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  1. Asked: February 2, 2024In:

    Are the events independent?

    acedstud
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    acedstud

    For event G (getting a total of 3 with two dice), there are two possible outcomes: (1, 2) and (2, 1). So, the probability of event Q is $\frac{2}{36}=\frac{1}{18}$. For event H (getting one odd and one even number), there are $2\times3\times3 = 18$ possible outcomes: (1, 2), (1, 4), (1, 6), (3, 2),Read more

    For event G (getting a total of 3 with two dice), there are two possible outcomes: (1, 2) and (2, 1). So, the probability of event Q is \frac{2}{36}=\frac{1}{18}.

    For event H (getting one odd and one even number), there are 2\times3\times3 = 18 possible outcomes: (1, 2), (1, 4), (1, 6), (3, 2), (3, 4), (3, 6), (5, 2), (5, 4), (5, 6), (2, 1), (2, 3), (2, 5), (4, 1), (4, 3), (4, 5), (6, 1), (6, 3), and (6, 5). So, the probability of event S is \frac{18}{36}=\frac{1}{2}.

    Now, consider P(G and H), the probability of both events happening. Two outcomes satisfy both events: (1, 2) and (2, 1). So, P(G and H) =\frac{2}{36}=\frac{1}{18}.

    Using the formula for independence, P(G and H) =P(G)\times P(H)=\frac{1}{18}\times\frac{1}{2}=\frac{1}{36}.

    Since, the probability of both events happening is not equal to the product of the probabilities of the individual events, we conclude that events G and H are not independent.

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  2. Asked: January 10, 2021In:

    Find \dfrac{dy}{dx}, if 3xy+x^{2}=7y

    acedstud
    Best Answer
    acedstud

    To find $\frac{dy}{dx}$ given $3xy + x^2 = 7y$, you can use implicit differentiation. The equation is not given in the form $y = f(x)$, so we will differentiate both sides of the equation with respect to $x$ using implicit differentiation. Start by differentiating both sides of the equation: $\frac{Read more

    To find \frac{dy}{dx} given 3xy + x^2 = 7y, you can use implicit differentiation. The equation is not given in the form y = f(x), so we will differentiate both sides of the equation with respect to x using implicit differentiation.

    Start by differentiating both sides of the equation:

    \frac{d}{dx}(3xy + x^2) = \frac{d}{dx}(7y).

    Now, differentiate each term on the left side of the equation separately:

    1. \frac{d}{dx}(3xy) = 3x\frac{dy}{dx} + 3y.
    2. \frac{d}{dx}(x^2) = 2x.

    Now, differentiate the right side of the equation:

    \frac{d}{dx}(7y) = 7\frac{dy}{dx}.

    Now, you can rewrite the equation with these derivatives:

    3x\frac{dy}{dx} + 3y + 2x = 7\frac{dy}{dx}.

    Now, isolate the term with \frac{dy}{dx}:

    3x\frac{dy}{dx} - 7\frac{dy}{dx} = -3y - 2x.

    Factor out \frac{dy}{dx}:

    (3x - 7)\frac{dy}{dx} = -3y - 2x.

    Now, solve for \frac{dy}{dx} by dividing both sides by (3x - 7):

    \frac{dy}{dx} = \frac{-3y - 2x}{3x - 7}.

    So, \frac{dy}{dx} = \frac{-3y - 2x}{3x - 7} is the derivative with respect to x of the equation 3xy + x^2 = 7y.

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  3. Asked: January 10, 2021In:

    Find \dfrac{dy}{dx}, if xy+x^{2}=3

    acedstud
    Best Answer
    acedstud

    To find $\frac{dy}{dx}$ given $xy + x^2 = 3$, you can use implicit differentiation. The equation is not given in the form $y = f(x)$, so we will differentiate both sides of the equation with respect to $x$ using implicit differentiation. Start by differentiating both sides of the equation: $\frac{d}Read more

    To find \frac{dy}{dx} given xy + x^2 = 3, you can use implicit differentiation. The equation is not given in the form y = f(x), so we will differentiate both sides of the equation with respect to x using implicit differentiation.

    Start by differentiating both sides of the equation:

    \frac{d}{dx}(xy + x^2) = \frac{d}{dx}(3).

    Now, differentiate each term on the left side of the equation separately:

    1. \frac{d}{dx}(xy) = x\frac{dy}{dx} + y.
    2. \frac{d}{dx}(x^2) = 2x.

    Now, differentiate the right side of the equation:

    \frac{d}{dx}(3) = 0 because the derivative of a constant is 0.

