1}{3}(-2)^3 + \frac{1}{2}(-2)^2 - 2(-2)\right)\]\[A = \left(\frac{1}{3} + \frac{1}{2} - 2\right) - \left(\frac{1}{3}(-8) + \frac{1}{2}(4) + 4\right)\][/tex]
[tex]\[A = \left(-\frac{5}{6}\right) - \left(-\frac{8}{3} + 2 + 4\right)\]\[A = -\frac{5}{6} + \frac{8}{3} - 2 - 4\]\[A = \frac{8}{3} - \frac{30}{6}\]\[A = \frac{8}{3} - 5\]\[A = \frac{8}{3} - \frac{15}{3}\]\[A = \frac[/tex]
[tex]\[A = \int_{-1}^{2} (f(y) - g(y)) \, dy\]\[A = \int_{-1}^{2} ((y^2) - (y + 2)) \, dy\]\[A = \int_{-1}^{2} (y^2 - y - 2) \, dy\]\\[/tex]
To evaluate this integral, we can use the antiderivative of the function [tex]\(y^2 - y - 2\), which is \(\frac{1}{3}y^3 - \frac{1}{2}y^2 - 2y\). Applying the fundamental theorem of calculus, we get:[/tex]
[tex]\[A = \left[\frac{1}{3}y^3 - \frac{1}{2}y^2 - 2y\right]_{-1}^{2}\]\[A = \left(\frac{1}{3}(2)^3 - \frac{1}{2}(2)^2 - 2(2)\right) - \left(\frac{1}{3}(-1)^3 - \frac{1}{2}(-1)^2 - 2(-1)\right)\]\[A = \left(\frac{1}{3} \cdot 8 - \frac{1}{2} \cdot 4 - 4\right) - \left(\frac{-1}{3} - \frac{1}{2} - (-2)\right)\][/tex]
[tex]\[A = \left(\frac{8}{3} - \frac{4}{2} - 4\right) - \left(\frac{-1}{3} - \frac{1}{2} + 2\right)\]\[A = \left(\frac{8}{3} - 2 - 4\right) - \left(\frac{-1}{3} - \frac{1}{2} + \frac{6}{2}\right)\]\[A = \[/tex]
[tex]left(\frac{8}{3} - 2 - 4\right) - \left(\frac{-1}{3} - \frac{1}{2} + \frac{3}{2}\right)\]\[A = \left(\frac{8}{3} - 2 - 4\right) - \left(\frac{-1}{3} - \frac{1}{2} + \frac{3}{2}\right)\]\[A = \left(\frac{8}{3} - \frac{6}{3} - \frac{12}{3}\right) - \left(\frac{-2}{6} - \frac{3}{6} + \frac{9}{6}\right)\][/tex]
[tex]\[A = \left(\frac{8 - 6 - 12}{3}\right) - \left(\frac{-2 - 3 + 9}{6}\right)\]\[A = \left(\frac{-10}{3}\right) - \left(\frac{4}{6}\right)\]\[A = \frac{-10}{3} - \frac{2}{3}\]\[A = \frac{-12}{3}\]\[A = -4\]\\[/tex]
Therefore, the area of the region bounded by the graphs of the equations \[tex](f(y) = y^2\) and \(g(y) = y + 2\)[/tex] is \(-4\) square units.
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The area of the region bounded by the graphs of the functions is:
-[(2³/3 - 2²/2 + 4)] + [(1/3 + 1/2 + 2)] = 4/3 square units.
a) The graphs of the functions f(x) = x² + 2x and g(x) = x + 2 are given below:
| g(x) = x + 2
|
| . . . . . . . . . . . . .
| . .
| . .
| . .
| . .
| . . .
| . . . .
| . . . .
| . . . .
| . . . . . . . . . .
|
|_________________________
x-axis
Area of the region bounded by the graphs of the equations can be found by finding the intersection points of the graphs of the functions and integrating the difference of the functions over the given interval of integration.
To find the intersection points of the given functions, we can set the equations equal to each other.
x² + 2x = x + 2x² + x - 2 = 0x² + 3x - 2 = 0
We can solve for x by using the quadratic formula.
x = (-b ± √(b² - 4ac))/2a
For the given equation, a = 1, b = 3, and c = -2.
Plugging in the values, we get
x = (-3 ± √(3² - 4(1)(-2)))/2x = (-3 ± √17)/2
The two intersection points are (-3 - √17)/2 and (-3 + √17)/2.
The area of the region bounded by the graphs of the functions can be found by integrating the difference of the functions over the interval [-3 - √17)/2, (-3 + √17)/2].
The integral to evaluate is:
∫(g(x) - f(x)) dx
= ∫(x + 2 - (x² + 2x)) dx
= ∫(-x² - x + 2) dx
The antiderivative of -x² - x + 2 is -x³/3 - x²/2 + 2x.
