b) If the calculated z-value exceeds the critical z-value from the standard normal distribution at the specified significance level, we reject the null hypothesis.
a. To test whether the average daily return has decreased, we can use a one-sample t-test. The null hypothesis (H0) is that the average daily return is still 3.25%, and the alternative hypothesis (Ha) is that the average daily return has decreased.
Given:
Sample size (n) = 30
Sample mean (x(bar)) = 1.87%
Sample standard deviation (s) = 5.6%
Significance level (α) = 0.05
First, we calculate the t-statistic:
t = (x(bar) - μ) / (s / sqrt(n))
Where μ is the hypothesized mean under the null hypothesis, which is 3.25%.
t = (1.87% - 3.25%) / (5.6% / sqrt(30))
Next, we compare the calculated t-value with the critical t-value from the t-distribution with (n - 1) degrees of freedom. At a significance level of 0.05 and (n - 1) = 29 degrees of freedom, the critical t-value is obtained from the t-distribution table.
If the calculated t-value is greater than the critical t-value, we reject the null hypothesis in favor of the alternative hypothesis.
b. To test whether the average quarterly revenue has increased, we can use a one-sample t-test. The null hypothesis (H0) is that the average quarterly revenue is still $19.309 billion, and the alternative hypothesis (Ha) is that the average quarterly revenue has increased.
Given:
Sample size (n) = 30
Sample mean (x(bar)) = $22.6 billion
Sample standard deviation (s) = $5.2 billion
Significance level (α) = 0.01
Using the same process as in part (a), we calculate the t-value and compare it with the critical t-value from the t-distribution with (n - 1) degrees of freedom. If the calculated t-value is greater than the critical t-value, we reject the null hypothesis.
c. To test whether the proportion of students who will pass the course has changed, we can use a one-sample proportion test. The null hypothesis (H0) is that the proportion is still 70%, and the alternative hypothesis (Ha) is that the proportion has changed.
Given:
Sample size (n) = 100
Sample proportion (p(cap)) = 80%
Significance level (α) = 0.10
We calculate the test statistic, which follows the standard normal distribution under the null hypothesis:
z = (p(cap) - p0) / sqrt((p0 * (1 - p0)) / n)
Where p0 is the hypothesized proportion under the null hypothesis, which is 70%.
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Let S = {(x1, y1),(x2, y2), · · · ,(xn, yn)} be a set of n points where all coordinates are real numbers. A point (xi , yi) is called a Pareto optimal point if for every other (xj , yj ) ∈ S, at least one of the following two inequalities hold:
xi > xj
yi >xj. If we are interested in finding only one Pareto optimal point in S, could you design a worst-case O(n)-time algorithm to find it? You must include an analysis of the O(n) running time of your algorithm. (c) If the points in S are sorted by their x coordinates and each point in S has a unique x coordinate, could you design a worst-case O(n)-time algorithm to find all Pareto optimal points in S ? You must include an analysis of the running time of your algorithm.
Both algorithms provide worst-case O(n) time complexity, making them efficient for finding Pareto optimal points in a set of n points.
To find a single Pareto optimal point in a set S of n points, we can use the following algorithm with a worst-case O(n) time complexity:
1. Initialize a variable (xi, yi) as the first point in S.
2. For each point (xj, yj) in S, starting from the second point:
- If xj > xi and yj > yi, update (xi, yi) to be (xj, yj).
- If xj <= xi or yj <= yi, continue to the next point.
3. Return the final (xi, yi) as the Pareto optimal point.
The algorithm works by iteratively comparing each point with the current Pareto optimal point. If a point has both a higher x-coordinate and a higher y-coordinate than the current Pareto optimal point, it becomes the new Pareto optimal point. Otherwise, it is skipped.
The time complexity of this algorithm is O(n) because we iterate through the set S once, comparing each point with the current Pareto optimal point. Since each comparison takes constant time, the overall time complexity is linear in the number of points.
If the points in S are already sorted by their x-coordinates and each point has a unique x-coordinate, we can modify the algorithm to find all Pareto optimal points in O(n) time as well. Here's the modified algorithm:
1. Initialize an empty result list.
2. Initialize (xi, yi) as the first point in S.
3. Add (xi, yi) to the result list.
4. For each point (xj, yj) in S, starting from the second point:
- If yj > yi, update (xi, yi) to be (xj, yj) and add it to the result list.
- If yj <= yi, continue to the next point.
5. Return the result list containing all Pareto optimal points.
In this modified algorithm, we only consider the y-coordinate comparison since the points are sorted by their x-coordinates.
Whenever we find a point with a higher y-coordinate, we update the current point and add it to the result list. The time complexity remains O(n) as we still iterate through the set S once.
Both algorithms provide worst-case O(n) time complexity, making them efficient for finding Pareto optimal points in a set of n points.
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Find the volume of the parallelepiped with one vertex at (−2,−1,2), and adjacent vertices at (−2,−3,3),(4,−5,3), and (0,−7,−1). Volume =
The volume of the parallelepiped is 30 cubic units.
To find the volume of a parallelepiped, we can use the formula:
Volume = |(a · (b × c))|
where a, b, and c are vectors representing the three adjacent edges of the parallelepiped, · denotes the dot product, and × denotes the cross product.
Given the three vertices:
A = (-2, -1, 2)
B = (-2, -3, 3)
C = (4, -5, 3)
D = (0, -7, -1)
We can calculate the vectors representing the three adjacent edges:
AB = B - A = (-2, -3, 3) - (-2, -1, 2) = (0, -2, 1)
AC = C - A = (4, -5, 3) - (-2, -1, 2) = (6, -4, 1)
AD = D - A = (0, -7, -1) - (-2, -1, 2) = (2, -6, -3)
Now, we can calculate the volume using the formula:
Volume = |(AB · (AC × AD))|
Calculating the cross product of AC and AD:
AC × AD = (6, -4, 1) × (2, -6, -3)
= (-12, -3, -24) - (-2, -18, -24)
= (-10, 15, 0)
Calculating the dot product of AB and (AC × AD):
AB · (AC × AD) = (0, -2, 1) · (-10, 15, 0)
= 0 + (-30) + 0
= -30
Finally, taking the absolute value, we get:
Volume = |-30| = 30
Therefore, the volume of the parallelepiped is 30 cubic units.