    Now, you can rewrite the equation with these derivatives:

    x\frac{dy}{dx} + y + 2x = 0.

    Now, isolate the term with \frac{dy}{dx}:

    x\frac{dy}{dx} = -y - 2x.

    Now, solve for \frac{dy}{dx} by dividing both sides by x:

    \frac{dy}{dx} = \frac{-y}{x} - 2.

    So, \frac{dy}{dx} = -\frac{y}{x} - 2 is the derivative with respect to x of the equation xy + x^2 = 3.

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  4. Asked: January 10, 2021In:

    Find \dfrac{dy}{dx}, if x^{2}+y^{2}-2x+y=6

    acedstud
    Best Answer
    acedstud

    To find $\frac{dy}{dx}$ given $x^2 + y^2 - 2x + y = 6$, you can use implicit differentiation. The equation is not given in the form $y = f(x)$, so we will differentiate both sides of the equation with respect to $x$ using implicit differentiation. Start by differentiating both sides of the equation:Read more

    To find \frac{dy}{dx} given x^2 + y^2 - 2x + y = 6, you can use implicit differentiation. The equation is not given in the form y = f(x), so we will differentiate both sides of the equation with respect to x using implicit differentiation.

    Start by differentiating both sides of the equation:

    \frac{d}{dx}(x^2 + y^2 - 2x + y) = \frac{d}{dx}(6).

    Now, differentiate each term on the left side of the equation separately:

    1. \frac{d}{dx}(x^2) = 2x.
    2. \frac{d}{dx}(y^2) is a bit more involved because it’s a composite function. You can use the chain rule, which states that \frac{d}{dx}[f(g(x))] = f'(g(x)) \cdot g'(x). In this case, f(u) = u^2 and g(x) = y(x). So, applying the chain rule:

    \frac{d}{dx}(y^2) = 2y \frac{dy}{dx}.

    3. \frac{d}{dx}(-2x) = -2.

    4. \frac{d}{dx}(y) = \frac{dy}{dx}.

    Now, differentiate the right side of the equation:

    \frac{d}{dx}(6) = 0 because the derivative of a constant is 0.

    Now, you can rewrite the equation with these derivatives:

    2x + 2y \frac{dy}{dx} - 2 + \frac{dy}{dx} = 0.

    Now, isolate the term with \frac{dy}{dx}:

    2x - 2 + \frac{dy}{dx}(2y + 1) = 0.

    Factor out \frac{dy}{dx}:

    \frac{dy}{dx}(2y + 1) = 2 - 2x.

    Now, solve for \frac{dy}{dx} by dividing both sides by 2y + 1:

    \frac{dy}{dx} = \frac{2 - 2x}{2y + 1}.

    So, \frac{dy}{dx} = \frac{2 - 2x}{2y + 1} is the derivative with respect to x of the equation x^2 + y^2 - 2x + y = 6.

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  5. Asked: January 10, 2021In:

    Find \dfrac{dy}{dx}, if x^{3}-y^{3}=3xy

    acedstud
    Best Answer
    acedstud
    This answer was edited.

    To find $\frac{dy}{dx}$ given $x^3 - y^3 = 3xy$, you can use implicit differentiation. The equation is not given in the form $y = f(x)$, so we will differentiate both sides of the equation with respect to $x$ using implicit differentiation. Start by differentiating both sides of the equation: $\fracRead more

    To find \frac{dy}{dx} given x^3 - y^3 = 3xy, you can use implicit differentiation. The equation is not given in the form y = f(x), so we will differentiate both sides of the equation with respect to x using implicit differentiation.

    Start by differentiating both sides of the equation:

    \frac{d}{dx}(x^3 - y^3) = \frac{d}{dx}(3xy).

    Now, differentiate each term on the left side of the equation separately:

    1. \frac{d}{dx}(x^3) = 3x^2.
    2. \frac{d}{dx}(y^3) is a bit more involved because it’s a composite function. You can use the chain rule, which states that \frac{d}{dx}[f(g(x))] = f'(g(x)) \cdot g'(x). In this case, f(u) = u^3 and g(x) = y(x). So, applying the chain rule:

    \frac{d}{dx}(y^3) = 3y^2 \cdot \frac{dy}{dx}.

    Now, differentiate the right side of the equation:

    \frac{d}{dx}(3xy) = 3x\frac{dy}{dx} + 3y.

    Now, you can rewrite the equation with these derivatives:

    3x^2 - 3y^2 \cdot \frac{dy}{dx} = 3x\frac{dy}{dx} + 3y.

    Now, isolate the term with \frac{dy}{dx}:

    3x^2 - 3y = 3x\frac{dy}{dx} + 3y^2\frac{dy}{dx}.

