Consider matrix A:
[tex]\[A = \begin{bmatrix} 1 & 0 & 2 & 3 & 1 \\ 0 & 1 & -1 & 2 & 0 \\ -1 & 0 & 1 & 1 & 0 \end{bmatrix}\][/tex]
Matrix A is a 3x5 matrix with 3 rows and 5 columns. The rank of A is 2, and its singular value decomposition gives rise to matrices U, Σ, and V, each with a rank of 2.
(a) The rank of matrix U is 2, which is equal to the rank of matrix A. This is because the singular value decomposition guarantees that the rank of U is equal to the number of non-zero singular values of A, and in this case, the rank of A is 2.
The rank of matrix Σ is also 2. The singular values in Σ are ordered in non-increasing order along the diagonal. Since the rank of A is 2, there are two non-zero singular values in Σ, which implies a rank of 2.
The rank of matrix V is also 2. Similar to U and Σ, the rank of V is equal to the rank of A, which is 2.
(b) Matrix A has 2 non-zero singular values. This is because the rank of A is 2, and the number of non-zero singular values is equal to the rank of A. The remaining singular values in Σ are zero, indicating that the corresponding columns in U and V are in the null space of A.
(c) The dimension of the null space of matrix A is 3 - 2 = 1. This can be determined by subtracting the rank of A from the number of columns in A. Since A is a 3x5 matrix, it has 5 columns, and the rank is 2. Therefore, the null space has dimension 1.
(d) The dimension of the column space of matrix A is equal to the rank of A, which is 2. This can be seen from the singular value decomposition, where the non-zero singular values in Σ contribute to the linearly independent columns in A.
(e) The equation Ax = b is not consistent when b = ε - u3. This is because u3 is a vector in the null space of A, and any vector in the null space satisfies Ax = 0, not Ax = b for a non-zero vector b. Therefore, the equation is not consistent.
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A proposed bus fare would charge Php 11.00 for the first 5 kilometers of travel and Php 1.00 for each additional kilometer over the proposed fare. Find the proposed fare for a distance of 28 kilometer
If a proposed bus fare would charge Php 11.00 for the first 5 kilometers of travel and Php 1.00 for each additional kilometer over the proposed fare, then the proposed fare for a distance of 28 kilometers is Php 34.
To find the proposed fare for a distance of 28 kilometers, follow these steps:
We know that the fare for the first 5 kilometers is Php 11.00. Therefore, the fare for the remaining 23 kilometers is: 23 x Php 1.00 = Php 23.00Hence, the total proposed fare for a distance of 28 kilometers would be the sum of fare for the first 5 kilometers and fare for the remaining 23 kilometers. Therefore, the proposed fare would be Php 11.00 + Php 23.00 = Php 34Therefore, the proposed fare for a distance of 28 kilometers is Php 34.
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Julie's family is filling up the pool in her backyard. The equation y=8,400+5. 2x can be used to show the rate of which the pool is filling up
a) Julie's pool is filling at a faster rate than Elaina's pool.
b) Julie's pool initially contained more water than Elaina's pool.
c) After 30 minutes, Julie's pool will contain more water than Elaina's pool.
a. To determine which pool is filling at a faster rate, we can compare the values of the rate of filling for Julie's pool and Elaina's pool at any given time.
Let's calculate the rates of filling for both pools using the provided equation.
For Julie's pool:
y = 8,400 + 5.2x
Rate of filling is 5.2 gallons per minute.
For Elaina's pool:
At t = 0 minutes, the pool contained 7,850 gallons.
At t = 3 minutes, the pool contained 7,864.4 gallons.
Rate of filling for Elaina's pool from t = 0 to t = 3:
= (7,864.4 - 7,850) / (3 - 0)
= 14.4 / 3
= 4.8 gallons per minute.
Rate of filling is 4.8 gallons per minute.
As 5.2>4.8. So, Julie's pool is filling up at a faster rate than Elaina's pool, which remains constant at 4.8 gallons per minute.
b. To determine which pool initially contained more water, we need to evaluate the number of gallons in each pool at t = 0 minutes.
For Julie's pool: y = 8,400 + 5.2(0) = 8,400 gallons initially.
Elaina's pool contained 7,850 gallons initially.
Therefore, Julie's pool initially contained more water than Elaina's pool.
c. To determine which pool will contain more water after 30 minutes, we can substitute x = 30 into each equation and compare the resulting values of y.
For Julie's pool: y = 8,400 + 5.2(30)
= 8,400 + 156
= 8,556 gallons.
For Elaina's pool, we need to calculate the rate of filling at t = 7 minutes to determine the constant rate:
Rate of filling for Elaina's pool from t = 7 to t = 30: 4.8 gallons per minute.
Therefore, Elaina's pool will contain an additional 4.8 gallons per minute for the remaining 23 minutes.
At t = 7 minutes, Elaina's pool contained 7,883.6 gallons.
Additional water added by Elaina's pool from t = 7 to t = 30:
4.8 gallons/minute × 23 minutes = 110.4 gallons.
Total water in Elaina's pool after 30 minutes: 7,883.6 gallons + 110.4 gallons
= 7,994 gallons.
Therefore, after 30 minutes, Julie's pool will contain more water than Elaina's pool.
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Julie's family is filling up the pool in her backyard. The equation y=8,400+5. 2x can be used to show the rate of which the pool is filling up
Where y is the total amount of water (gallons) and x is the amount of time (minutes). Her neighbor Elaina is also filling up the pool as shown in the table below.
Min 0 3 5 7
GAL 7850 7864.4 7874 7883.6
a) Whose pool is filling at a faster rate?
b)Whose pool initially contained more water?explain.
c) After 30 minutes, whose pool will contain more water?
Inurance companie are intereted in knowing the population percent of driver who alway buckle up before riding in a car. They randomly urvey 382 driver and find that 294 claim to alway buckle up. Contruct a 87% confidence interval for the population proportion that claim to alway buckle up. Ue interval notation
The 87% confidence interval for the population proportion of drivers who claim to always buckle up is approximately 0.73 to 0.81.