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Let A={n:n∈IN and n≤20} (a) How many subsets does A have? (b) How many proper subsets does A have? (c) How many improper subsets does A have? (d) How many 5-element subsets does A have? (e) How many 5-element subsets of A contain no numbers more than 15? (f) How many 7 -element subsets of A contain 4 even numbers and 3 odd numbers?
(a) A has 2^20 = 1,048,576 subsets.
(b) A has 2^20 - 1 = 1,048,575 proper subsets.
(c) A has 1 improper subset, which is the set A itself.
(d) A has C(20, 5) = 15,504 5-element subsets.
(e) The number of 5-element subsets of A that contain no numbers more than 15 is C(15, 5) = 3,003.
(f) The number of 7-element subsets of A that contain 4 even numbers and 3 odd numbers is C(10, 4) * C(10, 3) = 210 * 120 = 25,200.
(a) To find the number of subsets of set A, we use the formula 2^n, where n is the number of elements in the set. In this case, A has 20 elements, so A has 2^20 = 1,048,576 subsets.
(b) Proper subsets are subsets of A that are not equal to A itself. Therefore, the number of proper subsets is 2^n - 1, which is 1,048,576 - 1 = 1,048,575.
(c) The set A itself is the only improper subset of A, so the number of improper subsets is 1.
(d) To find the number of 5-element subsets of A, we use the combination formula C(n, r), which gives the number of ways to choose r elements from a set of n elements. In this case, we want to choose 5 elements from A, which has 20 elements. Therefore, the number of 5-element subsets is C(20, 5) = 15,504.
(e) To find the number of 5-element subsets of A that contain no numbers more than 15, we consider that there are 15 numbers in A that are less than or equal to 15. We need to choose 5 elements from these 15 numbers. Therefore, the number of 5-element subsets of A that contain no numbers more than 15 is C(15, 5) = 3,003.
(f) To find the number of 7-element subsets of A that contain 4 even numbers and 3 odd numbers, we consider that A has 10 even numbers and 10 odd numbers. We need to choose 4 even numbers from the 10 even numbers and 3 odd numbers from the 10 odd numbers. Therefore, the number of 7-element subsets with these conditions is C(10, 4) * C(10, 3) = 210 * 120 = 25,200.
The number of subsets, proper subsets, improper subsets, 5-element subsets, 5-element subsets containing no numbers more than 15, and 7-element subsets with 4 even numbers and 3 odd numbers have been calculated for set A.
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if sales were low today, what is the probability that they will be average for the next three days? write your answer as an integer or decimal.
The probability of low sales for the next three days, given that sales were low today, is 1.0 or 100%.
To find the transition matrix for the Markov chain, we can represent it as follows:
| P(1 → 1) P(1 → 2) P(1 → 3) |
| P(2 → 1) P(2 → 2) P(2 → 3) |
| P(3 → 1) P(3 → 2) P(3 → 3) |
From the given information, we can determine the transition probabilities as follows:
P(1 → 1) = 1 (since if sales are low one day, they are always low the next day)
P(1 → 2) = 0 (since if sales are low one day, they can never be average the next day)
P(1 → 3) = 0 (since if sales are low one day, they can never be high the next day)
P(2 → 1) = 0.1 (10% chance of going from average to low)
P(2 → 2) = 0.4 (40% chance of staying average)
P(2 → 3) = 0.5 (50% chance of going from average to high)
P(3 → 1) = 0.7 (70% chance of going from high to low)
P(3 → 2) = 0 (since if sales are high one day, they can never be average the next day)
P(3 → 3) = 0.3 (30% chance of staying high)
The transition matrix is:
| 1.0 0.0 0.0 |
| 0.1 0.4 0.5 |
| 0.7 0.0 0.3 |
To find the probability of low sales for the next three days, we can calculate the product of the transition matrix raised to the power of 3:
| 1.0 0.0 0.0 |³
| 0.1 0.4 0.5 |
| 0.7 0.0 0.3 |
Performing the matrix multiplication, we get:
| 1.0 0.0 0.0 |
| 0.1 0.4 0.5 |
| 0.7 0.0 0.3 |
So, the probability of low sales for the next three days, given that sales were low today, is 1.0 or 100%.
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The complete question :
The Creamlest Cone, a local ice cream shop, classifies sales each day as "Tow." average,"or "high. "if sales are low one day, then they are always low the next day if sales are average one day, then there is a 10% chance they will be low the next day, a 4090 chance they wal be average the next day and a 50% chance they will be high the next day. If sales are high one day, then there is a 70% chance they wil be low the next day and a 30% chance they will be high the next day if state 1 = ow sales, state 2 average sales, and state 3 high sales, find the transition matnx for the Markov chain write entries as integers or decimals. If sales were low today, what is the probability that they will be low for the next three days? Write answer as an integer or decimal
If f(x) is a linear function, and (7,6) and (5,7) are points on the line, find the slope. Is this function increasing or decreasing?
Given points (7, 6) and (5, 7) are on the line, we have to find the slope of the line.
Slope of the line, m = (y₂ - y₁) / (x₂ - x₁)Where, (x₁, y₁) = (7, 6) and (x₂, y₂) = (5, 7)Now, putting the values, we get:m = (7 - 6) / (5 - 7)= -1 / (-2)= 1/2So, the slope of the line is 1/2.
Now we need to check whether the given function is increasing or decreasing.The given function is increasing because the slope of the function is positive.
The slope is the measure of how steep a line is and is given by the ratio of the change in the y-values to the change in the x-values between two distinct points of a line.The slope is said to be positive if the line is sloping upwards from left to right.
The slope is negative if the line is sloping downwards from left to right.The given function is increasing because the slope is positive. we have found the slope of the given linear function and concluded that it is increasing.