    Factor out \frac{dy}{dx}:

    3x^2 - 3y = \frac{dy}{dx}(3x + 3y^2).

    Now, solve for \frac{dy}{dx} by dividing both sides by 3x + 3y^2:

    \frac{3x^2 - 3y}{3x + 3y^2} = \frac{dy}{dx}.

    Simplify the expression:

    \frac{x^2 - y}{x + y^2} = \frac{dy}{dx}.

    So, \frac{dy}{dx} = \frac{x^2 - y}{x + y^2} if x^3 - y^3 = 3xy.

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  6. Asked: January 10, 2021In:

    Find \dfrac{dy}{dx}, if x^{2}y-5x=3

    acedstud
    Best Answer
    acedstud
    This answer was edited.

    To find $\frac{dy}{dx}$ given $x^2y - 5x = 3$, you can use implicit differentiation. The equation is not given in the form $y = f(x)$, so we will differentiate both sides of the equation with respect to $x$ using implicit differentiation. Start by differentiating both sides of the equation: $$\frac{Read more

    To find \frac{dy}{dx} given x^2y - 5x = 3, you can use implicit differentiation. The equation is not given in the form y = f(x), so we will differentiate both sides of the equation with respect to x using implicit differentiation.

    Start by differentiating both sides of the equation:

        \[\frac{d}{dx}(x^2y - 5x) = \frac{d}{dx}(3)\]

    which can be further simplified as:

        \[\frac{d}{dx}(x^2y) - \frac{d}{dx}(5x) = 0\]

    (the derivative of a constant, 3, is 0).

    Now, apply the product rule to the first term of the equation. Recall that the product rule states:

        \[\frac{d}{dx}(u(x)v(x)) = u(x)\frac{dv}{dx} + v(x)\frac{du}{dx}\]

    In our case, let u(x) = x^2 and v(x) = y. We also need to differentiate y with respect to x, which is \frac{dy}{dx}.

    Now, we can differentiate each term, applying the product rule to the first term:

    1. \frac{d}{dx}(x^2y) = x^2\frac{dy}{dx} + 2xy,
    2. \frac{d}{dx}(5x) = 5.

    Now, substitute these results back into the equation:

        \[x^2\frac{dy}{dx} + 2xy - 5 = 0\]

    Now, isolate the term with \frac{dy}{dx}:

        \[x^2\frac{dy}{dx} = 5 - 2xy\]

    Finally, divide by x^2 to solve for \frac{dy}{dx}:

        \[\frac{dy}{dx} = \frac{5 - 2xy}{x^2}\]

    So, \frac{dy}{dx} = \frac{5 - 2xy}{x^2} if x^2y - 5x = 3.

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  7. Asked: January 10, 2021In:

    Differentiate y=x\sqrt{x+1} wrt x

    acedstud
    Best Answer
    acedstud

    To differentiate the function $y = x\sqrt{x+1}$ with respect to $x$, you can use the product rule. The product rule states that if you have a function of the form $u(x)v(x)$, the derivative is given by: $\frac{d}{dx}[u(x)v(x)] = u(x)v'(x) + u'(x)v(x)$ In this case, let $u(x) = x$ and $v(x) = \sqrt{xRead more

    To differentiate the function y = x\sqrt{x+1} with respect to x, you can use the product rule. The product rule states that if you have a function of the form u(x)v(x), the derivative is given by:

    \frac{d}{dx}[u(x)v(x)] = u(x)v'(x) + u'(x)v(x)

    In this case, let u(x) = x and v(x) = \sqrt{x+1}. We need to find their derivatives:

    1. u'(x) = \frac{d}{dx}(x) = 1 (the derivative of x is 1).

    2. To find v'(x), you can use the chain rule. Let w(x) = x+1, so v(x) = \sqrt{w(x)}. Now, we can find v'(x):

        \[v'(x) &= \frac{d}{dx}[\sqrt{w(x)}] \\ &= \frac{1}{2\sqrt{w(x)}}\frac{d}{dx}[w(x)]\]

    Now, find \frac{d}{dx}[w(x)]:

        \[\frac{d}{dx}[w(x)] &= \frac{d}{dx}(x+1) \\ &= 1\]

    Now, we can compute v'(x):

        \[v'(x) &= \frac{1}{2\sqrt{w(x)}}\cdot 1 \\ &= \frac{1}{2\sqrt{x+1}}\]

