Therefore, we can conclude that f(x) = x² is not uniformly continuous on (0,1].
To show that the function f(x) = x² is not uniformly continuous on the interval (0,1], we can use a proof by contradiction.
Assume that f(x) is uniformly continuous on (0,1].
According to the definition of uniform continuity, for any given ε > 0, there exists a δ > 0 such that for any x,y in (0,1] satisfying |x-y| < δ, we have |f(x)-f(y)| < ε.
Let's choose ε = 1.
Now, for any δ > 0, consider the points x = δ/2 and y = δ/4 in (0,1].
We have |x-y| = |δ/2 - δ/4|
= δ/4 < δ.
Using the function f(x) = x², we find that |f(x)-f(y)|
= |(δ/2)² - (δ/4)²|
= δ²/4.
Now, we need to find a δ such that δ²/4 ≥ 1 for all δ > 0. However, this is not possible because as δ approaches 0,
δ²/4 approaches 0 as well. Hence, there is no δ that satisfies the condition |f(x)-f(y)| < 1 for all x,y in (0,1] with |x-y| < δ.
This contradicts our assumption that f(x) is uniformly continuous on (0,1].
Therefore, we can conclude that f(x) = x² is not uniformly continuous on (0,1].
To know more about continuous visit:
https://brainly.com/question/31523914
#SPJ11
El Joan ha comprado 3 kg de naranjas en la mañana y 5 kilos en la tarde cuanto a pagado por todas las naranjas si un kilogramo le ah costado 2 euros
According to the question El Joan ha comprado 3 kg de naranjas en la mañana y 5 kilos Joan has paid 16 euros for all the oranges.
To find out how much Joan has paid for all the oranges, we need to calculate the total cost by multiplying the total weight of oranges by the cost per kilogram.
In the morning, Joan bought 3 kg of oranges, so the cost of the oranges in the morning is 3 kg * 2 euros/kg = 6 euros.
In the afternoon, Joan bought 5 kg of oranges, so the cost of the oranges in the afternoon is 5 kg * 2 euros/kg = 10 euros.
To find the total cost, we add the cost of the oranges in the morning and the cost of the oranges in the afternoon:
Total cost = Cost of morning oranges + Cost of afternoon oranges
= 6 euros + 10 euros
= 16 euros
Therefore, Joan has paid 16 euros for all the oranges.
To know more about cost visit -
brainly.com/question/23341920
#SPJ11
quadrilateral vwxy is a parallelogram. find the measure of ∠v. question 4 options: a) 52° b) 120° c) 38° d) 60°
The measure of ∠V in parallelogram VWXY is 120°. This is determined by the fact that opposite angles in a parallelogram are congruent. In the given parallelogram, the opposite angle to ∠V is ∠X, which is calculated to be 90°. Therefore, ∠V must also measure 90°.
In a parallelogram, opposite angles are congruent. Therefore, ∠V and ∠X are equal. To find the measure of ∠V, we can first find the measure of ∠X.
Since VWXY is a parallelogram, we know that the opposite sides are parallel. This means that the transversal VY forms corresponding angles with VW and XY. Corresponding angles are congruent, so we can set up the following equation:
∠X = ∠VY
Now, we need to find the measure of ∠VY. Since VWXY is a parallelogram, we know that opposite sides are equal in length. This means that VY is equal in length to WX.
We also know that the sum of the interior angles in a quadrilateral is 360°. Thus, we can set up the following equation:
∠VY + ∠YX + ∠XW + ∠WV = 360°
Since ∠YX and ∠XW are opposite angles and congruent, and ∠VY and ∠WV are also opposite angles and congruent, we can rewrite the equation as:
∠VY + ∠VY + ∠X + ∠X = 360°
Simplifying, we get:
2∠VY + 2∠X = 360°
Dividing by 2, we obtain:
∠VY + ∠X = 180°
Since ∠X = ∠VY, we can substitute and solve for ∠X:
∠X + ∠X = 180°
2∠X = 180°
∠X = 90°
Now that we know ∠X, we can conclude that ∠V is also 90° because ∠V and ∠X are opposite angles in a parallelogram.
To know more about parallelogram, visit
https://brainly.com/question/970600
#SPJ11
a frustum of a right circular cone is formed by cutting a small cone off of the top of a larger cone. if a particular frustum has an altitude of $24$ centimeters, the area of its lower base is $225\pi$ sq cm and the area of its upper base is $25\pi$ sq cm, what is the altitude of the small cone that was cut off?
The altitude of the small cone that was cut off is 8 centimeters.
The altitude of the small cone that was cut off can be found using similar triangles.
Let's denote the altitude of the small cone as h.
We know that the ratio of the altitudes of two similar cones is equal to the ratio of the radii of their bases.
Therefore, we can set up the following proportion:
h / 24 = √(25π) / √(225π)
Simplifying this proportion, we get:
h / 24 = √25 / √225
Since √25 = 5 and √225 = 15, we can substitute these values into the proportion:
h / 24 = 5 / 15
Cross-multiplying, we get:
15h = 5 * 24
Simplifying further:
15h = 120
Dividing both sides by 15, we find that the altitude of the small cone, h, is:
h = 120 / 15 = 8 centimeters.
Therefore, the altitude of the small cone that was cut off is 8 centimeters.
To know more about cone refer here:
https://brainly.com/question/28202499
#SPJ11
(a) Show by definition that the interval (0,1) is not compact. That is, find an open cover of (0,1) that does not have a finite subcover. (b) We know the Heine-Borel Theorem holds for (Rn,d2 ). Is it true that for every metric space, every closed and bounded set is compact? Prove your result.
a. {Un} is an open cover of (0,1) that does not have a finite subcover, which shows that (0,1) is not compact.
b. No, it is not true that every closed and bounded set is compact in every metric space.
How to show that the interval (0,1) is not compacta. To show that (0,1) is not compact, we must find an open cover of (0,1) that does not have a finite subcover.
A typical example is the set of open intervals {Un} where Un = (1/n, 1 - 1/n) for n ≥ 2.
To show that this is an open cover of (0,1), we must show that every point in (0,1) is contained in at least one of the sets Un.
Let x be any point in (0,1). Then there exists some positive integer N such that N > 1/x.
It follows that x > 1/N, so x is greater than the left endpoint of the interval Un = (1/n, 1 - 1/n) for n ≥ N.
Since N > 1/x,
N ≤ xN < xN + 1 ≤ N + 1.
