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  1. Asked: March 22, 2021In: ,

    Question of the Day (22/03/2021)

    acedstud
    acedstud

    Let $X$ be a random variable representing the number of accidents that occur on the stretch of the road in a certain month, then since $X$ follows a Poisson distribution, $P(X=x)=\dfrac{\lambda^xe^{-\lambda}}{x!}$, where $x=0,1,2,3,...; \lambda$ is the average number of accidents on the road per monRead more

    Let X be a random variable representing the number of accidents that occur on the stretch of the road in a certain month, then since X follows a Poisson distribution, P(X=x)=\dfrac{\lambda^xe^{-\lambda}}{x!}, where x=0,1,2,3,...; \lambda is the average number of accidents on the road per month and e is the exponential constant which is approximately 2.71828.

    a.) Given: \lambda=2

    (i) P(X=4)=\dfrac{2^4e^{-2}}{4!}=\dfrac{16\times0.135335}{4\times3\times2\times1}=\dfrac{2.16536}{24}=0.0902

    (ii) The probability of at least 1 accident means the probability of 1 or more accidents. That is, P(X\ge 1)=P(X=1)+P(X=2)+P(X=3)+.... Using the compliment law of probabilities, we have P(X\ge 1)=1-P(X=0).

    P(X=0)=\dfrac{2^0e^{-2}}{0!}=\dfrac{1\times0.135335}{1}=0.135335

    Therefore, P(X\ge 1)=1-P(X=0)=1-0.135335=0.864665

     

    b.) Since the average number (expected number) of accidents per month is 2, the expected number of accidents in three months is 2\times3=6 accidents.

     

    c.) Let Y be a random variable representing the number of accidents that occur on the stretch of the road in a certain three months. In three months, the average number (expected number) of accidents is 6, so \lambda=6

    P(X=10)=\dfrac{6^{10}e^{-6}}{10!}=0.0413

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  2. Asked: January 13, 2021In:

    The demand equations for related products A and B are given by q_A=10\sqrt{\frac{p_B}{p_A}} …

    acedstud
    acedstud
    This answer was edited.

    i.) Given: $q_A=10\sqrt{\dfrac{P_B}{P_A}}=10P_B^{\frac{1}{2}}P_A^{-\frac{1}{2}}$; $\dfrac{dq_A}{dP_A}=-\dfrac{1}{2}\cdot10P_B^{\frac{1}{2}}P_A^{-\frac{3}{2}}=-5P_B^{\frac{1}{2}}P_A^{-\frac{3}{2}}$ When $P_A=9$, and $P_B=16$: $\dfrac{dq_A}{dP_A}=-5\times(16)^{\frac{1}{2}}(9)^{-\frac{3}{2}}=-5\times4\Read more

    i.) Given: q_A=10\sqrt{\dfrac{P_B}{P_A}}=10P_B^{\frac{1}{2}}P_A^{-\frac{1}{2}};
    \dfrac{dq_A}{dP_A}=-\dfrac{1}{2}\cdot10P_B^{\frac{1}{2}}P_A^{-\frac{3}{2}}=-5P_B^{\frac{1}{2}}P_A^{-\frac{3}{2}}

    When P_A=9, and P_B=16:
    \dfrac{dq_A}{dP_A}=-5\times(16)^{\frac{1}{2}}(9)^{-\frac{3}{2}}=-5\times4\times\dfrac{1}{27}=-\dfrac{20}{27}

    Given: q_B=3\sqrt[3]{\dfrac{P_A}{P_B}}=3P_A^{\frac{1}{3}}P_B^{-\frac{1}{3}}
    \dfrac{dq_B}{dP_A}=\dfrac{1}{3}\cdot3P_A^{-\frac{2}{3}}P_B^{-\frac{1}{3}}=P_A^{-\frac{2}{3}}P_B^{-\frac{1}{3}}

    When P_A=9, and P_B=16:
    \dfrac{dq_A}{dP_A}=(9)^{-\frac{2}{3}}(16)^{-\frac{1}{3}}
    =\dfrac{1}{9^\frac{2}{3}}\times\dfrac{1}{16^\frac{1}{3}}
    =\dfrac{1}{81^{\frac{1}{3}}}\times16^{\frac{1}{3}}
    =\sqrt[3]{\dfrac{1}{81\times16}}=\sqrt[3]{\dfrac{1}{1296}}\approx0.09172

    ii.) \dfrac{dq_A}{dP_B}=\dfrac{1}{2}\cdot10P_B^{-\frac{1}{2}}P_A^{-\frac{1}{2}}
    =5\times(14)^{-\frac{1}{2}}\times(9)^{-\frac{1}{2}}
    =5\times\dfrac{1}{14^{\frac{1}{2}}}\times3
    =15\times0.267261\approx4

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  3. Asked: March 16, 2021In:

    Question of the Day (16/03/2021)

    acedstud
    acedstud

    Given that the relationship between the amount spent, A, and the number of hours, h, is $A=9h+13$. Given that he spent \$238 last month, that is $A=\$238$. Thus, $238=9h+13$ Subtracting 13 from both sides, we have: $238 - 13 = 9h + 13 - 13 \Rightarrow 225 = 9h$ Dividing both sides by 9, we have: $\dRead more

    Given that the relationship between the amount spent, A, and the number of hours, h, is A=9h+13.