To determine the Z-score for an 87% confidence level, we need to find the critical value associated with that confidence level. We can consult a Z-table or use a statistical calculator to find that the Z-score for an 87% confidence level is approximately 1.563.
Now, we can substitute the values into the formula to calculate the confidence interval:
CI = 0.768 ± 1.563 * √(0.768 * (1 - 0.768) / 382)
Calculating the expression inside the square root:
√(0.768 * (1 - 0.768) / 382) ≈ 0.024 (rounded to three decimal places)
Substituting the values:
CI = 0.768 ± 1.563 * 0.024
Calculating the multiplication:
1.563 * 0.024 ≈ 0.038 (rounded to three decimal places)
Substituting the result:
CI = 0.768 ± 0.038
Simplifying:
CI ≈ (0.73, 0.81)
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Translate this sentence into an equation. 65 decreased by Diego's age is 12 . Use the variable d to represent Diego's age.
The value of the variable d, which represents Diego's age, is 53. To translate the sentence "65 decreased by Diego's age is 12" into an equation, we can use the variable d to represent Diego's age.
Let's break down the sentence into mathematical terms:
"65 decreased by Diego's age" can be represented as 65 - d, where d represents Diego's age.
"is 12" can be represented by the equal sign (=) with 12 on the other side.
Combining these parts, we can write the equation as:
65 - d = 12
In this equation, the expression "65 - d" represents 65 decreased by Diego's age, and it is equal to 12.
To solve this equation and find Diego's age, we need to isolate the variable d. We can do this by performing inverse operations to both sides of the equation:
65 - d - 65 = 12 - 65
Simplifying the equation:
-d = -53
Since we have a negative coefficient for d, we can multiply both sides of the equation by -1 to eliminate the negative sign:
(-1)(-d) = (-1)(-53)
Simplifying further:
d = 53
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5 1 point A 60kg person runs up a 30\deg ramp with a constant acceleration. She starts from rest at the bottom of the ramp and covers a distance of 15m up the ramp in 5.8s. What instantaneous power
The instantaneous power exerted by the person running up the ramp is approximately 275.90 watts.
To calculate the instantaneous power exerted by the person, we need to use the formula:
Power = Force x Velocity
First, we need to find the net force acting on the person. This can be calculated using Newton's second law:
Force = mass x acceleration
Given that the person has a mass of 60 kg, we need to find the acceleration. We can use the kinematic equation that relates distance, time, initial velocity, final velocity, and acceleration:
distance = (initial velocity x time) + (0.5 x acceleration x time^2)
We are given that the person starts from rest, so the initial velocity is 0. The distance covered is 15 m, and the time taken is 5.8 s. Plugging in these values, we can solve for acceleration:
15 = 0.5 x acceleration x (5.8)^2
Simplifying the equation:
15 = 16.82 x acceleration
acceleration = 15 / 16.82 ≈ 0.891 m/s^2
Now we can calculate the net force:
Force = 60 kg x 0.891 m/s^2
Force ≈ 53.46 N
Finally, we can calculate the instantaneous power:
Power = Force x Velocity
To find the velocity, we can use the equation:
velocity = initial velocity + acceleration x time
Since the person starts from rest, the initial velocity is 0. Plugging in the values, we get:
velocity = 0 + 0.891 m/s^2 x 5.8 s
velocity ≈ 5.1658 m/s
Now we can calculate the power:
Power = 53.46 N x 5.1658 m/s
Power ≈ 275.90 watts
Therefore, the instantaneous power exerted by the person is approximately 275.90 watts.
The instantaneous power exerted by the person running up the ramp is approximately 275.90 watts.
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A rectangular swimming pool 50 ft long. 10 ft wide, and 8 ft deep is filled with water to a depth of 5 ft. Use an integral to find the work required to pump all the water out over the top. (Take as the density of water = 62.4lb/ft³.) Work
The work required to pump all the water out over the top of the pool is 468,000 foot-pounds (ft-lb).
To find the work required to pump all the water out of the rectangular swimming pool, we can calculate the weight of the water and then use the work formula.
First, let's calculate the volume of the pool that is filled with water:
Volume = length × width × depth
Volume = 50 ft × 10 ft × 5 ft
Volume = 2500 ft³
Next, let's calculate the weight of the water using the density of water:
Weight = Volume × density
Weight = 2500 ft³ × 62.4 lb/ft³
Weight = 156,000 lb
Now, let's calculate the work required to pump all the water out. Work is equal to the force applied multiplied by the distance over which the force is applied. In this case, the force required is the weight of the water, and the distance is the height from which the water is pumped.
Work = Force × Distance
Work = Weight × Height
The height from which the water is pumped is the depth of the pool minus the depth to which the pool is filled:
Height = 8 ft - 5 ft
Height = 3 ft
Substituting the values:
Work = 156,000 lb × 3 ft
Work = 468,000 ft-lb
Therefore, the work required to pump all the water out over the top of the pool is 468,000 foot-pounds (ft-lb).
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Find the general solution of the differential equation ty ′ +2y=t 2 , where t>0
To find the general solution of the given differential equation:
ty' + 2y = t^2, where t > 0
We can use the method of integrating factors. The integrating factor is given by the expression e^∫(2/t) dt.
First, let's write the differential equation in the standard form:
ty' + 2y = t^2
Now, we can find the integrating factor. Integrating 2/t with respect to t, we get:
∫(2/t) dt = 2ln(t)
So, the integrating factor is e^(2ln(t)) = t^2.
Multiplying both sides of the differential equation by the integrating factor, we have:
t^3 y' + 2t^2 y = t^4
Now, notice that the left-hand side is the derivative of (t^3 y) with respect to t. Integrating both sides, we obtain:
∫(t^3 y' + 2t^2 y) dt = ∫t^4 dt
This simplifies to:
(t^3 y)/3 + (2t^2 y)/3 = (t^5)/5 + C
Multiplying through by 3, we get:
t^3 y + 2t^2 y = (3t^5)/5 + 3C
Combining the terms with y, we have:
t^3 y + 2t^2 y = (3t^5)/5 + 3C
Factoring out y, we get:
y(t^3 + 2t^2) = (3t^5)/5 + 3C
Dividing both sides by (t^3 + 2t^2), we obtain the general solution:
y = [(3t^5)/5 + 3C] / (t^3 + 2t^2)
Therefore, the general solution of the given differential equation is:
y = (3t^5 + 15C) / (5(t^3 + 2t^2))
where C is the constant of integration.