Hence, xN + 1 > 1, which implies that x < 1 - 1/N.
Therefore, x is less than the right endpoint of the interval Un = (1/n, 1 - 1/n) for n ≥ N.
Thus, x is contained in the set Un, and so {Un} is an open cover of (0,1).
To show that {Un} does not have a finite subcover.
Suppose for contradiction that {U1, U2, ..., Um} is a finite subcover of {Un}. Then there exists some positive integer N such that N > max{1, m, 1/ε}, where ε is the minimum distance between any two distinct points in {U1, U2, ..., Um}.
N exists because there are only finitely many sets in the subcover. For any n ≥ N, we have 1/n < ε/2, which implies that (1/n, 1 - 1/n) is contained in some Ui for i ≤ m. But then there is some x ∈ (0,ε/2) that is not covered by any of the sets {U1, U2, ..., Um}, which contradicts the assumption that {U1, U2, ..., Um} is a cover of (0,1).
Therefore, {Un} is an open cover of (0,1) that does not have a finite subcover, which shows that (0,1) is not compact.
b. No, it is not true that every closed and bounded set is compact in every metric space.
A counter example is the metric space (0,1) with the usual metric. The set [0,1] is closed and bounded in (0,1), but it is not compact because it is not complete.
In fact, [0,1] is not even compact in the larger metric space R with the usual metric, because it is not bounded above .
Learn more on metric space on https://brainly.com/question/32412222
#SPJ1
Show that e
n
=Ω(n
2
). b) [10 marks] Show that n
2
+n+logn=θ(n
2
). c) [10 marks] Let f(x)=
2x
2
+3
x
4
+2x
2
and g(x)=x
2
. Is f(x)=O(g(x)) ? Justify your answer. d) [10 marks] Let f(n) and g(n) be non-negative functions. Show that f(n)=θ(g(n)) if and only if g(n)=θ(f(n)).
a) To show that e^n = Ω(n^2), we need to demonstrate that there exist positive constants c and k such that for all sufficiently large values of n, e^n ≥ c * n^2.
We can rewrite the expression e^n as (e^(n/2))^2. Since e^(n/2) is an increasing function for n ≥ 0, we can set c = 1 and k = 1, and observe that for all n ≥ 0, e^(n/2) ≥ 1 * n^2.
Therefore, e^n = Ω(n^2).
b) To show that n^2 + n + log(n) = θ(n^2), we need to prove both Ω and O bounds.
For the Ω bound, we need to show that there exist positive constants c1 and k1 such that for all sufficiently large n, n^2 + n + log(n) ≥ c1 * n^2. Since n^2 is the dominant term, we can choose c1 = 1 and k1 = 1, and observe that for all n ≥ 1, n^2 + n + log(n) ≥ 1 * n^2.
For the O bound, we need to show that there exist positive constants c2 and k2 such that for all sufficiently large n, n^2 + n + log(n) ≤ c2 * n^2. Again, since n^2 is the dominant term, we can choose c2 = 2 and k2 = 1, and observe that for all n ≥ 1, n^2 + n + log(n) ≤ 2 * n^2.
Therefore, n^2 + n + log(n) = θ(n^2).
c) To determine if f(x) = O(g(x)), we need to show that there exist positive constants c and k such that for all sufficiently large values of x, f(x) ≤ c * g(x).
Given f(x) = 2x^2 + 3x^4 + 2x^2 and g(x) = x^2, we can see that for x ≥ 1, f(x) = 2x^2 + 3x^4 + 2x^2 ≤ 7x^4 ≤ 7x^4 = c * g(x), where c = 7 and k = 1.
Therefore, f(x) = O(g(x)).
d) To prove f(n) = θ(g(n)), we need to demonstrate that both f(n) = O(g(n)) and f(n) = Ω(g(n)).
For the O bound, we need to show that there exist positive constants c1 and k1 such that for all sufficiently large n, f(n) ≤ c1 * g(n).
For the Ω bound, we need to show that there exist positive constants c2 and k2 such that for all sufficiently large n, f(n) ≥ c2 * g(n).
By satisfying both conditions, we can conclude that f(n) = θ(g(n)).
However, since the provided functions f(n) and g(n) are not specified, it is not possible to determine the exact values of c1, k1, c2, and k2 or provide a complete proof without further information about the functions.
To learn more about positive constants click here : brainly.com/question/29002309
#SPJ11
This is the second part of a three-part problem. Consider the systom of differential equations
y
1
′
=y
1
+3y
21
y
2
′
=3y
1
+y
2
+
Verify that for any constants c
1
and c
2
, the functions
y
1
(t)=c
1
e
u
+c
2
e
−2t
y
2
(t)=c
1
c
u
−c
2
e
−2t
satisfy the system of differential equations. Enter c
1
as of and c
2
as c2. a. Find the value of each term in the equation y
1
′
=y
1
+3y
2
in terms of the variable f (Enter the terms in the order given.) b. Find the value of each term in the equation y
2
′
=3y
1
+y
2
in terms of the variable f (Enter the terms in the order given.)
The given system of differential equations is:
y1' = y1 + 3y2
y2' = 3y1 + y2
We need to verify that the functions:
y1(t) = c1e^(ut) + c2e^(-2t)
y2(t) = c1ue^(ut) - c2e^(-2t)
satisfy the system. In part (a), we find the value of each term in the equation y1' = y1 + 3y2 in terms of the variable f. In part (b), we find the value of each term in the equation y2' = 3y1 + y2 in terms of the variable f.
(a) To find the value of each term in y1' = y1 + 3y2, we differentiate y1(t) with respect to t. The derivative of c1e^(ut) is c1ue^(ut), and the derivative of c2e^(-2t) is -2c2e^(-2t). Thus, we have:
y1' = c1ue^(ut) - 2c2e^(-2t) + 3(c1ue^(ut) - c2e^(-2t))
Combining like terms, we get:
y1' = (2c1u + 3c1u)e^(ut) + (-2c2 - 3c2)e^(-2t)
(b) Similarly, we differentiate y2(t) with respect to t. The derivative of c1ue^(ut) is c1u^2e^(ut), and the derivative of c2e^(-2t) is -2c2e^(-2t). Thus, we have:
y2' = c1u^2e^(ut) - 2c2e^(-2t) + 3(c1e^(ut) + c2e^(-2t))
Combining like terms, we get:
y2' = (c1u^2 + 3c1)e^(ut) + (-2c2 + 3c2)e^(-2t)
Therefore, the value of each term in y1' = y1 + 3y2 is given by:
Term 1: (2c1u + 3c1)e^(ut)
Term 2: (-2c2 - 3c2)e^(-2t)
And the value of each term in y2' = 3y1 + y2 is given by:
Term 1: (c1u^2 + 3c1)e^(ut)
Term 2: (-2c2 + 3c2)e^(-2t)
These results verify that the functions y1(t) and y2(t) satisfy the given system of differential equations.