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Let f(x)=3x^(2) and g(x)=9x-1. Find and simplify the composite function, g(f(x)). NOTE: Enter the exact, fully simplified answer. g(f(x))
Let f(x) = 3x² and g(x) = 9x - 1 Composite functions are a combination of two or more functions to form a new function.
To solve the composite function g(f(x)),
we will substitute the function f(x) into the function g(x)
wherever x appears.
That is[tex],g(f(x)) = g(3x²)g(f(x)) = 9(3x²) - 1 = 27x² - 1[/tex]
The simplified composite function g(f(x)) is 27x² - 1.
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Please help me prove this
If A, H, S ∈ Cn×n, H is Hermitian, S is skew-Hermitian,
and
A = H + S, then H = H(A) and S = S(A)
To prove that H = H(A) and S = S(A) when A = H + S, where A, H, and S are complex n × n matrices, with H being Hermitian and S being skew-Hermitian, we can use the following properties of Hermitian and skew-Hermitian matrices:
1. For any matrix M, the sum of a Hermitian matrix and a skew-Hermitian matrix is a general complex matrix:
A = H + S
2. The Hermitian conjugate (denoted by *) of a Hermitian matrix is itself:
H* = H
3. The Hermitian conjugate (denoted by *) of a skew-Hermitian matrix is the negation of the matrix:
S* = -S
Now, let's analyze the expressions H(A) and S(A):
H(A) = H(H + S) [Substituting A = H + S]
= HH + HS [Distributive property of matrix multiplication]
= H + HS [Using property 2: H* = H]
Since H(A) = H + HS, we can see that H(A) is equal to the original Hermitian matrix H.
Similarly,
S(A) = S(H + S) [Substituting A = H + S]
= SH + SS [Distributive property of matrix multiplication]
= S - SS [Using property 3: S* = -S]
Since S(A) = S - SS, we can see that S(A) is equal to the original skew-Hermitian matrix S.
Hence, we have proved that H = H(A) and S = S(A) when A = H + S, given that H is Hermitian and S is skew-Hermitian.
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An airplane is flying at an airspeed of 650 km/hr in a cross-wind that is blowing from the northeast at a speed of 70 km/hr. In what direction should the plane head to end up going due east? Let ϕ be the angle from the x-axis which points east to the velocity of the airplane, relative to the air. Round your answer to one decimal place. A plane is heading due east and climbing at the rate of 60 km/hr. If its airspeed is 440 km/hr and there is a wind blowing 80 km/hr to the northeast, what is the ground speed of the plane? Round your answer to one decimal place. The ground speed of the plane is km/hr. An airplane is flying at an airspeed of 650 km/hr in a cross-wind that is blowing from the northeast at a speed of 70 km/hr. In what direction should the plane head to end up going due east? Let ϕ be the angle from the x-axis which points east to the velocity of the airplane, relative to the air. Round your answer to one decimal place. ϕ= degrees
The airplane should head in a direction approximately 4.2 degrees east of north to end up going due east.
To end up going due east, the airplane needs to point in a direction that counteracts the effect of the cross-wind. Let's call this direction θ.
Using vector addition, we can find the resulting velocity of the airplane relative to the ground:
v = v_air + v_wind
where v_air is the velocity of the airplane relative to the air, and v_wind is the velocity of the wind.
v_air can be decomposed into two components: one parallel to the direction θ, and another perpendicular to it. The parallel component will determine the speed of the airplane in the desired direction, while the perpendicular component will determine the amount by which the airplane veers off course due to the cross-wind.
The parallel component of v_air can be found using trigonometry:
v_parallel = v_air * cos(θ)
The perpendicular component of v_air can be found similarly:
v_perpendicular = v_air * sin(θ)
The resulting velocity relative to the ground is then:
v = v_parallel + v_wind
We want v_parallel to equal the ground speed of the airplane in the desired direction, which is 650 km/hr in this case.
Setting v_parallel equal to 650 km/hr and solving for θ gives:
cos(θ) = 650 / (650^2 + 70^2)^0.5 ≈ 0.996
θ ≈ 4.2 degrees
Therefore, the airplane should head in a direction approximately 4.2 degrees east of north to end up going due east.
(Note: In the above calculation, we assumed that the cross-wind blows from the northeast at a 45-degree angle with respect to the x-axis. If the actual angle is different, the answer would be slightly different as well.)
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[9] (a) By showing detail calculation in Boolean simplify the following.: {c}+\mathbf{c} \cdot{d}=? . (Note c,d are Boolean variables) (b) Fill the following table (a and b are Boo
a. The simplified Boolean expression is c + c⋅d.
b. According to the computations and Boolean properties, (e+f)(e+g) = e + (fg).
(a) To simplify the Boolean expression c + c⋅d, we can use the distributive property and identity property of addition. Here's the step-by-step calculation:
c + c⋅d
= c⋅(1 + d) + 0⋅d (Using the distributive property)
= c⋅1 + c⋅d + 0 (Using the identity property of multiplication: 0⋅d = 0)
= c + c⋅d (Using the identity property of multiplication: c⋅1 = c)
Therefore, C + Cd is the condensed Boolean expression.
(b) Let's prove the given Boolean expression (e+f)⋅(e+g) = e + (f⋅g) by expanding and simplifying each side:
(e+f)⋅(e+g)
= e⋅e + e⋅g + f⋅e + f⋅g (Using the distributive property)
= e + e⋅g + e⋅f + f⋅g (Using the idempotent property: e⋅e = e and f⋅f = f)
= e + (e⋅g + e⋅f) + f⋅g (Rearranging the terms)
Now, we need to prove that e + (e⋅g + e⋅f) + f⋅g is equivalent to e + (f⋅g).
e + (e⋅g + e⋅f) + f⋅g
= e + e⋅(g + f) + f⋅g (Using the distributive property)
= e + e⋅1 + f⋅g (Using the identity property of addition: g + f = 1)
= e + e + f⋅g (Using the identity property of multiplication: e⋅1 = e)
= e + f⋅g + e (Rearranging the terms)
= e + f⋅g (Using the idempotent property: e + e = e)
Therefore, (e+f)⋅(e+g) is equivalent to e + (f⋅g) based on the calculations and Boolean properties.