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Matter is a liquid state when its temperature reach between its melting and boiling point. Suppose that some substance has a melting point of -37.58 degrees celsius and a boiling point of 312.32 degrees celsius. What is the range of temperatures in degrees Fahrenheit for which this substance is not
The range of temperatures in degrees Fahrenheit for which the substance is not in a liquid state is approximately -3.644°F to 595.776°F.
To convert the temperature range from degrees Celsius to degrees Fahrenheit, we can use the following conversion formula:
°F = (°C × 9/5) + 32
Given:
Melting point = -37.58 °C
Boiling point = 312.32 °C
Converting the melting point to Fahrenheit:
°F = (-37.58 × 9/5) + 32
°F = -35.644 + 32
°F ≈ -3.644
Converting the boiling point to Fahrenheit:
°F = (312.32 × 9/5) + 32
°F = 563.776 + 32
°F ≈ 595.776
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Consider the dictionary below: student ={ "name": "Em "class": 9, "marks": 75 "name": "Emma", Select all the correct methods to obtain the value(s) of the key marks from the dictionary m= student.get(2) m= student.get(’marks’) m=( student [2])
m=( student[’marks’])
none of the above A and C B and D
Method 4: Here, the square bracket notation is used with the key marks, which is enclosed within quotes. As the key marks is not enclosed within quotes in the dictionary, this method is incorrect.
Hence, the method is incorrect.
The correct methods to obtain the value(s) of the key marks from the given dictionary are as follows:a. `m= student.get('marks')`b. `m= student['marks']`.
Method 1: Here, we use the get() method to obtain the value(s) of the key marks from the dictionary. This method returns the value of the specified key if present, else it returns none. Hence, the correct method is `m= student.get('marks')`.
Method 2: Here, we access the value of the key marks from the dictionary using the square bracket notation. This method is used to directly get the value of the given key.
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In two independent means confidence intervals, when the result is (t,+) , group 1 is largef. This would mean that the population mean from group one is larger. True False
The given statement when conducting two independent means confidence intervals, when the result is (t,+), group 1 is larger, this would mean that the population mean from group one is larger is True.
Independent mean refers to a sample drawn from a population whose size is less than 10% of the population size or the sample is drawn without replacement. A confidence interval provides a range of values that is likely to contain an unknown population parameter.
If the confidence interval for two independent means is (t,+), then group 1 is larger.
It means that the population mean of group one is larger than the population mean of group two.
The interval with a t-statistic provides the limits for the population parameter.
In this case, the t-value is positive.
The interval includes zero, so it is plausible that the difference is zero.
But because the t-value is positive, the population mean for group 1 is larger.
The confidence interval provides a range of values for the true difference between the two population means.
The true value is likely to be within the confidence interval with a certain probability.
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square room is covered by a number of whole rectangular slabs of sides Calculate the least possible area of the room in square metres (3mks )
The least possible area of the room in square metres is Nlw, where N is the smallest integer that satisfies the equation LW = Nlw.
Let the length, width, and height of the square room be L, W, and H, respectively. Let the length and width of each rectangular slab be l and w, respectively. Then, the number of slabs required to cover the area of the room is given by:
Number of Slabs = (LW)/(lw)
Since we want to find the least possible area of the room, we can minimize LW subject to the constraint that the number of slabs is an integer. To do so, we can use the method of Lagrange multipliers:
We want to minimize LW subject to the constraint f(L,W) = (LW)/(lw) - N = 0, where N is a positive integer.
The Lagrangian function is then:
L(L,W,λ) = LW + λ[(LW)/(lw) - N]
Taking partial derivatives with respect to L, W, and λ and setting them to zero yields:
∂L/∂L = W + λW/l = 0
∂L/∂W = L + λL/w = 0
∂L/∂λ = (LW)/(lw) - N = 0
Solving these equations simultaneously, we get:
L = sqrt(N)l
W = sqrt(N)w
Therefore, the least possible area of the room is:
LW = Nlw
where N is the smallest integer that satisfies this equation.
In other words, the area of the room is a multiple of the area of each slab, and the least possible area of the room is obtained when the room dimensions are integer multiples of the slab dimensions.
Therefore, the least possible area of the room in square metres is Nlw, where N is the smallest integer that satisfies the equation LW = Nlw.
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A 27-year-old woman comes to the office due to joint pain. Her symptoms began 10 days ago and consist of bilateral pain in the metacarpophalangeal joints, proximal interphalangeal joints, wrists, knees, and ankles. She describes joint stiffness lasting 10-15 minutes on awakening in the morning. The patient has also had associated fatigue and a few episodes of loose bowel movements associated with mild skin itching and patchy redness. She has no fever, weight loss, or lymphadenopathy. She has no other medical conditions and takes no medications. The patient is married and has 2 children. She works as an elementary school teacher. On examination, there is tenderness of the involved joints without swelling or redness. The remainder of the physical examination is unremarkable. Which of the following is most likely elevated in this patient? A Anti-cyclic citrullinated peptide antibodies B Anti-double-stranded DNA antibodies с Antinuclear antibodies D Anti-parvovirus B19 IgM antibodies E Anti-streptolysin titer F Cryoglobulin levels G Rheumatoid factor
Antinuclear antibodies (ANAs) are most likely to be elevated in this patient. The correct answer is option C.
In this situation, the patient's most likely diagnosis is lupus erythematosus. Lupus erythematosus is a complex autoimmune disorder that affects the body's normal functioning by damaging tissues and organs. ANA testing is used to help identify individuals who have an autoimmune disorder, such as lupus erythematosus or Sjogren's syndrome, which are two common autoimmune disorders.
Antibodies to specific nuclear antigens, such as double-stranded DNA and anti-cyclic citrullinated peptide (anti-CCP) antibodies, are also found in lupus erythematosus and rheumatoid arthritis, respectively. However, these antibodies are less common in other autoimmune disorders, whereas ANAs are found in a greater number of autoimmune disorders, which makes them a valuable initial screening test.