To learn more about derivative click here : brainly.com/question/32963989
#SPJ11
Find an equation for the line that passes through the given point and satisfies the given conditions. P=(8,8); perpendicular to v=4i+3j
4x+3y=25
4x+3y=56
3x−4y=−8
y−8=
4
5
(x−4)
The equation for the line that passes through the point P=(8,8) and is perpendicular to the vector v=4i+3j is 4x - 3y = 40.
To find the equation of a line perpendicular to a given vector, we can use the fact that the dot product of two perpendicular vectors is zero. The given vector v=4i+3j has components (4, 3), so the slope of the line perpendicular to it can be obtained by taking the negative reciprocal of the slope of v.
The slope of v is 3/4, so the slope of the perpendicular line is -4/3. We can use the point-slope form of a linear equation to find the equation of the line. Substituting the coordinates of the point P=(8,8) and the slope -4/3 into the point-slope form, we get:
y - 8 = (-4/3)(x - 8)
Simplifying the equation, we have:
3y - 24 = -4x + 32
Rearranging the terms, we obtain:
4x - 3y = 40
This is the equation for the line that passes through the point P=(8,8) and is perpendicular to the vector v=4i+3j.
Learn more about Equation of a line perpendicular to a given vector
brainly.com/question/32293743
#SPJ11
Let T:R
2
→R
2
be a linear transformation such that T(x
1
,x
2
)=(x
1
+x
2
,4x
1
+5x
2
). Find x such that T(x)= (3,8).
The vector x such that T(x) = (3, 8) is x = (7, -4).
To find the vector x such that T(x) = (3, 8), we need to solve the equation T(x) = (3, 8) using the given linear transformation T.
Let's write out the equation using the components of T(x):
T(x1, x2) = (x1 + x2, 4x1 + 5x2)
Setting this equal to (3, 8), we have:
(x1 + x2, 4x1 + 5x2) = (3, 8)
This gives us the following system of equations:
x1 + x2 = 3 ...(1)
4x1 + 5x2 = 8 ...(2)
We can solve this system of equations to find the values of x1 and x2.
Multiplying equation (1) by 4, we get:
4x1 + 4x2 = 12 ...(3)
Subtracting equation (3) from equation (2), we eliminate x1:
(4x1 + 5x2) - (4x1 + 4x2) = 8 - 12
x2 = -4
Substituting this value of x2 into equation (1), we can solve for x1:
x1 + (-4) = 3
x1 = 3 + 4
x1 = 7
Therefore, the vector x such that T(x) = (3, 8) is x = (7, -4).
To know more about vector click-
https://brainly.com/question/12949818
#SPJ11
Exercise Price Call Price Put Price
120 15.40 9.25
125 13.50 11.50
130 11.35 14.25
The underlying stock is currently price at $125.94
For each option strategy make sure to graph the position and label breakeven stock prices, maximum profit, and minimum profit.
Construct a butterfly spread using the three calls available.
The butterfly spread is constructed using the following three calls: - Buy 1 call with an exercise price of 120 at a price of $15.40.- Sell 2 calls with an exercise price of 125 at a price of $13.50 each.- Buy 1 call with an exercise price of 130 at a price of $11.35.
A butterfly spread involves buying one option with a lower exercise price, selling two options with a middle exercise price, and buying one option with a higher exercise price. In this case, we buy the call with an exercise price of 120, sell two calls with an exercise price of 125, and buy the call with an exercise price of 130.
To analyze the strategy, we need to consider the stock price at expiration. The breakeven stock prices are calculated by adding and subtracting the net debit or credit from the middle exercise price. In this case, the net debit is $1.60 ($15.40 – 2*$13.50 + $11.35), so the breakeven prices are $121.40 ($125 - $1.60) and $128.60 ($125 + $1.60).
The maximum profit of $1.60 occurs when the stock price is exactly at the middle exercise price of 125 at expiration. In this case, the two sold calls expire worthless, and the two bought calls have intrinsic value of $5 each. The minimum profit of -$1.40 occurs when the stock price is below $121.40 or above $128.60 at expiration. In this situation, all the options expire worthless, resulting in a loss of the net debit.
Learn more about Butterfly spread here: brainly.com/question/32201277
#SPJ11
Let h be a differentiable function defined on the interval [0,3] and assume that h(0)=1,h(1)=2, and h(3)=2, then (a) SHOW THAT there exists a point d∈[0,3] such that h(d)=d (5 points) (b) SHOW THAT there exists c∈[0,3] such that h
′
(c)=
3
1
. (5 points) (c) SHOW THAT there exists e∈[0,3] such that h
′
(e)=
4
1
. (5 points)
Since h(3)-h(0) is equal to 2-1=1, and 3-0=3, the average rate of change is 41. Therefore, there exists e∈[0,3] such that h'(e)=41.
To show that there exists a point d∈[0,3] such that h(d)=d, we can use the Intermediate Value Theorem.
Since h(0)=1 and h(3)=2, and h(x) is continuous on [0,3], there must exist a value d∈[0,3] such that h(d)=d.
However, a detailed calculation is not necessary for this proof.
To show that there exists c∈[0,3] such that h'(c)=31, we can use the Mean Value Theorem.
Since h(x) is differentiable on [0,3], there must exist a value c∈(0,3) such that h'(c) is equal to the average rate of change of h(x) over [0,3].
Since h(3)-h(0) is equal to 2-1=1, and 3-0=3, the average rate of change is 31.
Therefore, there exists c∈[0,3] such that h'(c)=31.
To show that there exists e∈[0,3] such that h'(e)=41, we can use the Mean Value Theorem again.
Since h(x) is differentiable on [0,3], there must exist a value e∈(0,3) such that h'(e) is equal to the average rate of change of h(x) over [0,3].
Since h(3)-h(0) is equal to 2-1=1, and 3-0=3, the average rate of change is 41.
Therefore, there exists e∈[0,3] such that h'(e)=41.