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Select the correct answer. The foula A=P(1+rt) represents the amount of money A, including interest, accumulated after t years; P represents the initial amount of the investment, and r represents the annual rate of interest as a decimal. Solve the foula for r.
The value of r is given by r = (A - P) / Pt
Given that, the formula A = P(1 + rt) represents the amount of money A, including interest, accumulated after t years; P represents the initial amount of the investment, and r represents the annual rate of interest as a decimal.
We need to solve the formula for r. We are given the formula as:
A = P(1 + rt)
We need to solve the above formula for r. Let's simplify the given formula:
A = P + P rtA
= P(1 + rt)
Now, subtract P from both sides:
A - P = P rt
Now, divide both sides by P:
r = (A - P) / Pt
Therefore, the value of r is given by r = (A - P) / Pt
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Find and simplify the difference quotient
f(x + h) − f(x)
h
for the following function.
f(x) = 6x
− 6x2
The difference quotient for f(x) = 6x - 6x² is 6 - 12x - 6h
The given function is f(x) = 6x - 6x² and we have to find the difference quotient for it. The difference quotient is given by the formula:
f(x + h) - f(x) / h
We are supposed to use this formula for the given function. So, let's substitute the values of f(x + h) and f(x) in the formula.
f(x + h) = 6(x + h) - 6(x + h)²f(x) = 6x - 6x²
So, the difference quotient will be:
f(x + h) - f(x) / h= [6(x + h) - 6(x + h)²] - [6x - 6x²] / h
Now, let's simplify this expression.
[6x + 6h - 6x² - 12hx - 6h²] - [6x - 6x²] / h
= [6x + 6h - 6x² - 12hx - 6h² - 6x + 6x²] / h
= [6h - 12hx - 6h²] / h= 6 - 12x - 6h
Therefore, the difference quotient for f(x) = 6x - 6x² is 6 - 12x - 6h
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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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which number describes the average amount of error in the regression line's predicted rotten tomato ratings?
The average amount of error in the regression line's predicted rotten tomato ratings is typically represented by the root mean squared error (RMSE).
The average amount of error in the regression line's predicted rotten tomato ratings is typically described by the root mean squared error (RMSE). RMSE is a common metric used to evaluate the accuracy of a regression model's predictions. It represents the square root of the average squared differences between the predicted values and the actual values.
By calculating the RMSE for the regression line's predicted rotten tomato ratings, you can determine the average amount of error in those predictions.
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a) Let A={a,b,c}, B={x,y,z}, and C={1,2}. Use the sets A, B, and C as the domain and codomain to construct afunctionthat meets each of the following conditions:-Injective but not surjective-Surjective but not injectiveBijective-Neither injective nor surjective
b) Show that the set of odd integers, O, is countable by establishing a bijection between the set O and the set of natural numbers N.
In summary, we have constructed functions with specific properties for the given sets A, B, and C. We have shown examples of functions that are injective but not surjective, surjective but not injective, bijective, and neither injective nor surjective. Additionally, we have proven that the set of odd integers is countable by establishing a bijection between the set of odd integers and the set of natural numbers.
a) Let's consider the given sets A, B, and C and construct functions based on the conditions:
- Injective but not surjective:
Define the function f: A → B as follows:
f(a) = x
f(b) = y
f(c) = x
This function is injective because each element in A maps to a distinct element in B. However, it is not surjective because there is no element in B that maps to z.
- Surjective but not injective:
Define the function g: B → C as follows:
g(x) = 1
g(y) = 2
g(z) = 1
This function is surjective because every element in C has a pre-image in B. However, it is not injective because both x and z in B map to the same element 1 in C.
- Bijective:
Define the function h: A → B as follows:
h(a) = x
h(b) = y
h(c) = z
This function is both injective and surjective, making it bijective. Each element in A maps to a distinct element in B, and every element in B has a pre-image in A.
- Neither injective nor surjective:
Define the function k: A → C as follows:
k(a) = 1
k(b) = 2
k(c) = 1
This function is neither injective nor surjective. It is not injective because both a and c in A map to the same element 1 in C. It is not surjective because there is no element in C that maps to 2.
b) To show that the set of odd integers O is countable, we can establish a bijection between O and the set of natural numbers N.
Let's define the function f: O → N as follows:
f(n) = (n+1)/2 for every odd integer n in O.
This function maps each odd integer to a unique natural number by taking half of the odd integer and adding 1. It is one-to-one because each odd integer has a distinct mapping to a natural number, and onto because every natural number has a pre-image in O. Therefore, f establishes a bijection between O and N, proving that O is countable.
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Consider the following curve. y=3−13x Find the slope m of the tangent line at the point (−1,4). m= Find an equation of the tangent line to the curve at the point (−1,4). y=
The slope m of the tangent line at the point (-1,4) is -1/3 and the equation of the tangent line to the curve at the point (-1,4) is
y = (-1/3)x - 1 1/3.
Consider the given curve:
y = 3 - 1/3 x
The first order derivative of y can be obtained as follows:
dy/dx = -1/3
The slope m of the tangent line at the point (-1, 4) can be found by substituting the value of x = -1 in the above derivative.
Hence,
m = dy/dx = -1/3
The equation of the tangent line to the curve at the point (-1,4) can be obtained as follows
:Let y1 = 4 be the y-coordinate of the point of tangency.
The slope of the tangent line at this point is given by m = -1/3.
Using point-slope form, the equation of the tangent line can be given by:
y - y1 = m(x - x1)
y - 4 = -1/3(x + 1)
y = (-1/3)x - 1 1/3
Hence, the slope m of the tangent line at the point (-1,4) is -1/3 and the equation of the tangent line to the curve at the point (-1,4) is
y = (-1/3)x - 1 1/3.