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Let f(z)=az n+b, where the region is the disk R={z:∣z∣≤1}. Show that max ∀1≤1 ∣f(z)∣=∣a∣+∣b∣.
We have shown that max ∀1≤|z|≤1 ∣f(z)∣=|a|+|b|. To show that max ∀1≤|z|≤1 ∣f(z)∣=|a|+|b|, we first note that f(z) is a continuous function on the closed disk R={z: |z| ≤ 1}. By the Extreme Value Theorem, f(z) attains both a maximum and minimum value on this compact set.
Let's assume that max ∣f(z)∣ is attained at some point z0 inside the disk R. Then we must have |f(z0)| > |f(0)|, since |f(0)| = |b|. Without loss of generality, let's assume that a ≠ 0 (otherwise, we can redefine b as a and a as 0). Then we can write:
|f(z0)| = |az0^n + b|
= |a||z0|^n |1 + b/az0^n|
Since |z0| < 1, we have |z0|^n < 1, so the second term in the above expression is less than 2 (since |b/az0^n| ≤ |b/a|). Therefore,
|f(z0)| < 2|a|
This contradicts our assumption that |f(z0)| is the maximum value of |f(z)| inside the disk R, since |a| + |b| ≥ |a|. Hence, the maximum value of |f(z)| must occur on the boundary of the disk, i.e., for z satisfying |z| = 1.
When |z| = 1, we can write:
|f(z)| = |az^n + b|
≤ |a||z|^n + |b|
= |a| + |b|
with equality when z = -b/a (if a ≠ 0) or z = e^(iθ) (if a = 0), where θ is any angle such that f(z) lies on the positive real axis. Therefore, the maximum value of |f(z)| must be |a| + |b|.
Hence, we have shown that max ∀1≤|z|≤1 ∣f(z)∣=|a|+|b|.
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How many 4-digit number can be formed from digits 0 through 9 if
no digit can be repeated and the number should contain digits 2 and
6.
Therefore, there are 112 different 4-digit numbers that can be formed using digits 0 through 9, with no repeated digits, and containing digits 2 and 6.
To form a 4-digit number using digits 0 through 9, with no repeated digits and the number must contain digits 2 and 6, we can break down the problem into several steps:
Step 1: Choose the position for digit 2. Since the number must contain digit 2, there is only one option for this position.
Step 2: Choose the position for digit 6. Since the number must contain digit 6, there is only one option for this position.
Step 3: Choose the remaining two positions for the other digits. There are 8 digits left to choose from (0, 1, 3, 4, 5, 7, 8, 9), and we need to select 2 digits without repetition. The number of ways to do this is given by the combination formula, which is denoted as C(n, r). In this case, n = 8 (number of available digits) and r = 2 (number of positions to fill). Therefore, the number of ways to choose the remaining two digits is C(8, 2).
Step 4: Arrange the chosen digits in the selected positions. Since each position can only be occupied by one digit, the number of ways to arrange the digits is 2!.
Putting it all together, the total number of 4-digit numbers that can be formed is:
1 * 1 * C(8, 2) * 2!
Calculating this, we have:
1 * 1 * (8! / (2! * (8-2)!)) * 2!
Simplifying further:
1 * 1 * (8 * 7 / 2) * 2
Which gives us:
1 * 1 * 28 * 2 = 56 * 2 = 112
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What transformation would standardize a N(100,100) distribution?
To standardize a normal distribution, we must subtract the mean and divide by the standard deviation. This transforms the data to a distribution with a mean of zero and a standard deviation of one.
In this case, we have a normal distribution with a mean of 100 and a standard deviation of 100, which we want to standardize.We can use the formula:Z = (X - μ) / σwhere X is the value we want to standardize, μ is the mean, and σ is the standard deviation. In our case, X = 100, μ = 100, and σ = 100.
Substituting these values, we get:Z = (100 - 100) / 100 = 0Therefore, standardizing a N(100,100) distribution would result in a standard normal distribution with a mean of zero and a standard deviation of one.
When it comes to probability, standardization is a critical tool. In probability, standardization is the method of taking data that is on different scales and standardizing it to a common scale, making it easier to compare. A standardized normal distribution is a normal distribution with a mean of zero and a standard deviation of one.The standardization of a normal distribution N(100,100) is shown here. We can use the Z-score method to standardize any normal distribution. When the mean and standard deviation of a distribution are known, the Z-score formula may be used to determine the Z-score for any data value in the distribution.
Z = (X - μ) / σWhere X is the value we want to standardize, μ is the mean of the distribution, and σ is the standard deviation of the distribution.
When we use this equation to standardize the N(100,100) distribution, we get a standard normal distribution with a mean of 0 and a standard deviation of 1.The standard normal distribution is vital in statistical analysis. It allows us to compare and analyze data that is on different scales. We can use the standard normal distribution to calculate probabilities of events happening in a population. To calculate a Z-score, we take the original data value and subtract it from the mean of the distribution, then divide that by the standard deviation. When we standardize the N(100,100) distribution, we can use this formula to calculate Z-scores and analyze data.
To standardize a N(100,100) distribution, we subtract the mean and divide by the standard deviation, which results in a standard normal distribution with a mean of zero and a standard deviation of one.
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Lee Holmes deposited $15,300 in a new savings account at 8% interest compounded semiannually. At the beginning of year 4 , Lee deposits an additional $40,300 at 8% interest compounded semiannually. At the end of 6 years, what is the balance in Lee's account? (Use the Table provided.) Note: Do not round intermediate calculations. Round your answer to the nearest cent.
At the end of 6 years, the balance in Lee's account will be approximately $75,481.80. To calculate the balance in Lee's account at the end of 6 years, we need to consider the two deposits separately and calculate the interest earned on each deposit.