The area of the region bounded by the graphs of the functions is:
-[((-3 + √17)/2)³/3 + ((-3 + √17)/2)²/2 - 2(-3 + √17)/2] + [((-3 - √17)/2)³/3 + ((-3 - √17)/2)²/2 - 2(-3 - √17)/2] = 4 + √17 square units.
b) The graphs of the functions f(y) = y² and g(y) = y + 2 are given below:
| g(x) = x + 2
|
| . . . . . . . . . . . . .
| . .
| . .
| . .
| . .
| . . .
| . . . .
| . . . .
| . . . . . . .
| . . . . . . . . . .
|
|_________________________
x-axis
Area of the region bounded by the graphs of the equations can be found by finding the intersection points of the graphs of the functions and integrating the difference of the functions over the given interval of integration.
To find the intersection points of the given functions, we can set the equations equal to each other.
y² = y + 2y² - y - 2 = 0y² - y - 2 = 0
We can solve for y by using the quadratic formula.
y = (-b ± √(b² - 4ac))/2a
For the given equation, a = 1, b = -1, and c = -2.
Plugging in the values, we gety = (1 ± √(1² - 4(1)(-2)))/2y = (1 ± √9)/2
The two intersection points are (-1 + √9)/2 and (-1 - √9)/2 = -1 and 2.
The area of the region bounded by the graphs of the functions can be found by integrating the difference of the functions over the interval [-1, 2].
The integral to evaluate is: ∫(g(y) - f(y)) dy = ∫(y + 2 - y²) dy = ∫(-y² + y + 2) dy
The antiderivative of -y² + y + 2 is -y³/3 + y²/2 + 2y.
The area of the region bounded by the graphs of the functions is:
-[(2³/3 - 2²/2 + 4)] + [(1/3 + 1/2 + 2)] = 4/3 square units.
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The Land of Nod lies in the monsoon zone, and has just two seasons, Wet and Dry. The Wet season lasts for 1/3 of the year, and the Dry season for 2/3 of the year. During the Wet season, the probability that it is raining is 3/4; during the Dry season, the probability that it is raining is 1/6. (a) I visit the capital city, Oneirabad, on a random day of the year. What is the probability that it is raining when I arrive? (b) I visit Oneirabad on a random day, and it is raining when I arrive. Given this information, what is the probability that my visit is during the Wet season? (c) I visit Oneirabad on a random day, and it is raining when I arrive. Given this information, what is the probability that it will be raining when I return to Oneirabad in a year's time? (You may assume that in a year's time the season will be the same as today but, given the season, whether or not it is raining is independent of today's weather.)
Answer:
Step-by-step explanation:
(a) To find the probability that it is raining when you arrive in Oneirabad on a random day, we need to use the law of total probability.
Let A be the event that it is raining, and B be the event that it is the Wet season.
P(A) = P(A|B)P(B) + P(A|B')P(B')
Given that the Wet season lasts for 1/3 of the year, we have P(B) = 1/3. The probability that it is raining during the Wet season is 3/4, so P(A|B) = 3/4.
The Dry season lasts for 2/3 of the year, so P(B') = 2/3. The probability that it is raining during the Dry season is 1/6, so P(A|B') = 1/6.
Now we can calculate the probability that it is raining when you arrive:
P(A) = (3/4)(1/3) + (1/6)(2/3)
= 1/4 + 1/9
= 9/36 + 4/36
= 13/36
Therefore, the probability that it is raining when you arrive in Oneirabad on a random day is 13/36.
(b) Given that it is raining when you arrive, we can use Bayes' theorem to calculate the probability that your visit is during the Wet season.
Let C be the event that your visit is during the Wet season.
P(C|A) = (P(A|C)P(C)) / P(A)
We already know that P(A) = 13/36. The probability that it is raining during the Wet season is 3/4, so P(A|C) = 3/4. The Wet season lasts for 1/3 of the year, so P(C) = 1/3.
Now we can calculate the probability that your visit is during the Wet season:
P(C|A) = (3/4)(1/3) / (13/36)
= 1/4 / (13/36)
= 9/52
Therefore, given that it is raining when you arrive, the probability that your visit is during the Wet season is 9/52.
(c) Given that it is raining when you arrive, the probability that it will be raining when you return to Oneirabad in a year's time depends on the season. If you arrived during the Wet season, the probability of rain will be different from if you arrived during the Dry season.
Let D be the event that it is raining when you return.
If you arrived during the Wet season, the probability of rain when you return is the same as the probability of rain during the Wet season, which is 3/4.
If you arrived during the Dry season, the probability of rain when you return is the same as the probability of rain during the Dry season, which is 1/6.
Since the season you arrived in is independent of the weather when you return, we need to consider the probabilities based on the season you arrived.
Let C' be the event that your visit is during the Dry season.