To know more about average visit:
https://brainly.com/question/24057012
#SPJ11
Let X={2,3,4} and let P(X) be the power set of X. The subset relation is defined on P(X) as follows: For all S,T∈P(X), SUT ⇔S⊆T. Draw the Hasse Diagram for this relation. Make sure to show all intermediate steps. Scan or photograph your answer and upload the file.
To draw the Hasse Diagram for the subset relation on the power set P(X), we start by listing all the elements of P(X). In this case, P(X) consists of the empty set {}, the singleton sets {2}, {3}, {4}, and the sets {2,3}, {3, 4}, {2, 4}, {2, 3, 4}.
Next, we draw a vertex for each set in P(X). We then draw a directed line from set A to set B if A is a subset of B. In other words, if A ⊆ B.
Learn more about Hasse Diagram from the given link:
https://brainly.com/question/33109915
#SPJ11
Patricia throws a football in the air and the path of the football in the air is given by the following function: h(x)=−0.018(x−20)
2
+8 where h(x) is the number of feet the football is above the ground, and x is the distance the football has traveled. (a) What transformations are taking place from the parent function f(x)=x
2
? Describe all that are happening. (b) Determine the x intercept of this function. Interpret your solution in a complete sentence with units.
(a) The given function, h(x) = -0.018(x - 20)^2 + 8, is a transformation of the parent function f(x) = x^2. Let's describe the transformations that are taking place:
1. Horizontal translation: The function h(x) is shifted 20 units to the right compared to the parent function f(x). This is because the term (x - 20) in the equation represents the distance the football has traveled.
2. Vertical translation: The function h(x) is shifted 8 units upwards compared to the parent function f(x). This is because the constant term 8 in the equation represents the initial height of the football above the ground.
3. Vertical compression: The coefficient -0.018 in front of the squared term (x - 20)^2 causes the function h(x) to be compressed vertically compared to the parent function f(x). This means that the graph of h(x) is narrower than the graph of f(x).
(b) To find the x-intercept of h(x), we need to find the value of x when h(x) = 0. Setting h(x) = 0 and solving for x, we get:
0 = -0.018(x - 20)^2 + 8
Rearranging the equation, we have:
0.018(x - 20)^2 = 8
Dividing both sides by 0.018, we get:
(x - 20)^2 = 8/0.018
Taking the square root of both sides, we have:
x - 20 = ±√(8/0.018)
Simplifying the expression under the square root, we get:
x - 20 = ± 33.6
Adding 20 to both sides, we get:
x = 20 ± 33.6
Therefore, the x-intercepts of the function h(x) are x = 20 + 33.6 and x = 20 - 33.6.
Interpreting the solution in a complete sentence with units: The x-intercepts of the function h(x) represent the distances the football has traveled when it is at ground level (h(x) = 0).
Learn more about Horizontal translation from the given link:
https://brainly.com/question/31938343
#SPJ11
Divided Difference) to construct the interpolating polynomial of degree three for the following data: f(1)=1.2,f(2)=1.6,f(4)=2, and f(5)=2.6, then f(1.8)≈
To construct the interpolating polynomial of degree three using the Divided Difference method, we can use the following steps:
1. Begin by creating a divided difference table. The first column will be the x-values (1, 2, 4, 5), and the second column will be the corresponding f(x) values (1.2, 1.6, 2, 2.6).
2. Calculate the divided differences. The divided differences are the differences between consecutive f(x) values. In the second column, the first divided difference is (1.6 - 1.2) = 0.4, the second divided difference is (2 - 1.6) = 0.4, and the third divided difference is (2.6 - 2) = 0.6.
3. Continue calculating the divided differences until you have a single value in the last column. In this case, the divided differences are constant, so the third column will be all 0.4.
4. Use the divided differences to construct the interpolating polynomial. The formula for the interpolating polynomial is:
f(x) = f[x₀] + (x - x₀)f[x₀, x₁] + (x - x₀)(x - x₁)f[x₀, x₁, x₂] + ...
Substituting the values, we get:
f(x) = 1.2 + (x - 1)(0.4) + (x - 1)(x - 2)(0.4) + (x - 1)(x - 2)(x - 4)(0.4)
5. Finally, substitute x = 1.8 into the interpolating polynomial to approximate f(1.8):
f(1.8) = 1.2 + (1.8 - 1)(0.4) + (1.8 - 1)(1.8 - 2)(0.4) + (1.8 - 1)(1.8 - 2)(1.8 - 4)(0.4)
Calculating this expression will give you the approximation for f(1.8) in terms of the given data points.
Learn more about interpolating polynomial here:
https://brainly.com/question/31494775?referrer=searchResults
#SPJ11
Assume that \( f \) and \( g \) are differentiable functions. Let \( h(x)=\frac{4 f(x)}{g(x)+2} \), find \( h^{\prime}(x) \) \( h^{\prime}(x)= \) help (formulas)
The derivative of \( h(x) = \frac{4f(x)}{g(x) + 2} \) is given by \( h'(x) = \frac{4f'(x)g(x) + 8f'(x) - 4f(x)g'(x)}{(g(x) + 2)^2} \), which is obtained using the quotient rule for differentiation.
To find the derivative of \( h(x) \), we can use the quotient rule, which states that if \( f(x) \) and \( g(x) \) are differentiable functions, then the derivative of their quotient \( \frac{f(x)}{g(x)} \) is given by:
\[ \left(\frac{f(x)}{g(x)}\right)' = \frac{f'(x)g(x) - f(x)g'(x)}{(g(x))^2} \]
In this case, \( h(x) = \frac{4f(x)}{g(x) + 2} \). Let's find its derivative \( h'(x) \) using the quotient rule:
\[ h'(x) = \frac{\left(4f(x)\right)'\left(g(x) + 2\right) - 4f(x)\left(g(x) + 2\right)'}{(g(x) + 2)^2} \]
To differentiate \( 4f(x) \), we can simply multiply the derivative of \( f(x) \) by 4. Similarly, to differentiate \( g(x) + 2 \), we differentiate \( g(x) \) and the constant term 2. Let's rewrite the expression:
\[ h'(x) = \frac{4f'(x)\left(g(x) + 2\right) - 4f(x)g'(x)}{(g(x) + 2)^2} \]
Simplifying further, we have:
\[ h'(x) = \frac{4f'(x)g(x) + 8f'(x) - 4f(x)g'(x)}{(g(x) + 2)^2} \]
Therefore, the derivative \( h'(x) \) of \( h(x) = \frac{4f(x)}{g(x) + 2} \) is given by:
\[ h'(x) = \frac{4f'(x)g(x) + 8f'(x) - 4f(x)g'(x)}{(g(x) + 2)^2} \]
To learn more about differentiation Click Here: brainly.com/question/31539041
#SPJ11
Eastons for Computing Control Chant Limits.(3 sigma) for this problem. this, Mogul executives take samples of eight Eagletrons at a time. For each sample, they determine the avorage maximum speed and the range of the maximum speeds within the sample. They repeat this with 35 samples to obtain 35 sample means and 35 ranges. They tind that the average sample mean is 103.50 miles per hour, and the average rage is 3.00 milies per heur. Uiing these results, the executives decide to establish an R-chart. They would like this chart to be establiahed so that when it shows that the range of a sample is rot wathin the contra litris. thire is cily approximately a 0.0027 probability that this is due to natural variation. The control limits for the chart based on the above requirement for the given information are: U
CL
R
= miles per hour (round the reponse to three decimal places).