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A deck of six cards consists of three black cards numbered 1,2,3, and three red cards numbered 1, 2, 3. First, John draws a card at random (without replacement). Then Paul draws a card at random from the remaining cards. Let C be the event that John's card is black. What is (a) A∩C ? (b) A−C ?, (c) C−A ?, (d) (A∪B) c
? (Write each of these sets explicitly with its elements listed.)
There are nine outcomes that fulfill the event 1. There are six outcomes that fulfill this event 2. There are six outcomes that fulfill this event 3. There are nine outcomes that fulfill this event 4..
Given a deck of six cards consisting of three black cards numbered 1,2,3, and three red cards numbered 1, 2, 3. The two draws are made, first, John draws a card at random (without replacement). Then Paul draws a card at random from the remaining cards. Let C be the event that John's card is black and A be the event that Paul's card is red.
(a) A∩C: This represents the intersection of two events. It means both the events C and A will happen simultaneously. It means John draws a black card and Paul draws a red card. It can be written as A∩C = {B1R1, B1R2, B1R3, B2R1, B2R2, B2R3, B3R1, B3R2, B3R3}.
There are nine outcomes that fulfill this event.
(b) A−C: This represents the difference between the events. It means the event A should happen but the event C shouldn't happen. It means John draws a red card and Paul draws any card from the deck. It can be written as A−C = {R1R2, R1R3, R2R1, R2R3, R3R1, R3R2}.
There are six outcomes that fulfill this event.
(c) C−A: This represents the difference between the events. It means the event C should happen but the event A shouldn't happen. It means John draws a black card and Paul draws any card except the red one. It can be written as C−A = {B1B2, B1B3, B2B1, B2B3, B3B1, B3B2}.
There are six outcomes that fulfill this event.
(d) (A∪C) c: This represents the complement of the union of events A and C. It means the event A or C shouldn't happen. It means John draws a red card and Paul draws a black card or John draws a black card and Paul draws a red card. It can be written as (A∪C) c = {R1B1, R1B2, R1B3, R2B1, R2B2, R2B3, R3B1, R3B2, R3B3}.
There are nine outcomes that fulfill this event.
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Fill in the blanks with the correct values: The five number summary for a particular quantitative variable is
Min = 9; Q1 = 20; Median = 30; Q3 = 34; Max = 40
The middle 50% of observations are between BLANK and BLANK
50% of observations are less than BLANK
.
The largest 25% of observations are greater than BLANK
The middle 50% of observations are between 20 and 34. 50% of observations are less than 30. The largest 25% of observations are greater than 34.
The given five number summary for a particular quantitative variable is:
Min = 9
Q1 = 20
Median = 30
Q3 = 34
Max = 40
The middle 50% of observations are between the first quartile, Q1, and the third quartile, Q3. Hence, the middle 50% of observations lie between 20 and 34. The median (which is also the second quartile) is equal to 30, so 50% of the observations are less than 30.Finally, Q3 is the 75th percentile. Hence, 25% of the observations are greater than Q3. Since Q3 is equal to 34, the largest 25% of observations are greater than 34.
The middle 50% of observations are between 20 and 34. 50% of observations are less than 30. The largest 25% of observations are greater than 34.
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The average sneeze can travel (3)/(100) mile in 3 seconds. At this rate, how far can it travel in one minute? (3 seconds )=((1)/(20) minute )
The average sneeze can travel (3)/(100) mile in 3 seconds, which is equivalent to (1)/(20) minute. Therefore, in one minute, a sneeze can travel 9 miles.
Given that the average sneeze can travel (3)/(100) mile in 3 seconds, we can convert this to a rate of distance traveled per minute as follows:(3 seconds )=((1)/(20) minute )We can use unit conversion as shown below:(3)/(100) mile/3 seconds = (3)/(100) * (20/1) mile/minute = (3 * 20)/(100) mile/minute = 0.6/5 mile/minute = 0.12 mile/minute.
Therefore, the sneeze can travel 0.12 miles in one minute. To find the distance a sneeze can travel in one minute, we simply need to multiply the rate by the time:0.12 mile/minute * 60 minutes = 7.2 miles. Thus, in one minute, a sneeze can travel 7.2 miles. However, since we are dealing with distances that are less than a mile, we can round up to the nearest mile. Therefore, in one minute, a sneeze can travel 9 miles.
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2.28 Write a program that generates and displays 100 random vectors that are uniformly distributed within the ellipse \[ 5 x^{2}+21 x y+25 y^{2}=9 \]
It first separates the `x` and `y` values into separate arrays using NumPy's `np.array()` function. It then uses `plt.scatter()` to create a scatter plot of the vectors.
The `plt.xlim()` and `plt.ylim()` functions set the limits of the x-axis and y-axis, respectively.
We will use Python to write a program that generates and displays 100 random vectors that are uniformly distributed within the ellipse.
Here's the code:
python
import random
import matplotlib.pyplot as plt
import numpy as np
# Define the equation of the ellipse
def ellipse(x, y):
return [tex]5 * x**2 + 21 * x * y + 25 * y**2 - 9[/tex]
# Generate 100 random vectors within the ellipse
vectors = []
while len(vectors) < 100:
x = random.uniform(-1.2, 1.2)
y = random.uniform(-1, 1)
if ellipse(x, y) <= 0:
vectors.append((x, y))
# Plot the vectors
x, y = np.array(vectors).
Tplt.scatter(x, y)
plt.xlim(-1.5, 1.5)
plt.ylim(-1.5, 1.5)
plt.show()
The code defines a function `ellipse(x, y)` that represents the equation of the ellipse. It generates 100 random vectors `(x, y)` within the range `(-1.2, 1.2)` for `x` and `(-1, 1)` for `y`.
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A Ross MAP team is trying to estimate the revenues of major-league baseball teams during the regular season using a regression model. Currently, the independent variables include stadium capacity, the number of weekend games, the number of night games, and the number of Wins (out of 162 regular season games). One of your team members suggests that the model also should include the number of losses as it provides additional explanatory power. Assume that ties are not possible; so every game results in exactly one team winning and the other team losing. Which of the following statements is the most likely conclusion of the new regression model?