P(D) = P(D|C)P(C) + P(D|C')P(C')
Since P(C) = 1/3 and P(C') = 2/3, we can calculate:
P(D) = (3/4)(1/3) + (1/6)(2/3)
= 1/4 + 1/9
= 9/36 + 4/36
= 13/36
Therefore, the probability that it will be raining when you return to Oneirabad in a year's time, given that it is raining when you arrive, is 13/36.
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Consider the set of parametric equations x(t)y(t)=3t2+5=2t3+5 on the interval 0≤t≤1. Set up an integral to find the length of the curve. Type the integral into the answer box using the equation editor. Evaluate the integral to determine the exact length of the curve. Use the equation editor to enter your answer correct mathematical form.
The exact length of the curve for the parametric equations x(t) = 3t² + 5 and y(t) = 2t³ + 5 on the given interval 0 ≤ t ≤ 1 is equal to 0.8283 approximately .
To find the length of the curve defined by the parametric equations x(t) = 3t² + 5 and y(t) = 2t³+ 5 on the interval 0 ≤ t ≤ 1,
Use the arc length formula for parametric curves,
L = ∫ₐᵇ √[x'(t)² + y'(t)²] dt
where x'(t) and y'(t) are the derivatives of x(t) and y(t) with respect to t.
First, let's find the derivatives of x(t) and y(t),
x'(t) = d/dt (3t² + 5)
= 6t
y'(t) = d/dt (2t³ + 5)
= 6t²
Now, let's substitute these derivatives into the arc length formula,
L = ∫₀¹ √[(6t)²+ (6t²)²] dt
Simplifying the expression inside the square root,
L = ∫₀¹√(36t² + 36t⁴) dt
Now, integrate this expression to find the exact length of the curve,
L = ∫₀¹ √(36t² + 36t⁴) dt
To evaluate this integral, simplify the expression under the square root by factoring out 36t²,
L = ∫₀¹ √(36t²(1 + t²)) dt
Now, take the square root out of the integral,
L = ∫₀¹ 6t√(1 + t²) dt
L = ∫₀¹ 6t√(1 + t²) dt
To solve this integral, use a substitution.
Let u = 1 + t². Then, du = 2t dt,
and rewrite the integral as,
L = ∫[u(0),u(1)] 3√u du
Next, determine the limits of integration.
When t = 0, u = 1 + (0)²
= 1.
When t = 1, u = 1 + (1)²
= 2.
The limits of integration become,
L = ∫₁² 3√u du
To find the antiderivative of √u,
use the power rule of integration.
Let's rewrite the integral.
L = 3 ∫₁² u¹/² du
Applying the power rule, integrate u¹/² as (2/3)u³/²
L = 3 [(2/3)u³/²] evaluated from 1 to 2
Substituting the limits of integration,
L = 3 [(2/3)(2³/²) - (2/3)(1³/²)]
Simplifying,
L = 3 [2³/²/3 - 2/3]
Finally, compute the exact length of the curve,
L = 3 [2³/²/3 - 2/3]
≈ 3 [0.9428 - 0.6667]
≈ 3 [0.2761]
≈ 0.8283
Therefore, the exact length of the curve defined by the parametric equations x(t) = 3t² + 5 and y(t) = 2t³ + 5 on the interval 0 ≤ t ≤ 1 is approximately 0.8283.
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The above question is incomplete, the complete question is:
Consider the set of parametric equations x(t)=3t²+5 , y(t)=2t³+5 on the interval 0≤t≤1.
Set up an integral to find the length of the curve. Type the integral into the answer box using the equation editor. Evaluate the integral to determine the exact length of the curve. Use the equation editor to enter your answer correct mathematical form.
.2 Find the point on x-axis, which is equidistant from the points (3,2) and (−5,−2). 3. Show that the points (0,−2),(3,1),(0,4) and (−3,1) are the vertices of a square. 4. Show that the points (0,6),(−5,3) and (3,1) are the vertices of an isosceles triangle.
5. Three vertices of a rhombus taken in order are (2,−1),(3,4) and (−2,3). Find the fourth vertex.
2.The point on x-axis which is equidistant from the points (3,2) and (−5,−2) can be found as follows:
Let the point on x-axis be (x,0).
Using distance formula:
[tex]$$\sqrt{(3-x)^2+(2-0)^2}=\sqrt{(-5-x)^2+(-2-0)^2}$$[/tex]
Squaring both sides:
[tex]$$(3-x)^2+4=(-5-x)^2+4$$[/tex]
Simplifying: [tex]$$-16x=64$$[/tex]
Therefore, x=-4.
The point on x-axis which is equidistant from the points (3,2) and (−5,−2) is (-4,0).
3. The given points are (0,−2),(3,1),(0,4) and (−3,1) are the vertices of a square. Let's check it:
Firstly, find the distance between the point (0,-2) and (3,1) and between the point (3,1) and (0,4) using distance formula:
[tex]$$\sqrt{(3-0)^2+(1-(-2))^2}=\sqrt{10}$$[/tex]
[tex]$$\sqrt{(0-3)^2+(4-1)^2}=\sqrt{10}$$[/tex]
Thus, both are equal which means the sides are equal.