The control limits for the R-chart, based on the given information, are UCL = 6.846 miles per hour and CL = 3.00 miles per hour.
To establish the control limits for the R-chart, we need to calculate the upper control limit (UCL) and the centerline (CL) for the range of maximum speeds.
Given:
Average sample mean (x) = 103.50 miles per hour
Average range (R) = 3.00 miles per hour
To calculate the control limits, we can use the following formulas:
UCL = D4 × R
CL = R
The constant D4 depends on the sample size. Since each sample contains 8 Eagletrons, we can refer to a statistical table to find the appropriate value of D4 for n = 8. For a sample size of 8, the value of D4 is approximately 2.282.
Plugging in the values, we can calculate the control limits:
UCL = D4 × R = 2.282 × 3.00 = 6.846 miles per hour (rounded to three decimal places)
CL = R = 3.00 miles per hour
Therefore, the control limits for the R-chart, based on the given information, are:
Upper Control Limit (UCL) = 6.846 miles per hour
Centerline (CL) = 3.00 miles per hour.
Learn more about Upper control limit here: brainly.com/question/14128732
#SPJ11
The gas tank of mr. sanchez's car is completely empty he feels it up with gasoline that cost three dollars per gallon the total cost to build a tank comes to $42 which equation can be used to find the number of gallons of gasoline that mr. stan chess is casting calls
The number of gallons of gasoline that Mr. Sanchez is purchasing is 14.
The equation 3x = 42 represents the relationship between the cost of gasoline and the number of gallons purchased.
In the equation, "x" represents the number of gallons of gasoline that Mr. Sanchez is purchasing. The cost per gallon of gasoline is $3, so when multiplied by the number of gallons (x), it gives the total cost of the gasoline.
On the right side of the equation, 42 represents the total cost of filling up the gas tank. It is given that the total cost comes to $42.
By solving the equation 3x = 42 for "x", we can find the value of "x", which represents the number of gallons of gasoline that Mr. Sanchez is purchasing. Dividing both sides of the equation by 3:
3x/3 = 42/3
x = 14
To know more about equation refer here
https://brainly.com/question/29514785#
#SPJ11
Maximally consistent sets Let Γ be a maximally consistent set of well formed formulas. Show that for all φ,ψ∈ WFF at least one of the following two statements is true: - Γ⊢(φ→ψ) - Γ⊢((¬φ)→ψ) Hint: Use parts of the last question on Assignment 2. (30 marks)
At least one of the following two statements is true:
Γ ⊢ (φ → ψ)
Γ ⊢ (¬φ → ψ).
This completes the proof.
To prove that for all φ,ψ∈WFF (well-formed formulas), at least one of the following two statements is true:
Γ ⊢ (φ → ψ)
Γ ⊢ (¬φ → ψ)
we can proceed with a proof by contradiction.
Assume that both statements are false, i.e., neither Γ ⊢ (φ → ψ) nor Γ ⊢ (¬φ → ψ) hold.
Since Γ is a maximally consistent set of well-formed formulas, it means that for any formula α, either α or its negation ¬α is in Γ, but not both.
Consider the formula φ. Since Γ ⊬ (φ → ψ), it means that ¬(φ → ψ) is in Γ.
By the definition of the implication, ¬(φ → ψ) is equivalent to φ ∧ ¬ψ.
Thus, φ ∧ ¬ψ is in Γ.
Similarly, consider the formula ¬φ.
Since Γ ⊬ (¬φ → ψ), it means that ¬(¬φ → ψ) is in Γ.
By the definition of the implication, ¬(¬φ → ψ) is equivalent to ¬φ ∧ ¬ψ.
Thus, ¬φ ∧ ¬ψ is in Γ.
Now, let's consider the conjunction of φ ∧ ¬ψ and ¬φ ∧ ¬ψ, which are both in Γ:
(φ ∧ ¬ψ) ∧ (¬φ ∧ ¬ψ).
By the distributive law of conjunction over conjunction, we have:
(φ ∧ ¬ψ) ∧ (¬φ ∧ ¬ψ) ≡ (φ ∧ (¬φ ∧ ¬ψ)) ∧ ¬ψ ≡ ⊥ ∧ ¬ψ ≡ ⊥.
Here, ⊥ represents a contradiction or a false statement.
Since we have arrived at a contradiction, our assumption that both statements are false must be incorrect.
Therefore, at least one of the following two statements is true:
Γ ⊢ (φ → ψ)
Γ ⊢ (¬φ → ψ).
This completes the proof.
Learn more about well-formed formulas from this link:
https://brainly.com/question/31456588
#SPJ11
simplify the following expression in terms of fractional exponents and write it in the form 10 to the power of a x to the power of b. cube root of 10 to the power of 4 x end root
The simplified expression in terms of fractional exponents and in the form 10 to the power of a times x to the power of b is 10^((4/3)x).
To simplify the given expression, we can use the properties of exponents and fractional exponents. Let's break down the expression step by step.
The given expression is the cube root of 10 to the power of (4x). We can rewrite this as:
(10^(4x))^(1/3)
Using the property (a^m)^n = a^(m*n), we can simplify further:
10^((4x)*(1/3))
Multiplying 4x and 1/3 gives us:
10^(4x/3)
Now, let's write this in the form 10 to the power of a times x to the power of b. To do this, we need to express 4x/3 as a sum of two terms, one involving a and the other involving b.