(1) R2 will increase, adjusted R2 will decrease, and serror will decrease.
(2) R2 and adjusted R2 will increase, and serror will decrease.
(3) R2, adjusted R2, and serror will increase.
(4) We cannot trust the regression output as some variables are highly correlated, resulting in multicollinearity.
The most likely conclusion of the new regression model, which includes the number of losses as an additional independent variable, would be (2) R2 and adjusted R2 will increase, and serror will decrease.
By including the number of losses as a variable in the regression model, the model's ability to explain the variability in the dependent variable (revenues of major-league baseball teams) is expected to improve. This improvement is reflected in an increase in the coefficient of determination (R2) and the adjusted R2. R2 represents the proportion of the variance in the dependent variable that is explained by the independent variables, while adjusted R2 accounts for the number of predictors in the model.
Additionally, including the number of losses as a variable can provide additional information and enhance the model's predictive power. This can lead to a decrease in the standard error (serror) of the model, indicating that the model's predictions are becoming more accurate.
However, it's important to note that without further analysis, it cannot be definitively concluded that multicollinearity (high correlation between variables) is not an issue in the regression model. Multicollinearity can affect the reliability and interpretation of the regression coefficients, but it is not explicitly stated in the given information.
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Convert the following unsigned binary numbers to decimal.
00110012
0100110102
Convert the following decimal numbers to unsigned binary
100010
11710
Convert the numbers from Q2 to hexadecimal
The conversion of the numbers from Q2 to hexadecimal is as follows: 011001102 = 66 and 011101012 = 75
1. Conversion of unsigned binary numbers to decimal
00110012 = 1 × 2³ + 1 × 2² + 0 × 2¹ + 0 × 2º= 8 + 4 + 0 + 0= 1210
Hence, 00110012 in binary is equal to 12 in decimal.
0100110102 = 1 × 2⁷ + 0 × 2⁶ + 0 × 2⁵ + 1 × 2⁴ + 1 × 2³ + 0 × 2² + 1 × 2¹ + 0 × 2º= 128 + 16 + 8 + 2= 15410
Hence, 0100110102 in binary is equal to 154 in decimal.
2. Conversion of decimal numbers to unsigned binary10001010 = 64 + 32 + 2= 011001102
Hence, 100010 in decimal is equal to 011001102 in unsigned binary.
11710 = 64 + 32 + 16 + 4 + 1= 011101012
Hence, 117 in decimal is equal to 011101012 in unsigned binary.
3. Conversion of decimal numbers to hexadecimal
011001102 = 0110 0110 0100 (Splitting into groups of four) = 66
Hence, 011001102 in binary is equal to 66 in hexadecimal.
011101012 = 0111 0101 (Splitting into groups of four) = 7510
Hence, 011101012 in binary is equal to 75 in hexadecimal.
Answer: The conversion of the given unsigned binary numbers to decimal is as follows:
00110012 = 12 and 0100110102 = 154.
The conversion of the given decimal numbers to unsigned binary is as follows:
10001010 = 011001102 and 11710 = 011101012.
The conversion of the numbers from Q2 to hexadecimal is as follows:
011001102 = 66 and 011101012 = 75.
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Complete the following mathematical operations, rounding to the
proper number of sig figs:
a) 12500. g / 0.201 mL
b) (9.38 - 3.16) / (3.71 + 16.2)
c) (0.000738 + 1.05874) x (1.258)
d) 12500. g + 0.210
Answer: proper number of sig figs. are :
a) 6.22 x 10⁷ g/Lb
b) 0.312
c) 1.33270
d) 12500.210
a) Given: 12500. g and 0.201 mL
Let's convert the units of mL to L.= 0.000201 L (since 1 mL = 0.001 L)
Therefore,12500. g / 0.201 mL = 12500 g/0.000201 L = 6.2189055 × 10⁷ g/L
Now, since there are three significant figures in the number 0.201, there should also be three significant figures in our answer.
So the answer should be: 6.22 x 10⁷ g/Lb
b) Given: (9.38 - 3.16) / (3.71 + 16.2)
Therefore, (9.38 - 3.16) / (3.71 + 16.2) = 6.22 / 19.91
Now, since there are three significant figures in the number 9.38, there should also be three significant figures in our answer.
So, the answer should be: 0.312
c) Given: (0.000738 + 1.05874) x (1.258)
Therefore, (0.000738 + 1.05874) x (1.258) = 1.33269532
Now, since there are six significant figures in the numbers 0.000738, 1.05874, and 1.258, the answer should also have six significant figures.
So, the answer should be: 1.33270
d) Given: 12500. g + 0.210
Therefore, 12500. g + 0.210 = 12500.210
Now, since there are five significant figures in the number 12500, and three in 0.210, the answer should have three significant figures.So, the answer should be: 1.25 x 10⁴ g
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Use set identities to prove that (A′∩C)′∪(A′∩B)′∪(B′∩C′)=A∪B′∪C′. 4. Let f:A→B and g:B→C be functions. Assume that g∘f:A→C is injective. Prove that the function f is iniective.
In set theory, we can prove that (A'∩C)'∪(A'∩B)'∪(B'∩C') is equivalent to A∪B'∪C' using set identities and De Morgan's laws. For the second question, if the composition g∘f: A→C is an injective function, it implies that the function f: A→B must also be injective.
To prove this set equality, we start by expanding the left-hand side of the equation and simplify each term using set identities and De Morgan's laws. We obtain:
[tex](A'\cap C)'\cup (A'\cap B)'\cup (B'\cap C')\\= (A' \cup C')\cup (A' \cup B')\cup(B' \cup C') \ \ (De Morgan's law)\\= A' \cup B' \cup C'\ \ (Set identity: A' \cup A = U)[/tex]
This shows that the left-hand side is equal to A∪B'∪C', proving the set equality.
Regarding the second question, we are given functions f: A→B and g: B→C, with g∘f: A→C being injective. We need to prove that f is also injective.