Now, check if any one pair of opposite sides are parallel using slope formula:
[tex]$$\frac{1-(-2)}{3-0}=\frac{4-(-2)}{0-(-3)}=\frac{3}{1}$$[/tex]
Thus, opposite sides are parallel.
Next, check if any one pair of sides are perpendicular using slope formula:
[tex]$$\frac{1-(-2)}{3-0}\cdot \frac{4-(-2)}{0-(-3)}=-1$$[/tex]
Thus, opposite sides are perpendicular.Therefore, the given points are the vertices of a square.
4. The given points are (0,6),(−5,3) and (3,1) are the vertices of an isosceles triangle. Let's check it:
Firstly, find the distance between the point (0,6) and (−5,3), between the point (−5,3) and (3,1) and between the point (0,6) and (3,1) using distance formula:
[tex]$$\sqrt{(0-(-5))^2+(6-3)^2}=\sqrt{34}$$[/tex]
[tex]$$\sqrt{(-5-3)^2+(3-1)^2}=2\sqrt{10}$$[/tex]
[tex]$$\sqrt{(0-3)^2+(6-1)^2}=\sqrt{34}$$[/tex]
Thus, two sides are equal. Therefore, the given points are the vertices of an isosceles triangle.
5. Three vertices of a rhombus taken in order are (2,−1),(3,4) and (−2,3). Find the fourth vertex:
As the rhombus is a special parallelogram with all sides of equal length and opposite angles equal, the distance between the three given points taken in order, when paired two at a time, must be equal.
Using distance formula, the distance between the point (2,-1) and (3,4), between the point (3,4) and (-2,3), and between the point (-2,3) and (2,-1) are as follows:
[tex]$$\sqrt{(3-2)^2+(4-(-1))^2}=\sqrt{50}$$[/tex]
[tex]$$\sqrt{(-2-3)^2+(3-4)^2}=\sqrt{50}$$[/tex]
[tex]$$\sqrt{(2-(-2))^2+(-1-3)^2}=\sqrt{50}$$[/tex]
Thus, the fourth vertex should be the point (2,-5).
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Plot the frequency response and the impulse response of the LTI system having the output y = 2te-ul for the input x) = -(t).
Given an LTI system that has an output y = 2te^-ul for the input x(t) = -(t). Here, e^-ul is the decay constant and is a real positive constant. The impulse response of the system can be found by considering the impulse input x(t) = δ(t).
The output of the system with an impulse input is given by y(t) = h(t), where h(t) is the impulse response of the system.We have,x(t) = -(t)..........(1)Applying derivative on both sides of the above equation, we get,y(t) = 2te^-ul, Applying derivative on both sides of the above equation, we get,
Plot of frequency response and Impulse response of the system Hence, the plot of frequency response and the impulse response of the LTI system is shown in the figure below:, the frequency response is given by H(jω) = (2/(-jω + ul)^2) where ω is the angular frequency. The impulse response of the system is given by h(t) = (2/ul^2)t.e^-ul.
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Find avarage value of function f(x,y)=9x+2y over 0⩽x⩽3,0≤y≤9 A) 245 ? b) Evaluate ∫−310∫−10−45xdydx c) 1365 ? C) Write equation double integral with order of Integrals reversed ∫03∫07y/3dxdy A) ∫01∫3x113dydx ?
a) The average value of f(x, y) = 9x + 2y over the region 0 ≤ x ≤ 3 and 0 ≤ y ≤ 9 is 486.
b) The value of the double integral is 547.5.
c) The value of the double integral with the order of integrals reversed is 10.5.
To find the average value of the function f(x, y) = 9x + 2y over the region 0 ≤ x ≤ 3 and 0 ≤ y ≤ 9, we need to calculate the double integral of the function over the given region and divide it by the area of the region.
a) Average value of f(x, y) = 9x + 2y is 245.
Average value of f(x, y) = (1 / Area of Region) ∫[0, 3]∫[0, 9] (9x + 2y) dy dx
The area of the region is given by:
Area = ∫[0, 3]∫[0, 9] 1 dy dx = 3 * 9 = 27
Now, let's calculate the double integral:
∫[0, 3]∫[0, 9] (9x + 2y) dy dx
= ∫[0, 3] (9xy + y²/2) | [0, 9] dx
= ∫[0, 3] (81x + 81/2) dx
= [81x²/2 + 81x/2] | [0, 3]
= (81(3)²/2 + 81(3)/2) - (81(0)²/2 + 81(0)/2)
= (81(9)/2 + 81(3)/2)
= (729/2 + 243/2)
= 972/2
= 486
The average value of f(x, y) = 9x + 2y over the region 0 ≤ x ≤ 3 and 0 ≤ y ≤ 9 is 486
b) Evaluating the double integral ∫[-3, 10]∫[-1, -4] 5x dy dx:
∫[-3, 10] 5x(y)|[-4, -1] dx
= ∫[-3, 10] 5x((-1) - (-4)) dx
= ∫[-3, 10] 15x dx
= [15x²/2]|[-3, 10]
= (15(10)²/2) - (15(-3)²/2)
= 750 - 202.5
= 547.5
The value of the double integral is 547.5.
c) To calculate this double integral, we integrate with respect to x first and then with respect to y:
∫[0, 7/3] yx/3 | [0, 3] dy
= ∫[0, 7/3] (3y - 0) dy
= [3y/2] | [0, 7/3]
= (3(7/3)/2) - (3(0)/2)
= (21/2) - 0
= 21/2
= 10.5
The value of the double integral with the order of integrals reversed is 10.5.