We can rewrite 4x/3 as (4/3)x. Therefore, our expression becomes:
10^((4/3)x)
To know more about fractional exponents refer here:
https://brainly.com/question/30069266#
#SPJ11
Mika hiked 4 mi in 77 min. use a proportion to find how many miles she will hike in 2 he if she hikes at the same rate.
She will hike 6.23 miles.
To find out how many miles Mika will hike in 2 hours if she hikes at the same rate, we can set up a proportion using the information given. It will compare the distance Mika hiked in 77 minutes to the time it took her to hike that distance, with the distance she will hike in 2 hours to the time it will take her to hike that distance.
We can set up the proportion as follows:
Suppose she hikes for x miles in 2 hrs (120 minutes)
Then, 4 miles / 77 minutes = x miles / 120 minutes
To solve for x, we can cross-multiply and then divide:
4 * 120 = 77 * x
480 = 77 * x
Dividing both sides by 77, we get:
480 / 77 = x
x ≈ 6.23
Therefore, Mika will hike approximately 6.23 miles in 2 hours if she hikes at the same rate.
Read more about Proportions: https://brainly.com/question/1496357
#SPJ11
Suppose that X is a CW complex with exactly one 0 -cell ∗∈X. Show that if Xn−1↪Xn is nullhomotopic rel ∗, for all n≥0, then X is contractible.
X is contractible.
To show that X is contractible, we need to show that there exists a homotopy between the identity map on X and a constant map.
Given that X_n-1↪X_n is nullhomotopic rel ∗ for all n≥0, we can construct a homotopy by induction.
Let's start with n = 0. Since X has only one 0-cell ∗, X_0 consists only of ∗. Thus, the inclusion map X_0↪X_0 is the identity map, which is already nullhomotopic rel ∗.
Now, assume that X_k-1↪X_k is nullhomotopic rel ∗ for some k≥1. We want to show that X_k is also nullhomotopic rel ∗.
Since X_k-1↪X_k is nullhomotopic, there exists a homotopy H: X_k × I → X_k such that H(x, 0) = x for all x in X_k and H(x, 1) = ∗ for all x in X_k-1.
Consider the inclusion map i: X_k-1↪Xk. We can define a new homotopy G: X_k × I → X_k by G(x, t) = H(i(x), t), where i(x) is the inclusion of x in X_k.
G satisfies G(x, 0) = H(i(x), 0) = x for all x in X_k, and G(x, 1) = H(i(x), 1) = ∗ for all x in X_k-1.
Therefore, we have constructed a homotopy between the inclusion map X_k-1↪X_k and the constant map ∗.
By induction, we can repeat this process for all n≥0. Thus, X is contractible.
Learn more about homotopy from the given link:
https://brainly.com/question/31120410
#SPJ11
Suppose the LU decomposition is already computed as A=
⎣
⎡
1
−2
3
0
1
5
0
0
1
⎦
⎤
⎣
⎡
4
0
0
1
2
0
−1
1
3
⎦
⎤
, then (for the row vectors, input numbers separated only by one comma) the 1st row of A is [ the 2 nd row of A is [ the 3rd row of A is [ To solve Ax=b, you need to utilize the LU factors and solve LUx=b in two steps: Step 1: Solve Ly=b for y; Step 2: Solve Ux=y for x. Now carry out the above two steps to solve Ax=b for the b given below
⎣
⎡
1
−2
3
0
1
5
0
0
1
⎦
⎤
⎣
⎡
4
0
0
1
2
0
−1
1
3
⎦
⎤
⎣
⎡
x
1
x
2
x
3
⎦
⎤
=
⎣
⎡
−6
13
−16
⎦
⎤
Step 1: Solve the lower triangular equation forwardly:
⎣
⎡
1
−2
3
0
1
5
0
0
1
⎦
⎤
⎣
⎡
y
1
y
2
y
3
⎦
⎤
=
⎣
⎡
−6
13
−16
⎦
⎤
y
1
=,y2=,y3= Step 2: Solve the upper triangular equation backwardly:
⎣
⎡
4
0
0
1
2
0
−1
1
3
⎦
⎤
⎣
⎡
x
1
x
2
x
3
⎦
⎤
=
⎣
⎡
y
1
y
2
y
3
⎦
⎤
x3=
If you are asked to solve Ax=−3+b for the same A,b as given above. Then, without doing any more equation solving. you know that the new solution should be the yector (input numbers separated only by a comma)
The new solution vector would be [-1; -10; -19].
To solve Ax=b using LU decomposition, we follow two steps:
Step 1: Solve Ly=b for y
Given the lower triangular equation:
[1, -2, 3; 0, 1, 5; 0, 0, 1] [y1; y2; y3]
= [-6; 13; -16]
Solving this equation, we find:
y1 = -6
y2 = 13
y3 = -16
Step 2: Solve Ux=y for x
Given the upper triangular equation:
[4, 0, 0; 1, 2, 0; -1, 1, 3] [x1; x2; x3]
= [y1; y2; y3]
Solving this equation, we find:
x3 = -16
x2 = -7
x1 = 2
If we are asked to solve Ax = -3+b for the same A and b, we can use the solution obtained from the previous steps.
The new solution vector would be
[y1; y2; y3]
= [-3+2; -3-7; -3-16]
= [-1; -10; -19].
To know more about vector visit:
https://brainly.com/question/24256726
#SPJ11
Ruby bought a $190,000 house, paying 18% down, and financing the rest at 6.2% interest for 30 years. Her monthly payments are $954.23. How much will he really pay for her $190,000 house? Ruby will pay a total of $ for the house.
a nurse mixes 60 cc of a 70% saline solution with 10% saline solution to produce a 35% saline solution. how much of the 10% solution should he use?
To determine how much of the 10% saline solution the nurse should use, let's denote the amount of the 10% solution as x cc.
Given:
Volume of the 70% saline solution: 60 cc
Concentration of the 70% saline solution: 70%
Concentration of the 10% saline solution: 10%
Desired concentration of the resulting solution: 35%
To find the amount of the 10% solution, we can set up an equation based on the principle of the concentration of solutions:
(0.7 * 60 + 0.1 * x) / (60 + x) = 0.35
In this equation, we multiply the concentration of each solution by its corresponding volume and divide by the total volume of the resulting solution.