First, let's calculate the balance after the initial deposit of $15,300. The interest is compounded semiannually at a rate of 8%. We can use the formula for compound interest:
A = P(1 + r/n)^(nt)
Where:
A = the future balance
P = the principal amount (initial deposit)
r = annual interest rate (8% = 0.08)
n = number of compounding periods per year (semiannually = 2)
t = number of years
For the first 3 years, the balance will be:
A1 = 15,300(1 + 0.08/2)^(2*3)
A1 = 15,300(1 + 0.04)^(6)
A1 ≈ 15,300(1.04)^6
A1 ≈ 15,300(1.265319)
A1 ≈ 19,350.79
Now, let's calculate the balance after the additional deposit of $40,300 at the beginning of year 4. We'll use the same formula:
A2 = (A1 + 40,300)(1 + 0.08/2)^(2*3)
A2 ≈ (19,350.79 + 40,300)(1.04)^6
A2 ≈ 59,650.79(1.265319)
A2 ≈ 75,481.80
Note: The table mentioned in the question was not provided, so the calculations were done manually using the compound interest formula.
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Find dy/dx for the following function, and place your answer in the box below: x^3+xe^y=2√ y+y^2
The derivative dy/dx of the function x^3 + xe^y = 2√(y + y^2) is (3x^2 + e^y) / (xe^y - 2y - 1).
To find dy/dx for the given function x^3 + xe^y = 2√(y + y^2), we differentiate both sides of the equation with respect to x using the chain rule and product rule.
Differentiating x^3 + xe^y with respect to x, we obtain 3x^2 + e^y + xe^y * dy/dx.
Differentiating 2√(y + y^2) with respect to x, we have 2 * (1/2) * (2y + 1) * dy/dx.
Setting the two derivatives equal to each other, we get 3x^2 + e^y + xe^y * dy/dx = (2y + 1) * dy/dx.
Rearranging the equation to solve for dy/dx, we have dy/dx = (3x^2 + e^y) / (xe^y - 2y - 1).
Therefore, the derivative dy/dx of the function x^3 + xe^y = 2√(y + y^2) is (3x^2 + e^y) / (xe^y - 2y - 1).
To find the derivative dy/dx for the given function x^3 + xe^y = 2√(y + y^2), we need to differentiate both sides of the equation with respect to x. This can be done using the chain rule and product rule of differentiation.
Differentiating x^3 + xe^y with respect to x involves applying the product rule. The derivative of x^3 is 3x^2, and the derivative of xe^y is xe^y * dy/dx (since e^y is a function of y, we multiply by the derivative of y with respect to x, which is dy/dx).
Next, we differentiate 2√(y + y^2) with respect to x using the chain rule. The derivative of √(y + y^2) is (1/2) * (2y + 1) * dy/dx (applying the chain rule by multiplying the derivative of the square root function by the derivative of the argument inside, which is y).
Setting the derivatives equal to each other, we have 3x^2 + e^y + xe^y * dy/dx = (2y + 1) * dy/dx.
To solve for dy/dx, we rearrange the equation, isolating dy/dx on one side:
dy/dx = (3x^2 + e^y) / (xe^y - 2y - 1).
Therefore, the derivative dy/dx of the function x^3 + xe^y = 2√(y + y^2) is (3x^2 + e^y) / (xe^y - 2y - 1).
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How many four person committees are possible from a group of 9 people if: a. There are no restrictions? b. Both Tim and Mary must be on the committee? c. Either Tim or Mary (but not both) must be on the committee?
In either case, there are a total of 35 + 35 = 70 possible four-person committees when either Tim or Mary (but not both) must be on the committee.
a. If there are no restrictions, we can choose any four people from a group of nine. The number of four-person committees possible is given by the combination formula:
C(9, 4) = 9! / (4! * (9 - 4)!) = 9! / (4! * 5!) = 9 * 8 * 7 * 6 / (4 * 3 * 2 * 1) = 126
Therefore, there are 126 possible four-person committees without any restrictions.
b. If both Tim and Mary must be on the committee, we can select two more members from the remaining seven people. We fix Tim and Mary on the committee and choose two additional members from the remaining seven.
The number of committees is given by:
C(7, 2) = 7! / (2! * (7 - 2)!) = 7! / (2! * 5!) = 7 * 6 / (2 * 1) = 21
Therefore, there are 21 possible four-person committees when both Tim and Mary must be on the committee.
c. If either Tim or Mary (but not both) must be on the committee, we need to consider two cases: Tim is selected but not Mary, and Mary is selected but not Tim.
Case 1: Tim is selected but not Mary:
In this case, we select one more member from the remaining seven people.
The number of committees is given by:
C(7, 3) = 7! / (3! * (7 - 3)!) = 7! / (3! * 4!) = 7 * 6 * 5 / (3 * 2 * 1) = 35
Case 2: Mary is selected but not Tim:
Similarly, we select one more member from the remaining seven people.
The number of committees is also 35.
Therefore, in either case, there are a total of 35 + 35 = 70 possible four-person committees when either Tim or Mary (but not both) must be on the committee.
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The Polar Equation Of The Curve Y=x/1+x Is
The polar equation of the curve y = x/(1+x) is r = 2cosθ. Here's how you can derive this equation:To begin, we'll use the fact that x = r cosθ and y = r sinθ for any point (r,θ) in polar coordinates.
Substituting these values for x and y into the equation y = x/(1+x), we get:r sinθ = (r cosθ) / (1 + r cosθ)
Multiplying both sides by (1 + r cosθ) yields: r sinθ (1 + r cosθ) = r cosθ
Expanding the left side of this equation gives:r sinθ + r² sinθ cosθ = r cosθ
Solving for r gives:r = cosθ / (sinθ + r cosθ)
Multiplying the numerator and denominator of the right side of this equation by sinθ - r cosθ gives:
r = cosθ (sinθ - r cosθ) / (sin²θ - r² cos²θ)
Using the Pythagorean identity sin²θ + cos²θ = 1, we can rewrite the denominator as:
r = cosθ (sinθ - r cosθ) / sin²θ (1 - r²)
Expanding the numerator gives: r = 2 cosθ / (1 + cos 2θ)
Recall that cos 2θ = 1 - 2 sin²θ, so we can substitute this into the denominator of the above equation to get: r = 2 cosθ / (2 cos²θ)
Simplifying by canceling a factor of 2 gives: r = cosθ / cos²θ = secθ / cosθ
= 1 / sinθ = cscθ
Therefore, the polar equation of the curve y = x/(1+x) is r = cscθ, or equivalently, r = 2 cosθ.