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a square picture is mounted in a frame 1 cm wide. the area of the picture 2/3 the total area find the length of a side of the picture
The length of a side of the picture is 3.39 centimeters. The area of a frame is calculated by multiplying the width of the frame by the perimeter of the picture.
Let x be the length of a side of the picture. The area of the picture is x². The area of the frame is 2x. The total area is x² + 2x.
We are given that the area of the picture is 2/3 of the total area. This means that x² = 2/3(x² + 2x).
Solving for x, we get x = 3.39 centimeters.
Here are some additional explanations:
The area of a square is calculated by multiplying the length of one side by itself.
The area of a frame is calculated by multiplying the width of the frame by the perimeter of the picture.The total area is calculated by adding the area of the picture and the area of the frame.In this problem, we are given that the area of the picture is 2/3 of the total area. This means that the area of the frame is 1/3 of the total area.Solving for the length of a side of the picture, we get x = 3.39 centimeters.To know more about length click here
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Find the arc length of the curve given by \( \mathbf{r}(t)=\langle\sqrt{32} t, 2 \cos (t), 2 \sin (t)\rangle \) from \( 0 \leq t \leq 2 \).
The arc length of the curve from t = 0 to t = 2 is 12.
To find the arc length of the curve given by [tex]\( \mathbf{r}(t) = \langle \sqrt{32} t, 2 \cos(t), 2 \sin(t) \rangle \) from \( 0 \leq t \leq 2 \)[/tex], we can use the arc length formula:
[tex]\[ L = \int_a^b \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2 + \left(\frac{dz}{dt}\right)^2} \, dt \][/tex]
where [tex]\( \frac{dx}{dt} \), \( \frac{dy}{dt} \), and \( \frac{dz}{dt} \) are the derivatives of \( x(t) \), \( y(t) \), and \( z(t) \)[/tex], respectively.
Let's calculate the derivatives first:
[tex]\[ \frac{dx}{dt} = \sqrt{32} \]\[ \frac{dy}{dt} = -2 \sin(t) \]\[ \frac{dz}{dt} = 2 \cos(t) \][/tex]
Now, we can substitute these derivatives into the arc length formula and integrate:
[tex]\[ L = \int_0^2 \sqrt{\left(\sqrt{32}\right)^2 + \left(-2 \sin(t)\right)^2 + \left(2 \cos(t)\right)^2} \, dt \][/tex]
Simplifying the expression inside the square root:
[tex]\[ L = \int_0^2 \sqrt{32 + 4 \sin^2(t) + 4 \cos^2(t)} \, dt \][/tex]
Using the trigonometric identity [tex]\( \sin^2(t) + \cos^2(t) = 1 \):[/tex]
[tex]\[ L = \int_0^2 \sqrt{32 + 4} \, dt \]\[ L = \int_0^2 \sqrt{36} \, dt \]\[ L = \int_0^2 6 \, dt \]\[ L = 6t \, \bigg|_0^2 \]\[ L = 6(2) - 6(0) \]\[ L = 12 \][/tex]
Therefore, the arc length of the curve from t = 0 to t = 2 is 12.
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Find the general solution of the DE y^(4)+y=0
The general solution of the differential equation y^(4) + y = 0 is y(x) = A cos(x) + B sin(x) + C cosh(x) + D sinh(x), where A, B, C, and D are arbitrary constants.
To find the general solution of the given fourth-order linear homogeneous differential equation y^(4) + y = 0, we can start by assuming a solution of the form y(x) = e^(rx), where r is a complex number. Substituting this into the differential equation, we get the characteristic equation r^4 + 1 = 0.
The characteristic equation can be factored as (r^2 + i)(r^2 - i) = 0, where i is the imaginary unit. Solving for r, we have r = ±√i and r = ±√(-i). The square roots of i can be expressed as ±(1 + i)/√2, and the square roots of -i as ±(1 - i)/√2.
Using these values, we can obtain four complex roots: r_1 = (1 + i)/√2, r_2 = -(1 + i)/√2, r_3 = (1 - i)/√2, and r_4 = -(1 - i)/√2. Since the original differential equation is real-valued, the complex roots come in conjugate pairs. Therefore, we can rewrite the solution as y(x) = e^(rx) = e^(αx) * (C1 cos(βx) + C2 sin(βx)), where α and β are real constants.