Simplifying the equation, we get:
(42 + 0.1x) / (60 + x) = 0.35
Cross-multiplying, we have:
42 + 0.1x = 0.35 * (60 + x)
Expanding the right side of the equation, we get:
42 + 0.1x = 21 + 0.35x
Moving all the x terms to one side, we have:
0.35x - 0.1x = 42 - 21
0.25x = 21
Dividing both sides by 0.25, we get:
x = 21 / 0.25
x = 84
Therefore, the nurse should use 84 cc of the 10% saline solution to mix with the 60 cc of the 70% saline solution to produce a 35% saline solution.
melissa believes that the aa similarity theorem can prove that the triangles are similar. which fact would be necessary in the proof? ââ€""³abc is an acute triangle. ââ€""³trs is larger than ââ€""³abc. the sum of the measures of the interior angles of a triangle is 180°. the sum of the side lengths of two sides of a triangle is greater than the third side length.
To prove that two triangles are similar using the AA (angle-angle) similarity theorem, it is necessary to show that the two triangles have two corresponding angles that are congruent.
In this case, if Melissa believes that the AA similarity theorem can prove that the triangles are similar, she needs to ensure that the fact "ââ€""³abc is an acute triangle" is true. This fact is necessary because the AA similarity theorem states that if two angles of one triangle are congruent to two angles of another triangle, and the triangles are both acute, then the two triangles are similar.
Therefore, the fact "ââ€""³abc is an acute triangle" is necessary in the proof.
In conclusion, to use the AA similarity theorem to prove that two triangles are similar, it is necessary to show that the triangles are both acute and have two corresponding congruent angles.
To know more about triangles visit
https://brainly.com/question/2773823
#SPJ11
State the converse, inverse and contrapositive of each of the following statements: i. I will buy a house if I can save more than $1000 per month. ii. ¬p∧q→p∨q. b) Determine whether the following proposition is a tautology: (¬p∨¬(r→q))↔(p→(¬q∧r)) Q-2: [4+4 marks] a) Find the set of solutions of each of the linear congruence: i. x≡3(mod5). ii. 2x≡5(mod9). b) Suppose f:R→Z, where f(x)=⌈2x−1⌉. i. If A={x:1≤x≤4}, find f(A). ii. If B={3,4,5,6,7}, find f(B). iii. If C={−9,−8}, find f
−1
(C). iv. Is f one-to-one? Explain.
i. The converse of the statement "I will buy a house if I can save more than $1000 per month" is "If I buy a house, then I can save more than $1000 per month." The inverse is "I will not buy a house if I cannot save more than $1000 per month." The contrapositive is "If I cannot save more than $1000 per month, then I will not buy a house."
ii. The converse of the statement "¬p∧q→p∨q" is "If p∨q, then ¬p∧q." The inverse is "If ¬p∧q, then ¬(¬p∧q)." The contrapositive is "If ¬p∨¬q, then ¬(¬p∧q)."
b) To determine whether the proposition "(¬p∨¬(r→q))↔(p→(¬q∧r))" is a tautology, we can use truth tables to evaluate the statement for all possible truth values of p, q, r. If the statement is true for all possible combinations, then it is a tautology.
a) i. The set of solutions for the linear congruence x≡3(mod5) is {3, 8, 13, 18, ...}. ii. The set of solutions for the linear congruence 2x≡5(mod9) is {7, 16, 25, ...}.
b) i. f(A) = {1, 3, 5, 7}.
ii. f(B) = {5, 7, 9, 11, 13}.
iii. f^(-1)(C) = {-4, -5}.
iv. No, f is not one-to-one because multiple inputs from the domain can map to the same output in the codomain.
Learn more about contrapositive
https://brainly.com/question/12151500
#SPJ11
Solve the equation for X, given that A=[
1
3
2
4
] and B=[
−1
1
0
1
]. x−4A+5B=0 x=[1
The answer of the given equation is , the solution for x is x = [0, 0; 16, 42].
To solve the equation x - 4A + 5B = 0, we can substitute the given values for A and B and solve for x.
Given that A = [1, 3; 2, 4] and B = [-1, 1; 0, 1], we can substitute these values into the equation:
x - 4A + 5B = 0
x - 4[1, 3; 2, 4] + 5[-1, 1; 0, 1] = 0
x - [4, 12; 8, 16] + [-5, 5; 0, 5] = 0
x - [4-5, 12+5; 8, 16+5] = 0
x - [-1, 17; 8, 21] = 0
To simplify further, we can subtract the matrices:
x - [-1, 17; 8, 21] = 0
x + [1, -17; -8, -21] = 0
By adding the matrices, we get:
x = [-1+1, 17-17; 8+8, 21+21]
x = [0, 0; 16, 42]
Thus, the solution for x is x = [0, 0; 16, 42].
To know more about Equation visit:
https://brainly.com/question/29538993
#SPJ11
1a.construct triangle ABC in which|AB|=7cm,AC=9.5cm and ABC=120°.measure|BC|
Using the given information, construct triangle ABC with |AB| = 7cm, |AC| = 9.5cm, and ∠ABC = 120°. Measure |BC| using a ruler or measuring tape on the constructed triangle.
To construct triangle ABC with the given conditions, follow these steps:
Draw a line segment AB of length 7 cm.
At point A, draw a ray in the direction of AC.
With a compass, set the radius to 9.5 cm.
Place the compass tip at point A and draw an arc that intersects the ray from step 2. Label this point of intersection as C.
Draw a line segment BC connecting points B and C.
Now, triangle ABC is constructed with side lengths |AB| = 7 cm, |AC| = 9.5 cm, and angle ABC = 120°.
To measure |BC|, you can use a ruler or measuring tape to determine the length of the line segment BC directly on the constructed triangle.
for such more question on triangle
https://brainly.com/question/17335144
#SPJ8
If
PQ
⊥
QR
and T is on
PQ
so that
PQ
≅
TQ
, then △PQR≅△TQR by a. ASA Congruence b. SSS Congruence c. HL Congruence d. LL Congruence e. △PQR≅△TQR are not necessarily congruent 10. If △DEF is isosceles and
ET
bisects ∠DEF, then a.
ET
⊥
DF
b.