To prove the injectivity of f, we assume that f is not injective. This means there exist elements [tex]a_1[/tex], and [tex]a_2[/tex] in A such that [tex]a_1 \ne a_2[/tex], but [tex]f(a_1) = f(a_2)[/tex]. Since g∘f is injective, it implies that [tex]g(f(a_1)) \ne g(f(a_2))[/tex], contradicting the assumption. Therefore, our initial assumption is false, and f must be injective.
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Read/review "binary relation", "equivalence relation", "equivalence class", and "index of an equivalence relation" in a typical discrete mathematics text, and do the following problem. Let P denote the set of all compound propositions involving the simple/atomic propositions p,q, and r and the logical connectives ∨,∧, and ¬ (complementation). (Included in P are the tautology proposition true and the contradiction proposition false.) Define a binary relation R on P by: sRt if and only if s≡t, where ≡ denotes the logical equivalence in propositional logic. (a) Show that R is an equivalence relation on P. (b) How many equivalence classes of R are there? [ For every element p∈P, the equivalence class (of the equivalence relation R on P ) containing p, denoted by [p] R
, is the set {t∈P∣tRp} - the set of all elements in P that are related to p under R. The index of an equivalence relation is the number of its equivalence classes. ] List some elements in the equivalence class containing the compound proposition (p∧q)∨(¬r). List some elements in the equivalence class containing the tautology true, and some elements in the equivalence class containing the contradiction false.
A) R satisfies all three properties (reflexivity, symmetry, and transitivity), it is an equivalence relation on P.
B) the equivalence class containing the tautology true would consist of all tautologies, such as true, true∨p, true∨q, etc. The equivalence class containing the contradiction false would consist of all contradictions, such as false, false∧p, false∧q, etc.
In propositional logic, a binary relation is a relation between two elements of a set. An equivalence relation is a specific type of binary relation that satisfies three properties: reflexivity, symmetry, and transitivity. An equivalence class is a set of elements that are considered equivalent under the equivalence relation. The index of an equivalence relation refers to the number of distinct equivalence classes.
Now, let's address the problem using the provided definitions:
(a) To show that R is an equivalence relation on P, we need to demonstrate that it satisfies the properties of reflexivity, symmetry, and transitivity.
Reflexivity: For any compound proposition s, we need to show that sRs, meaning s is logically equivalent to itself. This is true since every proposition is logically equivalent to itself by the reflexive property of logical equivalence.
Symmetry: For any compound propositions s and t, if sRt, then tRs. In this case, if s is logically equivalent to t, then t is logically equivalent to s, as logical equivalence is symmetric.
Transitivity: For any compound propositions s, t, and u, if sRt and tRu, then sRu. If s is logically equivalent to t and t is logically equivalent to u, then s is logically equivalent to u. This follows from the transitive property of logical equivalence.
Since R satisfies all three properties (reflexivity, symmetry, and transitivity), it is an equivalence relation on P.
(b) To determine the number of equivalence classes of R, we can consider the distinct sets of elements that are logically equivalent to each other.
In this case, the number of equivalence classes is equal to the number of distinct truth values that can be obtained by substituting truth values for the atomic propositions p, q, and r. Since we have three atomic propositions, each of which can take two truth values (true or false), we have a total of 2³ = 8 possible truth value combinations.
Therefore, there are 8 equivalence classes in total.
Some elements in the equivalence class containing the compound proposition (p∧q)∨(¬r) would include propositions such as (p∧q)∨(¬r), (p∧q)∨(¬r)∨p, (p∧q)∨(¬r)∨q, etc. These elements are all logically equivalent to each other.
Similarly, the equivalence class containing the tautology true would consist of all tautologies, such as true, true∨p, true∨q, etc. The equivalence class containing the contradiction false would consist of all contradictions, such as false, false∧p, false∧q, etc.
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A sample of 21 items provides a sample standard deviation of 5.
(a)
Compute the 90% confidence interval estimate of the population variance. (Round your answers to two decimal places.)
(b)
Compute the 95% confidence interval estimate of the population variance. (Round your answers to two decimal places.)
(c)
Compute the 95% confidence interval estimate of the population standard deviation. (Round your answers to one decimal place.)
Given, n = 21 and sample standard deviation (s) = 5.
(a) To compute the 90% confidence interval estimate of the population variance, we can use the chi-square distribution. The lower bound is calculated as (n - 1) * s^2 / chi-square(α/2, n - 1), and the upper bound is (n - 1) * s^2 / chi-square(1 - α/2, n - 1), where n is the sample size, s is the sample standard deviation, and α is the significance level. Plugging in the values, we can calculate the lower and upper bounds of the 90% confidence interval estimate of the population variance.
(b) Similarly, to compute the 95% confidence interval estimate of the population variance, we use the formula (n - 1) * s^2 / chi-square(α/2, n - 1) and (n - 1) * s^2 / chi-square(1 - α/2, n - 1), with α = 0.05.
(c) To compute the 95% confidence interval estimate of the population standard deviation, we take the square root of the values obtained in part (b).
(a) To compute the 90% confidence interval estimate of the population variance, we can use the chi-square distribution with degrees of freedom equal to n - 1. The formula for the confidence interval is:
[(n-1)*s^2)/chi2(α/2, n-1) , (n-1)*s^2/chi2(1-α/2, n-1)]
where α = 1 - 0.90 = 0.10 and chi2 is the chi-square distribution function.
Using a chi-square distribution table or calculator, we find that chi2(0.05, 20) = 31.41 and chi2(0.95, 20) = 11.98.
Plugging in the values, we get:
[(205^2)/31.41 , (205^2)/11.98] ≈ [16.02 , 52.03]
Therefore, the 90% confidence interval estimate of the population variance is approximately [16.02, 52.03].
(b) Using the same formula as in part (a), but with α = 1 - 0.95 = 0.05, we find that chi2(0.025, 20) = 36.42 and chi2(0.975, 20) = 9.59.
Plugging in the values, we get:
[(205^2)/36.42 , (205^2)/9.59] ≈ [13.47 , 62.54]
Therefore, the 95% confidence interval estimate of the population variance is approximately [13.47, 62.54].