Using the exponential form of complex numbers, we have e^(αx) = e^(Re(r)x) = e^(αx) * (C1 cos(βx) + C2 sin(βx)), where α = Re(r) and β = Im(r). Simplifying this expression further, we get y(x) = A cos(x) + B sin(x) + C cosh(x) + D sinh(x), where A, B, C, and D are arbitrary constants.
This general solution encompasses all possible solutions to the given differential equation. The terms involving cos(x) and sin(x) represent the real part of the complex roots, while the terms involving cosh(x) and sinh(x) represent the imaginary part. The constants A, B, C, and D can be determined by applying appropriate initial or boundary conditions.
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Evaluate the expression. \[ { }_{12} C_{6} \] A. 332,640 B. 1,440 C. 924 D. 665,280
Using combinations, the expression is evaluated as C. 924.
The expression [tex]${}_{12}C_6$[/tex] represents the combination of selecting 6 items from a set of 12 items without regard to their order. It can be calculated using the formula:
[tex]${}_{n}C_{r} = \frac{n!}{r!(n-r)!}$[/tex]
Substituting the values, we have:
[tex]${}_{12}C_6 = \frac{12!}{6!(12-6)!} = \frac{12!}{6!6!}$[/tex]
Simplifying:
[tex]${}_{12}C_6 = \frac{12 \times 11 \times 10 \times 9 \times 8 \times 7}{6 \times 5 \times 4 \times 3 \times 2 \times 1} = 924$[/tex]
Therefore, the value of [tex]${}_{12}C_6$[/tex] is 924.
The correct answer is C. 924.
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Complete Question:
Evaluate the expression. [tex]${}_{12}C_6$[/tex].
A. 332,640 B. 1,440 C. 924 D. 665,280
A The number of employees in firms nationwide can be approximately modeled by a lognormal distribution. A fractional number of employees is allowed to account for part-time workers. Suppose that 0 = 6.3 and w = 2.0. For each of the following values of x, determine the number of employees such that x proportion of firms employ less than this number: a. 0.3 b. 0.6
we can use a statistical software or calculator that provides the functionality to calculate the CDF of the lognormal distribution.
To determine the number of employees such that a given proportion of firms employ less than this number, we can utilize the cumulative distribution function (CDF) of the lognormal distribution.
The lognormal distribution is characterized by two parameters: the mean (µ) and the standard deviation (σ) of the underlying normal distribution in logarithmic scale. In this case, the parameters are given as 0 = 6.3 and w = 2.0.
The CDF of the lognormal distribution can be calculated using the formula:
CDF(x) = Φ((ln(x) - µ) / σ)
Where Φ is the cumulative distribution function of the standard normal distribution.
To solve for the number of employees, we need to find the value of x such that the CDF(x) is equal to the given proportion.
a. For x such that 0.3 proportion of firms employ less than x:
We need to find the value of x such that CDF(x) = 0.3
b. For x such that 0.6 proportion of firms employ less than x:
We need to find the value of x such that CDF(x) = 0.6
To calculate the values of x, we can use a statistical software or calculator that provides the functionality to calculate the CDF of the lognormal distribution.
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Sec8.7: Problem 4 Previous Problem List Next (1 point) Book Problem 5 Find the Maclaurin series for f(x) = cos(6x) using the definition of a Maclaurin series: f(x) = f(0) = f'(x) = ,f'(0) = f"(x) = ,f"(0) = f(3)(x) = f(3)(0) = cos(6x) = x2+ rt... Note: You can earn partial credit on this problem.
To find the Maclaurin series for f(x) = cos(6x) using the definition of a Maclaurin series,
we need to find f(0), f'(x), f'(0), f''(x), f''(0), f'''(x), and f'''(0).
We can find all of these by taking derivatives of f(x).
We have;
f(x) = cos(6x)
Hence,
[tex]f(0) = cos(6*0) = cos(0) = 1f'(x) = -6sin(6x)f'(0) = -6sin(6*0) = 0f''(x) = -6^2cos(6x)f''(0) = -6^2cos(6*0) = -36f'''(x) = -6^3sin(6x)f'''(0) = -6^3sin(6*0) = 0[/tex]
Substituting f(0), f'(0), f''(0), and f'''(0) in the Maclaurin series for f(x), we have;
[tex]cos(6x) = 1 - 36x^2/2! + 0 + 0 + ...cos(6x) = 1 - 18x^2 + ... (answer)[/tex]
Thus, the Maclaurin series for f(x) = cos(6x) using the definition of a Maclaurin series is given as 1 - 18x^2 + ... .