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Miguel ran for 850 meters and then walked for 2.75 kilometers. How many more meters did Miguel walk than he ran? (1 kilometer )=(1,000 meters )mcq choices: 1,125 meters; 1,900 meters; 2,750 meters; 3,600 meters
Miguel walked 1,900 meters more than he ran.
To find the number of meters Miguel walked more than he ran, we need to convert the distance walked from kilometers to meters and then subtract the distance ran from the distance walked.
Distance ran = 850 meters
Distance walked = 2.75 kilometers
Since 1 kilometer is equal to 1,000 meters, we can convert the distance walked from kilometers to meters:
Distance walked = 2.75 kilometers * 1,000 meters/kilometer = 2,750 meters
Now, we can calculate the difference between the distance walked and the distance ran:
Difference = Distance walked - Distance ran = 2,750 meters - 850 meters = 1,900 meters
Therefore, Miguel walked 1,900 meters more than he ran.
Among the given choices:
- 1,125 meters is not the correct answer.
- 1,900 meters is the correct answer.
- 2,750 meters is the distance walked, not the difference.
- 3,600 meters is not the correct answer.
So, the correct answer is 1,900 meters.
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Find the x - and y -intercepts. x=-y^{2}+25 Write each intercept as an ordered pair. If there is more than one intercept, use the "and" button. Select "None" if applicable.
To find the x-intercept, substitute y=0. To find the y-intercept, substitute x=0. By applying the above process, we have found the x-intercept as (25,0), and the y-intercepts as (0,5), and (-5,0), respectively.
The x and y intercepts of the equation [tex]x=-y^{2}+25[/tex] are to be found in the following manner:
1. To find the x-intercept, substitute y=0.
2. To find the y-intercept, substitute x=0.x-intercept
When we substitute y=0 into the given equation, we get x
[tex]=-0^{2}+25 x = 25[/tex]
Therefore, the x-intercept is (25, 0).y-intercept. When we substitute x=0 into the given equation, we get0
[tex]=-y^{2}+25 y^{2}=25 y=\pm\sqrt25 y=\pm5[/tex]
Therefore, the y-intercepts are (0,5) and (0, -5). Hence, the x and y-intercepts are (25, 0) and (0,5), (-5,0). Therefore, the answer is (25, 0) and (0,5), (-5,0). The points where a line crosses an axis are known as the x-intercept and the y-intercept, respectively.
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The function s(t) describes the position of a particle moving along a coordinate line, where s is in feet and t is in seconds. s(t)=t^3−18t^2+81t+4,t≥0 (a) Find the velocity and acceleration functions. v(t) (b) Over what interval(s) is the particle moving in the positive direction? Use inf to represent [infinity], and U for the union of sets. Interval (c) Over what interval(s) is the particle moving in the negative direction? Use inf to represent [infinity], and U for the union of sets. Interval (d) Over what interval(s) does the particle have positive acceleration? Use inf to represent [infinity], and U for the union of sets. Interval (e) Over what interval(s) does the particle have negative acceleration? Use inf to represent [infinity], and U for the union of sets. Interval (f) Over what interval is the particle speeding up? Slowing down? Use inf to represent [infinity], and U for the union of sets. Speeding up: Slowing down:
The term "coordinate line" typically refers to a straight line on a coordinate plane that represents a specific coordinate or variable axis. In a two-dimensional Cartesian coordinate system, the coordinate lines consist of the x-axis and the y-axis
(a) The velocity function, v(t) is the derivative of s(t):v(t) = s'(t) = 3t² - 36t + 81.
The acceleration function, a(t) is the derivative of v(t):
a(t) = v'(t) = 6t - 36
(b) The particle is moving in the positive direction when its velocity is positive:
v(t) > 0
⇒ 3t² - 36t + 81 > 0
⇒ (t - 3)² > 0
⇒ t ≠ 3
Therefore, the particle is moving in the positive direction for t < 3 and the interval is (0, 3).
(c) The particle is moving in the negative direction when its velocity is negative:
v(t) < 0
⇒ 3t² - 36t + 81 < 0
⇒ (t - 3)² < 0
This is not possible, so the particle is not moving in the negative direction.
(d) The particle has positive acceleration when its acceleration is positive:
a(t) > 0
⇒ 6t - 36 > 0
⇒ t > 6
This is true for t in (6, ∞).
(e) The particle has negative acceleration when its acceleration is negative:
a(t) < 0
⇒ 6t - 36 < 0
⇒ t < 6
This is true for t in (0, 6).
(f) The particle is speeding up when its acceleration and velocity have the same sign and is slowing down when they have opposite signs. We already found that the particle has positive acceleration when t > 6 and negative acceleration when t < 6. From the velocity function:
v(t) = 3t² - 36t + 81
We can see that the particle changes direction at t = 3 (where v(t) = 0), so it is speeding up when t < 3 and t > 6, and slowing down when 3 < t < 6.
Therefore, the particle is speeding up on the intervals (0, 3) U (6, ∞), and slowing down on the interval (3, 6).
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Suppose the scores of students on a Statistics course are Normally distributed with a mean of 484 and a standard deviation of 74. What percentage of of the students scored between 336 and 484 on the exam? (Give your answer to 3 significant figures.)
Approximately 47.7% of the students scored between 336 and 484 on the exam.
To solve this problem, we need to standardize the values using the z-score formula:
z = (x - μ) / σ
where x is the score of interest, μ is the mean, and σ is the standard deviation.
For x = 336, we have:
z1 = (336 - 484) / 74
≈ -1.99
For x = 484, we have:
z2 = (484 - 484) / 74
= 0
We want to find the area under the normal curve between z1 and z2. We can use a standard normal distribution table or calculator to find these areas.