    Now, apply the product rule to find \frac{d}{dx}[x\sqrt{x+1}]:

        \[\frac{d}{dx}[x\sqrt{x+1}] &= u(x)v'(x) + u'(x)v(x) \\ &= x\cdot \frac{1}{2\sqrt{x+1}} + 1\cdot \sqrt{x+1}\]

    Now, simplify this expression:

        \[\frac{d}{dx}[x\sqrt{x+1}] &= \frac{x}{2\sqrt{x+1}} + \sqrt{x+1}\]

    So, the derivative of y = x\sqrt{x+1} with respect to x is:

        \[\frac{d}{dx}[x\sqrt{x+1}] &= \frac{x}{2\sqrt{x+1}} + \sqrt{x+1}\]

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  8. Asked: January 10, 2021In:

    Differentiate y=\sqrt{x}(x+3)^{2} wrt x

    acedstud
    Best Answer
    acedstud
    This answer was edited.

    To differentiate the function $y = \sqrt{x}(x+3)^2$ with respect to $x$, you can use the product rule. The product rule states that if you have a function of the form $u(x)v(x)$, the derivative is given by: $$\frac{d}{dx}[u(x)v(x)] = u(x)v'(x) + u'(x)v(x)$$ In this case, let $u(x) = \sqrt{x}$ and $vRead more

    To differentiate the function y = \sqrt{x}(x+3)^2 with respect to x, you can use the product rule. The product rule states that if you have a function of the form u(x)v(x), the derivative is given by:

        \[\frac{d}{dx}[u(x)v(x)] = u(x)v'(x) + u'(x)v(x)\]

    In this case, let u(x) = \sqrt{x} and v(x) = (x+3)^2. We need to find their derivatives:

    1. u'(x) = \frac{d}{dx}(\sqrt{x}) = \frac{1}{2\sqrt{x}} (using the power rule).

    2. v'(x) = \frac{d}{dx}((x+3)^2) = 2(x+3) (using the chain rule).

    Now, apply the product rule:

        \[\frac{d}{dx}[\sqrt{x}(x+3)^2] = u(x)v'(x) + u'(x)v(x) = \sqrt{x} \cdot 2(x+3) + \frac{1}{2\sqrt{x}} \cdot (x+3)^2\]

    Now, simplify this expression:

        \[2\sqrt{x}(x + 3) + \frac{1}{2\sqrt{x}}(x+3)^2\]

    So, the is the derivative of y = \sqrt{x}(x+3)^2 with respect to x:

        \[2\sqrt{x}(x + 3) + \frac{1}{2\sqrt{x}}(x+3)^2\]

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  9. Asked: October 28, 2023In:

    When working with sets in mathematics, which type of graphical representation can help you understand their relationships?

    acedstud
    acedstud

    A Venn diagram is a visual tool used to represent the relationship between sets. It consists of one or more circles that overlap or intersect, with each circle representing a specific set. The overlapping regions show where sets have elements in common, while the non-overlapping parts display elemenRead more

    A Venn diagram is a visual tool used to represent the relationship between sets. It consists of one or more circles that overlap or intersect, with each circle representing a specific set. The overlapping regions show where sets have elements in common, while the non-overlapping parts display elements unique to each set. The circles are enclosed inside a rectangle which represents the universal set.

    For example, consider the universal set containing positive integers less than 10, and consider two sets: Set A, which contains even numbers, and Set B, which contains multiples of 3.

    A Venn diagram for these sets would have one circle representing even numbers and another circle representing multiples of 3. The overlapping section would represent numbers that are both even and multiples of 3, which in this case is 6. The other numbers that are neither even numbers nor multiples of 3 are placed within the rectangle but outside the circles.

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  10. Asked: October 28, 2023In:

    To depict the connections between sets, which graphical representation would you use?

    acedstud
    Best Answer
    acedstud

    A Venn diagram is a visual tool used to represent the relationship between sets. It consists of one or more circles that overlap or intersect, with each circle representing a specific set. The overlapping regions show where sets have elements in common, while the non-overlapping parts display elemenRead more

    A Venn diagram is a visual tool used to represent the relationship between sets. It consists of one or more circles that overlap or intersect, with each circle representing a specific set. The overlapping regions show where sets have elements in common, while the non-overlapping parts display elements unique to each set. The circles are enclosed inside a rectangle which represents the universal set.

    For example, consider the universal set containing positive integers less than 10, and consider two sets: Set A, which contains even numbers, and Set B, which contains multiples of 3.

    A Venn diagram for these sets would have one circle representing even numbers and another circle representing multiples of 3. The overlapping section would represent numbers that are both even and multiples of 3, which in this case is 6. The other numbers that are neither even numbers nor multiples of 3 are placed within the rectangle but outside the circles.

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