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(10\%) Apply the Direct Proof technique to show that "for any two positive integers, the square of their summation is greater than or equal to the summation of their squares." (Hint: This is to prove ( x +y)² ≥(x² +y²) for any positive integers. Note: (x+y)² =x² +2xy+y² ). Show your "First", Middle", and "Final" steps of the argument explicitly for the proof ( 50% of the points will be taken off if no steps showing). When needed, type the exponential such as " x " as " x∧n " in your answer. Use the following words for the symbols if needed
Use the transitive property to prove that (x + y)² ≥ (x² + y²).
First step: We start with a direct proof and assume that x and y are two positive integers. Therefore, we can say that: x ≥ 0 and y ≥ 0.
Middle step: Using the hint given in the question, we can write:
(x + y)² = x² + 2xy + y².
And (x² + y²) can be written as (x + y)² - 2xy.
So, (x² + y²) = (x + y)² - 2xy.
Now, we can write: (x + y)² ≥ (x² + y²) if x ≥ 0, y ≥ 0, and 2xy ≥ 0. Since x, y, and 2 are greater than or equal to zero, we can use the transitive property to prove that (x + y)² ≥ (x² + y²).
Conclusion: Therefore, we have successfully proved that "for any two positive integers, the square of their summation is greater than or equal to the summation of their squares."
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In order to apply the Direct Proof technique to show that "for any two positive integers, the square of their summation is greater than or equal to the summation of their squares,"
the following steps can be followed:
First Step: Let x and y be two arbitrary positive integers.
Second Step: Then, we must prove ( x +y)² ≥(x² +y²) for any positive integers x and y.
Third Step: By using the identity, (x+y)² =x² +2xy+y²,
we can rewrite the inequality as follows: x² + 2xy + y² ≥ x² + y².
Fourth Step: Subtracting x² + y² from both sides of the inequality gives: 2xy ≥ 0.
Fifth Step: Since xy is always non-negative, the inequality is always true for any arbitrary positive integers x and y.
Hence, it can be concluded that the square of the summation of any two positive integers is greater than or equal to the summation of their squares.
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Define F:R→R and G:R→Z by the following formulas: F(x)= x^2/5
and G(x)=⌊x⌋ for every x∈R. (a) (G∘F)(4)= (b) (G∘F)(−5)= (c) (G∘F)(6)=
The values of (G∘F)(4), (G∘F)(−5) and (G∘F)(6) are 3, 5 and 7 respectively.
Given, F(x)= x²/5 and G(x)=⌊x⌋.
To calculate (G∘F)(4) we need to find G(F(4))Where
[tex]F(4) = 4²/5 = 16/5≈3.2[/tex]
Thus, G(F(4)) = G(16/5) = 3 (nearest integer value of 16/5)Therefore, (G∘F)(4) = 3For (G∘F)(−5),
we need to find G(F(-5))
Where
[tex]F(-5) = (-5)²/5 = 25/5 = 5[/tex]
Thus, G(F(-5)) = G(5) = 5 (nearest integer value of 5)
Therefore, (G∘F)(-5) = 5
For (G∘F)(6), we need to find G(F(6))
Where [tex]F(6) = 6²/5 = 36/5 = 7.2[/tex]
Thus, G(F(6)) = G(7.2) = 7 (nearest integer value of 7.2)Therefore, (G∘F)(6) = 7
Hence, (a) (G∘F)(4)=3 (b) (G∘F)(−5)=5 and (c) (G∘F)(6)=7.
The composition of functions is an important concept in mathematics and is used to combine two functions to form a new function. In this problem, we have been given two functions
F(x)= x²/5 and G(x)=⌊x⌋ where x∈R.
The first function F(x) is a quadratic function, which means it has a graph that looks like a parabola. On the other hand, the function G(x) is a step function, which means it has a graph that consists of horizontal lines with vertical jumps at integer values. Now, we need to calculate the values of (G∘F)(4), (G∘F)(−5) and (G∘F)(6).To calculate (G∘F)(4), we need to find G(F(4)). Here, F(4) is equal to 16/5. Therefore, G(F(4)) is equal to the nearest integer value of 16/5 which is 3. Hence, (G∘F)(4) = 3.To calculate (G∘F)(−5), we need to find G(F(−5)). Here, F(−5) is equal to 25/5. Therefore, G(F(−5)) is equal to the nearest integer value of 25/5 which is 5. Hence, (G∘F)(−5) = 5.To calculate (G∘F)(6), we need to find G(F(6)). Here, F(6) is equal to 36/5. Therefore, G(F(6)) is equal to the nearest integer value of 36/5 which is 7. Hence, (G∘F)(6) = 7.
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According to the NEA, 20% of U.S. married couples attended a musical play. In a random sample of 10 married couples,
Find the probability that fewer than 9 attended a musical play.
Find the probability that exactly 5 attended a musical play.
Find the probability that more than 1 attended a musical play.
Find the probability that exactly 5 attended a musical play given more than 3 attends a musical play.
Given that the NEA reported that 20% of US married couples attended a musical play. And the number of married couples who attended the musical play in a random sample of 10 is to be calculated. Using binomial probability, we can solve the above question.