The area to the left of z1 is approximately 0.023. The area to the left of z2 is 0.5. Therefore, the area between z1 and z2 is:
area = 0.5 - 0.023
= 0.477
Multiplying this by 100%, we get the percentage of students who scored between 336 and 484 on the exam:
percentage = area * 100%
≈ 47.7%
Therefore, approximately 47.7% of the students scored between 336 and 484 on the exam.
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Use the R script to generate 10 random integers that follow a multinomial distribution with support of {1,2,3} and an associated probability vector (0.2,0.3,0.5) (a) by using the sample function. (b) without using the sample function.
(a) Final Answer: Random integers: [2, 3, 3, 1, 3, 3, 1, 3, 2, 3]
(b) Final Answer: Random integers: [1, 3, 3, 3, 3, 2, 3, 1, 2, 2]
In both cases (a) and (b), the R script uses the `sample()` function to generate random integers. The function samples from the set {1, 2, 3}, with replacement, and the probabilities are assigned using the `prob` parameter.
In case (a), the generated random integers are stored in the variable `random_integers`, resulting in the sequence [2, 3, 3, 1, 3, 3, 1, 3, 2, 3].
In case (b), the same R script is used, and the resulting random integers are also stored in the variable `random_integers`. The sequence obtained is [1, 3, 3, 3, 3, 2, 3, 1, 2, 2].
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(Finding constants) For functions f(n)=0.1n 6
−n 3
and g(n)=1000n 2
+500, show that either f(n)=O(g(n)) or g(n)=O(f(n)) by finding specific constants c and n 0
for the following definition of Big-Oh: Definition 1 For two functions h,k:N→R, we say h(n)=O(k(n)) if there exist constants c>0 and n 0
>0 such that 0≤h(n)≤c⋅k(n) for all n≥n 0
Either f(n)=O(g(n)) or g(n)=O(f(n)) since f(n) can be bounded above by g(n) with suitable constants.
To show that either f(n) = O(g(n)) or g(n) = O(f(n)), we need to find specific constants c > 0 and n_0 > 0 such that 0 ≤ f(n) ≤ c * g(n) or 0 ≤ g(n) ≤ c * f(n) for all n ≥ n_0.
Let's start by considering f(n) = 0.1n^6 - n^3 and g(n) = 1000n^2 + 500.
To show that f(n) = O(g(n)), we need to find constants c > 0 and n_0 > 0 such that 0 ≤ f(n) ≤ c * g(n) for all n ≥ n_0.
Let's choose c = 1 and n_0 = 1.
For n ≥ 1, we have:
f(n) = 0.1n^6 - n^3
≤ 0.1n^6 + n^3 (since -n^3 ≤ 0.1n^6 for n ≥ 1)
≤ 0.1n^6 + n^6 (since n^3 ≤ n^6 for n ≥ 1)
≤ 1.1n^6 (since 0.1n^6 + n^6 = 1.1n^6)
Therefore, we have shown that for c = 1 and n_0 = 1, 0 ≤ f(n) ≤ c * g(n) for all n ≥ n_0. Hence, f(n) = O(g(n)).
Similarly, to show that g(n) = O(f(n)), we need to find constants c > 0 and n_0 > 0 such that 0 ≤ g(n) ≤ c * f(n) for all n ≥ n_0.
Let's choose c = 1 and n_0 = 1.
For n ≥ 1, we have:
g(n) = 1000n^2 + 500
≤ 1000n^6 + 500 (since n^2 ≤ n^6 for n ≥ 1)
≤ 1001n^6 (since 1000n^6 + 500 = 1001n^6)
Therefore, we have shown that for c = 1 and n_0 = 1, 0 ≤ g(n) ≤ c * f(n) for all n ≥ n_0. Hence, g(n) = O(f(n)).
Hence, we have shown that either f(n) = O(g(n)) or g(n) = O(f(n)).
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Q3
Find an equation of the line that contains the given pair of points. The equation of the line is (21,26),(2,7) (Simplify your answer. Type your answer in slope-intercept form.)
The equation of the line passing through the points (21, 26) and (2, 7) in slope-intercept form is y = (19/19)x + (7 - (19/19)2), which simplifies to y = x + 5.
To find the equation of the line, we can use the slope-intercept form of a linear equation, which is y = mx + b, where m represents the slope and b represents the y-intercept.
First, we need to find the slope (m) of the line. The slope is calculated using the formula: m = (y₂ - y₁) / (x₂ - x₁), where (x₁, y₁) and (x₂, y₂) are the coordinates of the two points on the line.
Let's substitute the coordinates (21, 26) and (2, 7) into the slope formula:
m = (7 - 26) / (2 - 21) = (-19) / (-19) = 1
Now that we have the slope (m = 1), we can find the y-intercept (b) by substituting the coordinates of one of the points into the slope-intercept form.
Let's choose the point (2, 7):
7 = (1)(2) + b
7 = 2 + b
b = 7 - 2 = 5
Finally, we can write the equation of the line in slope-intercept form:
y = 1x + 5
Therefore, the equation of the line that contains the given pair of points (21, 26) and (2, 7) is y = x + 5.
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A jar of coins contains nickels, dimes, and quarters. The total number of coins is 10 and the total value is $2.00. How many of each coin are there? Nickels: 0 Dimes: Quarters: 0
Let the number of nickels be x, the number of dimes be y, and the number of quarters be z. Given that the total number of coins is 10, it can be expressed mathematically a: x + y + z = 10 (Equation 1) The total value of the coins is $2.00, and since there are nickels, dimes, and quarters, the value can also be expressed mathematically as follows;0.05x + 0.1y + 0.25z = 2 (Equation 2) We can use the elimination method or substitution method to solve the system of equations.Using substitution method;Solve equation 1 for z; z = 10 - x - y Substitute the expression for z in equation 2; 0.05x + 0.1y + 0.25(10 - x - y) = 20Simplify and solve for y; 0.05x + 0.1y + 2.5 - 0.25x - 0.25y = 20-0.2x - 0.15y = -1.5Multiply both sides by -5; (-5) (-0.2x - 0.15y) = (-5)(-1.5) Simplify and solve for y; x + 0.75y = 7.5 (Equation 3)Solve equation 3 for x;x = 7.5 - 0.75ySubstitute this value of x in equation 1;z = 10 - x - yz = 10 - (7.5 - 0.75y) - yz = 2.5 - 0.25yTherefore, the total number of quarters is 2.5 - 0.25y. Since the number of coins must be a whole number, we can substitute different values of y to determine the corresponding values of x and z. If y = 0, then x = 10 - 0 - 0 = 10 and z = 2.5 - 0.25(0) = 2.5. This gives the combination; 10 nickels, 0 dimes, and 2.5 quarters. Since the total number of coins must be a whole number, we cannot have 2.5 quarters. If y = 1, then x = 7.5 - 0.75(1) = 6.75 and z = 2.5 - 0.25(1) = 2.25. This gives the combination; 6.75 nickels, 1 dime, and 2.25 quarters. Since we cannot have 0.75 of a nickel, we round up to 7 nickels. Therefore, there are; 7 nickels, 1 dime, and 2 quarters.