ET
bisects
DF
c. Neither
ET
⊥
DF
nor
ET
bisects
DF
is necessarily true d. Both
ET
⊥
DF
nor
ET
bisects
DF
are true e. △DEF≅△FET 11. The following can be used to tell if two lines are parallel. a. If a transversal forms a pair of congruent alternate interior angles b. If a transversal forms a pair of complimentary interior angles on the same side of the transversal. c. If a transversal forms a pair of congruent corresponding angles with the transversal d. Both a and c are correct e. a, b, and c are correct 12. If point Q is between P and R on
PR
and T is not on
PR
, then a. ∠PQT≅∠PTQ b. m∠TQR=m∠QTP+m∠PTQ c. m∠TQR=90
∘
d. m∠TQR=m∠QPT+m∠PQT e. None of the above
10. The correct answer is c. Neither ET ⊥ DF nor ET bisects DF is necessarily true. Just because ET bisects ∠DEF in an isosceles triangle △DEF, it does not mean that ET is perpendicular to DF. The perpendicularity of ET and DF depends on the specific angles of the triangle.
11. The correct answer is e. a, b, and c are correct. The three conditions listed in options a, b, and c can be used to determine if two lines are parallel. If a transversal forms a pair of congruent alternate interior angles (option a) or a pair of congruent corresponding angles with the transversal (option c), then the lines are parallel. Additionally, if a transversal forms a pair of complimentary interior angles on the same side of the transversal (option b), then the lines are also parallel.
12. The correct answer is d. m∠TQR = m∠QPT + m∠PQT. According to the Angle Addition Postulate, the measure of an angle formed by two adjacent angles is equal to the sum of their measures. In this case, m∠TQR is equal to the sum of m∠QPT and m∠PQT.
Learn more about isosceles triangle :
https://brainly.com/question/29774496
#SPJ11
How many rows appear in a truth table for the compound proposition? (q→¬p)∨(¬p→¬q) a) 4 b) 8 c) 16 d) 32 e) 64 8. R=[x≡y↔∃k∈Z∋y−x=3k] relation is given. Which of the followings are true.? I. R is an equivalence relation. II.
2
={…−1,2,5…} III. {
0
,
1
,
2
} is not a partition of set of integers. a)I, b)I, II c)III d)I, III e)I, II, III 9.I. ∀n(n
2
≥n) II. ∃n((n
2
=2) III. ∃n((n
2
<0)
(a) The truth table for the compound proposition (q→¬p)∨(¬p→¬q) has 4 rows.
(b) Statements I, II, and III are all true.
(a) To determine the number of rows in the truth table for the given compound proposition, we need to consider all possible combinations of truth values for the variables p and q. Since each variable can take two truth values (true or false), there will be 2^2 = 4 rows in the truth table.
Now let's evaluate the given statements:
(b) Statement I: R is an equivalence relation.
To determine if R is an equivalence relation, we need to check if it satisfies three properties: reflexivity, symmetry, and transitivity.
Reflexivity: For every element x in the set, (x, x) must be in R. In this case, if we substitute x = y, the equation y - y = 3k simplifies to 0 = 3k, which holds true. So, R satisfies reflexivity.
Symmetry: For every pair (x, y) in R, (y, x) must also be in R. The given relation R is defined as x ≡ y ↔ ∃k∈Z∋y−x=3k. Since the equivalence relation is defined by ↔, which implies both directions, R satisfies symmetry.
Transitivity: For every three elements x, y, and z in R, if (x, y) and (y, z) are in R, then (x, z) must also be in R. To test this, we can substitute the given equations and verify that if y - x = 3k_1 and z - y = 3k_2, then z - x = (z - y) + (y - x) = 3k_2 + 3k_1 = 3(k_1 + k_2), which satisfies the condition. Hence, R satisfies transitivity. Therefore, statement I is true.
Statement II: 2 = {…, -1, 2, 5, …}
The set 2 represents the integers that can be expressed in the form 2k, where k is an integer. However, the set mentioned in the statement includes additional elements like -1 and 5, which do not fit this pattern. Hence, statement II is false.
Statement III: {0, 1, 2} is not a partition of the set of integers.
To be a partition of a set, the subsets should be non-empty, pairwise disjoint, and their union should cover the entire set. In this case, the set {0, 1, 2} does not cover the entire set of integers. Therefore, statement III is true.
In conclusion, the truth table for the compound proposition has 4 rows, and statements I and III are true, while statement II is false.
Learn more about Compound proposition
brainly.com/question/17406951
#SPJ11
If A and B are two events such that P(A)=0.4,P(B)=0.5 and P(B\A)=0.3. Find P(Aˉ∩Bˉ).
The probability is P([tex]\overline A[/tex] ∩ [tex]\overline B[/tex]) is equal to 0.7.
To find P([tex]\overline A[/tex] ∩ [tex]\overline B[/tex]), we can use the complement rule and the conditional probability formula.
First, we know that P(B|A) = 1 - P([tex]\overline B[/tex]|A), where [tex]\overline B[/tex] represents the complement of event B and | represents conditional probability.
Using this relationship, we can rewrite P([tex]\overline B[/tex]|A) as 1 - P(B|A) = 1 - 0.3 = 0.7.
Next, we can use the formula for the intersection of two events to calculate P([tex]\overline A[/tex] ∩ [tex]\overline B[/tex]):
P([tex]\overline A[/tex] ∩ [tex]\overline B[/tex]) = 1 - P(A ∪ B)
Since A and B are not mutually exclusive, we need to use the inclusion-exclusion principle to calculate P(A ∪ B):
P(A ∪ B) = P(A) + P(B) - P(A ∩ B)
Since P(A) = 0.4, P(B) = 0.5, and P(B|A) = 0.3, we can rearrange the equation to find P(A ∩ B):
P(A ∩ B) = P(A) + P(B) - P(B|A)
P(A ∩ B) = 0.4 + 0.5 - 0.3
P(A ∩ B) = 0.6
Now, we can calculate P([tex]\overline A[/tex] ∩ [tex]\overline B[/tex]) using the formula mentioned earlier:
P([tex]\overline A[/tex] ∩ [tex]\overline B[/tex]) = 1 - P(A ∪ B)
P([tex]\overline A[/tex] ∩ [tex]\overline B[/tex]) = 1 - (P(A) + P(B) - P(A ∩ B))
P([tex]\overline A[/tex] ∩ [tex]\overline B[/tex]) = 1 - (0.4 + 0.5 - 0.6)
P([tex]\overline A[/tex] ∩ [tex]\overline B[/tex]) = 1 - 0.3
P([tex]\overline A[/tex] ∩ [tex]\overline B[/tex]) = 0.7
Therefore, P([tex]\overline A[/tex] ∩ [tex]\overline B[/tex]) is equal to 0.7.
To know more about probability:
https://brainly.com/question/31828911
#SPJ4