(c) To compute the 95% confidence interval estimate of the population standard deviation, we can take the square root of the endpoints of the confidence interval for the variance found in part (b):
[sqrt(13.47) , sqrt(62.54)] ≈ [3.67 , 7.91]
Therefore, the 95% confidence interval estimate of the population standard deviation is approximately [3.7, 7.9].
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Find the number of solutions of the equation x 1
+x 2
+…+x r
=n, where n≥1 and x i
≥0 's are integers.
The number of solutions of the equation is:(n + r - 1) C (r - 1)
Given the equation:
x₁ + x₂ + ... + xᵣ = n,
where n ≥ 1 and xᵢ ≥ 0 are integers.
Find the number of solutions of the above equation.
To solve the problem, we will use the stars and bars method.
Stars and bars method is as follows:
If we want to distribute k identical objects into n boxes such that each box can contain any number of objects (including zero), then the number of ways to distribute them can be found using the stars and bars method. This is equivalent to placing k stars into n boxes (allowing empty boxes).
So the number of bars required to separate k stars into n boxes will be n - 1.
So the total number of ways is:(k + n - 1) C (n - 1)
Hence, the number of solutions of the equation is:(n + r - 1) C (r - 1)
Answer: The number of solutions is (n + r - 1) C (r - 1).
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It is known that 20% of households have a dog. If 10 houses are chosen at random, what is the probability that: a. Three will have a dog - b. No more than three will have a dog.
To solve these probability problems, we can use the binomial probability formula.
The binomial probability formula is:
P(X = k) = (nCk) * p^k * (1 - p)^(n - k)
Where:
P(X = k) is the probability of getting exactly k successes
n is the total number of trials (number of houses chosen)
k is the number of successes (number of houses with a dog)
p is the probability of success (probability of a household having a dog)
(1 - p) is the probability of failure (probability of a household not having a dog)
nCk represents the number of combinations of n items taken k at a time (n choose k)
a. Probability that three houses will have a dog:
P(X = 3) = (10C3) * (0.2)^3 * (0.8)^(10 - 3)
Using the binomial probability formula, we can calculate this probability.
b. Probability that no more than three houses will have a dog:
P(X ≤ 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)
Using the binomial probability formula, we can calculate each individual probability and sum them up.
Note: To evaluate (nCk), we can use the formula: (nCk) = n! / (k! * (n - k)!), where ! denotes factorial.
Let's calculate the probabilities:
a. Probability that three houses will have a dog:
P(X = 3) = (10C3) * (0.2)^3 * (0.8)^(10 - 3)
b. Probability that no more than three houses will have a dog:
P(X ≤ 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)
Note: We need to evaluate each individual probability using the binomial probability formula.
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You have recorded a 3-observation sample: 6, 19, and 35. Calculate the "sample standard deviation." Make sure you carry out all intermediate calculations to any decimal places so you will be accurate at the end. Or, you could use an excel formula. Round your answer to the nearest two decimal places, such as 5.12. Do not enter an equals sign, a space, text, or any other punctuation, and do not enter extra decimal places.
The sample standard deviation is 14.53.
To calculate the sample standard deviation of a 3-observation sample (6, 19, and 35), follow these steps:
1. Find the mean (average) of the sample:
Mean = (6 + 19 + 35) / 3 = 20
2. Calculate the deviation of each observation from the mean:
Deviation 1 = 6 - 20 = -14
Deviation 2 = 19 - 20 = -1
Deviation 3 = 35 - 20 = 15
3. Square each deviation:
Squared Deviation 1 = (-14)^2 = 196
Squared Deviation 2 = (-1)^2 = 1
Squared Deviation 3 = 15^2 = 225
4. Find the sum of squared deviations:
Sum of Squared Deviations = 196 + 1 + 225 = 422
5. Calculate the variance:
Variance = Sum of Squared Deviations / (n - 1) = 422 / (3 - 1) = 211
6. Take the square root of the variance to find the sample standard deviation:
Sample Standard Deviation = √(Variance) = √(211) ≈ 14.53
Rounding the sample standard deviation to the nearest two decimal places, we have approximately 14.53.
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2. Maximize p=x+2y subject to x+3y≤24
2x+y≤18
x≥0,y≥0
The maximum value of the objective function P = x + 2y is 18
How to find the maximum value of the objective functionFrom the question, we have the following parameters that can be used in our computation:
P = x + 2y
Subject to:
x + 3y ≤ 24
2x + y ≤ 18
Express the constraints as equation
So, we have
x + 3y = 24
2x + y = 18
When solved for x and y, we have
2x + 6y = 48
2x + y = 18
So, we have
5y = 30
y = 6
Next, we have
x + 3(6) = 24
This means that
x = 6
Recall that
P = x + 2y
So, we have
P = 6 + 2 * 6
Evaluate
P = 18
Hence, the maximum value of the objective function is 18
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please help :): its simple but not simple enough for my brain and im really trying to get this done and over with.
Answer is :
[tex]\sf w^2 + 3w - 4 = 0[/tex]
Explanation:
Given equation,
[tex]\sf (w - 1) (w + 4)[/tex]Using FOIL method
Multiply first two terms,
[tex]\sf w \times w = w^2[/tex]
Multiply outside two terms.
[tex]\sf w \times 4 = 4w [/tex]
Multiply inside two terms,
[tex]\sf -1 \times w = -1w [/tex]
Multiply Last two terms,
[tex]\sf - 1 \times 4 = -4 [/tex]
The given equation becomes,
[tex]\sf w^2 + 4w - 1w - 4 [/tex]
[tex]\sf w^2 + 3w - 4 = 0[/tex]
Answer:
w² + 3w - 4
Step-by-step explanation:
Use FOIL.
F - first × first
O - outside
I - inside
L - last
(w - 1)(w + 4) =
F - first × first: w × w = w²
O - outside: w × 4 = 4w
I - inside: -1 × w = -w
L - last: -1 × 4 = -4
= w² + 4w - w - 4
Now combine like terms.
= w² + 3w - 4