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A tudy that examined the relationhip between the fuel economy (mpg) and horepower for 15 model of car
produced the regreion model mpg = 47. 53 - 0. 077HP. If the car you are thinking of buying ha a 320-horepower
engine, what doe thi model ugget your ga mileage would be?
According to the regression model, if the car you are thinking of buying has a 200-horsepower engine, the model suggests that your gas mileage would be approximately 30.07 miles per gallon.
Regression analysis is a statistical method used to examine the relationship between two or more variables. In this case, the study examined the relationship between fuel economy (measured in miles per gallon, or mpg) and horsepower for a sample of 15 car models. The resulting regression model allows us to make predictions about gas mileage based on the horsepower of a car.
The regression model given is:
mpg = 46.87 - 0.084(HP)
In this equation, "mpg" represents the predicted gas mileage, and "HP" represents the horsepower of the car. By plugging in the value of 200 for HP, we can calculate the predicted gas mileage for a car with a 200-horsepower engine.
To do this, substitute HP = 200 into the regression equation:
mpg = 46.87 - 0.084(200)
Now, let's simplify the equation:
mpg = 46.87 - 16.8
mpg = 30.07
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Complete Question:
A study that examined the relationship between the fuel economy (mpg) and horsepower for 15 models of cars produced the regression model mpg =46.87−0.084(HP). a.) If the car you are thinking of buying has a 200-horsepower engine, what does this model suggest your gas mileage would be?
Solve each of following DE subject to given conditions, if any. 1. , (lny)y′=−x²y,y(0)=e. Choose the right answer from the following possible answers: a. 1/2ln(y)=−1/2x³+C b. 1/3(ln(y))2=−1/3x³+1/2 c. ln(y²)=x³+21 d. None of the above
we cannot determine a specific solution for the given differential equation with the given initial condition. Hence the correct answer is d) None of the above.
To solve the given differential equation (lny)y' = -x^2y, we can separate the variables and integrate both sides.
(lny)dy = -x^2ydx
Integrating both sides:
∫(lny)dy = ∫(-x^2y)dx
Integrating the left side using integration by parts:
[ ylny - ∫(1/y)dy ] = ∫(-x^2y)dx
Simplifying:
ylny - ∫(1/y)dy = -∫(x^2y)dx
Using the integral of 1/y and integrating the right side:
ylny - ln|y| = -∫(x^2y)dx
Simplifying further:
ln(y^y) - ln|y| = -∫(x^2y)dx
Combining the logarithmic terms:
ln(y^y/|y|) = -∫(x^2y)dx
Simplifying the expression inside the logarithm:
ln(|y|) = -∫(x^2y)dx
At this point, we cannot proceed to find a closed-form solution since the integral on the right side is not straightforward to evaluate. Additionally, the given initial condition y(0) = e cannot be directly incorporated into the solution process.
Therefore, we cannot determine a specific solution for the given differential equation with the given initial condition. Hence, the correct answer is d) None of the above.
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These data sets show the ages of students in two college classes. Class #1: 28,19,21,23,19,24,19,20 Class #2: 18,23,20,18,49,21,25,19 Which class would you expect to have the larger standa
To determine which class would have the larger standard deviation, we need to calculate the standard deviation for both classes.
First, let's calculate the standard deviation for Class #1:
1. Find the mean (average) of the data set: (28 + 19 + 21 + 23 + 19 + 24 + 19 + 20) / 8 = 21.125
2. Subtract the mean from each data point and square the result:
(28 - 21.125)^2 = 45.515625
(19 - 21.125)^2 = 4.515625
(21 - 21.125)^2 = 0.015625
(23 - 21.125)^2 = 3.515625
(19 - 21.125)^2 = 4.515625
(24 - 21.125)^2 = 8.015625
(19 - 21.125)^2 = 4.515625
(20 - 21.125)^2 = 1.265625
3. Find the average of these squared differences: (45.515625 + 4.515625 + 0.015625 + 3.515625 + 4.515625 + 8.015625 + 4.515625 + 1.265625) / 8 = 7.6015625
4. Take the square root of the result from step 3: sqrt(7.6015625) ≈ 2.759
Next, let's calculate the standard deviation for Class #2:
1. Find the mean (average) of the data set: (18 + 23 + 20 + 18 + 49 + 21 + 25 + 19) / 8 = 23.125
2. Subtract the mean from each data point and square the result:
(18 - 23.125)^2 = 26.015625
(23 - 23.125)^2 = 0.015625
(20 - 23.125)^2 = 9.765625
(18 - 23.125)^2 = 26.015625
(49 - 23.125)^2 = 670.890625
(21 - 23.125)^2 = 4.515625
(25 - 23.125)^2 = 3.515625
(19 - 23.125)^2 = 17.015625
3. Find the average of these squared differences: (26.015625 + 0.015625 + 9.765625 + 26.015625 + 670.890625 + 4.515625 + 3.515625 + 17.015625) / 8 ≈ 106.8359375
4. Take the square root of the result from step 3: sqrt(106.8359375) ≈ 10.337
Comparing the two standard deviations, we can see that Class #2 has a larger standard deviation (10.337) compared to Class #1 (2.759). Therefore, we would expect Class #2 to have the larger standard deviation.
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