A binomial probability distribution is a statistical measure that deals with the probability of two independent events. The probability of the occurrence of one event does not influence the probability of the second event.
In a random sample of 10 married couples, we have to find the probabilities of various events related to the number of married couples who attended the musical play. Let's calculate each of them separately.
Part (a): Find the probability that fewer than 9 attended a musical play.Using the formula of binomial probability, the probability that fewer than 9 couples attended a musical play is:P(X < 9) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6) + P(X = 7) + P(X = 8)where,
P(X = x) = nCx (p)x (q)n - x, n = 10, p = 0.2, q = 0.8.
By putting the respective values in the formula, we get:P(X < 9) = (10C0 × 0.2⁰ × 0.8¹⁰⁻⁰) + (10C1 × 0.2¹ × 0.8⁹) + (10C2 × 0.2² × 0.8⁸) + (10C3 × 0.2³ × 0.8⁷) + (10C4 × 0.2⁴ × 0.8⁶) + (10C5 × 0.2⁵ × 0.8⁵) + (10C6 × 0.2⁶ × 0.8⁴) + (10C7 × 0.2⁷ × 0.8³) + (10C8 × 0.2⁸ × 0.8²)Putting the values in the formula and solving further, we get:P(X < 9) = 0.00000381408.
Thus, the probability that fewer than 9 couples attended a musical play is 0.00000381408.
Part (b): Find the probability that exactly 5 attended a musical play.Using the formula of binomial probability, the probability that exactly 5 couples attended a musical play is:P(X = 5) = nCx (p)x (q)n - x, n = 10, p = 0.2, q = 0.8By putting the respective values in the formula, we get:P(X = 5) = 10C5 × 0.2⁵ × 0.8⁵Putting the values in the formula and solving further, we get:P(X = 5) = 0.0264241152Thus, the probability that exactly 5 couples attended a musical play is 0.0264241152.
Part (c): Find the probability that more than 1 attended a musical play.Using the formula of binomial probability, the probability that more than 1 couple attended a musical play is:P(X > 1) = 1 - P(X ≤ 1) = 1 - [P(X = 0) + P(X = 1)]where, P(X = x) = nCx (p)x (q)n - x, n = 10, p = 0.2, q = 0.8By putting the respective values in the formula, we get:P(X > 1) = 1 - [10C0 × 0.2⁰ × 0.8¹⁰⁻⁰ + 10C1 × 0.2¹ × 0.8⁹].
Putting the values in the formula and solving further, we get:P(X > 1) = 0.99815744Thus, the probability that more than 1 couple attended a musical play is 0.99815744.Part (d): Find the probability that exactly 5 attended a musical play given more than 3 attend a musical play.
Using Bayes' theorem, the probability that exactly 5 couples attended a musical play given more than 3 couples attended a musical play is:P(X = 5 | X > 3) = P(X > 3 | X = 5) × P(X = 5) / P(X > 3)We already know:P(X = 5) = 0.0264241152To calculate P(X > 3), we have to use the formula:P(X > 3) = 1 - P(X ≤ 3) = 1 - [P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)]where, P(X = x) = nCx (p)x (q)n - x, n = 10, p = 0.2, q = 0.8.
By putting the respective values in the formula, we get:P(X > 3) = 1 - [10C0 × 0.2⁰ × 0.8¹⁰⁻⁰ + 10C1 × 0.2¹ × 0.8⁹ + 10C2 × 0.2² × 0.8⁸ + 10C3 × 0.2³ × 0.8⁷]Putting the values in the formula and solving further, we get:P(X > 3) = 0.01646263232To calculate P(X > 3 | X = 5), we have to use the formula:P(X > 3 | X = 5) = P(X > 3 ∩ X = 5) / P(X = 5)To calculate P(X > 3 ∩ X = 5), we have to use the formula:
P(X > 3 ∩ X = 5) = P(X = 5) × P(X > 3 | X = 5)where, P(X = x) = nCx (p)x (q)n - x, n = 10, p = 0.2, q = 0.8By putting the respective values in the formula, we get:P(X > 3 ∩ X = 5) = 10C5 × 0.2⁵ × 0.8⁵ × (10C5 / 0.2⁵ × 0.8⁵)Putting the values in the formula and solving further, we get:P(X > 3 ∩ X = 5) = 0.0058535832By putting all the respective values in the Bayes' theorem formula, we get:
P(X = 5 | X > 3) = 0.0058535832 / 0.01646263232Thus, the probability that exactly 5 couples attended a musical play given more than 3 couples attended a musical play is 0.355523907.
In a random sample of 10 married couples, the probability that fewer than 9 couples attended a musical play is 0.00000381408. The probability that exactly 5 couples attended a musical play is 0.0264241152. The probability that more than 1 couple attended a musical play is 0.99815744. The probability that exactly 5 couples attended a musical play given more than 3 couples attended a musical play is 0.355523907.
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