    Given that he spent $238 last month, that is A=$238.

    Thus, 238=9h+13

    Subtracting 13 from both sides, we have:

    238 - 13 = 9h + 13 - 13 \Rightarrow 225 = 9h

    Dividing both sides by 9, we have:

    \dfrac{225}{9} = \dfrac{9h}{9} \Rightarrow 25 = h

    Therefore, Edward spent 25 hours exercising at the gym last month.

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  4. Asked: March 15, 2021In: ,

    Question of the Day (15/03/2021)

    acedstud
    acedstud
    This answer was edited.

    The volume of the box is the sum of the volumes of the rectangular prism (cuboid) and the triangular prism. Volume of the rectangular prism = length $\times$ width $\times$ height $=18\times7\times2=252 \text{in}^3$. Volume of the triangular prism = $\dfrac{1}{2}\times$ length $\times$ width $\timesRead more

    The volume of the box is the sum of the volumes of the rectangular prism (cuboid) and the triangular prism.

    Volume of the rectangular prism = length \times width \times height =18\times7\times2=252 \text{in}^3.

    Volume of the triangular prism = \dfrac{1}{2}\times length \times width \times height =\dfrac{1}{2}\times18\times7\times2=126 \text{in}^3.

    Volume of the box = Volume of the rectangular prism + Volume of the triangular prism = 252 \text{in}^3+126 \text{in}^3=378 \text{in}^3.

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  5. Asked: March 14, 2021In: ,

    Question of the Day (14/03/2021)

    acedstud
    acedstud

    First, we find the slope of the tangent line by finding the derivative of $f(x)$ at $x=-1$. Given: $f(x)=-3x-x^4$. Then, $f'(x)=-3-4x^3$ and $f'(-1)=-3-4(-1)^3=-3-4(-1)=-3+4=1$. Thus, the slope of the tangent line is 1. Next, we find the y-coordinate of the function $f(x)$ at $x=-1$. At $x=-1, y=f(-Read more

    First, we find the slope of the tangent line by finding the derivative of f(x) at x=-1.

    Given: f(x)=-3x-x^4.

    Then, f'(x)=-3-4x^3 and f'(-1)=-3-4(-1)^3=-3-4(-1)=-3+4=1.

    Thus, the slope of the tangent line is 1.

    Next, we find the y-coordinate of the function f(x) at x=-1.

    At x=-1, y=f(-1)=-3(-1)-(-1)^4=3-1=2.

    Thus, the point of tangency is (-1, 2).

    Finally, the equation of a line with a slope of m=1 and passing through the point (x_1, y_1) = (-1, 2) is given as follows:

    y-y_1=m(x-x_1)

    \Rightarrow y-2=1(x-(-1))

    \Rightarrow y-2=x+1

    \Rightarrow y=x+1+2

    \Rightarrow y=x+3

    Therefore, the required equation is y=x+3.

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  6. Asked: March 10, 2021In:

    Question of the Day (10/03/2021)

    acedstud
    acedstud

    We can solve this question by evaluating the option. Option A: 67 + 31 = 98.  (Not correct)   Option B: 31 + 16 = 47.  (Not correct)   Option C: 13 + 68 = 81.  (Correct)   Option D: 24 + 81 = 105.  (Not correct)   There option C (13 + 68) will give 81, and hence, the correct answRead more

    We can solve this question by evaluating the option.

    Option A:

    67 + 31 = 98.  (Not correct)

     

    Option B:

    31 + 16 = 47.  (Not correct)

     

    Option C:

    13 + 68 = 81.  (Correct)

     

    Option D:

    24 + 81 = 105.  (Not correct)

     

    There option C (13 + 68) will give 81, and hence, the correct answer.

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  7. Asked: March 9, 2021In: ,

    Question of the Day (09/03/2021)

    acedstud
    acedstud

    A term is a single number or variable or a product of number(s) and variables(s) in a mathematical expression or equation, separated by + or - signs. Given $4+5x^4+x^3+7x^2-7$. Simplifying the like terms gives: $5x^4+x^3+7x^2-3$. Thus, there are 4 terms after simplification. Therefore, the number ofRead more

    A term is a single number or variable or a product of number(s) and variables(s) in a mathematical expression or equation, separated by + or – signs.

    Given 4+5x^4+x^3+7x^2-7.

    Simplifying the like terms gives: 5x^4+x^3+7x^2-3.

    Thus, there are 4 terms after simplification.

    Therefore, the number of terms is 4.

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  8. Asked: March 8, 2021In: ,

    Question of the Day (08/03/2021)

    acedstud
    acedstud

    Since the roots are at $x=1$ and $x=-17$, the equation is of the form $y=a(x-1)(x+17)$ . . . . . . . .(1) Also, since the vertex of the graph is at the point (-8, -81), the vertex form of the quadratic equation is $y=a(x+8)^2-81$ . . . . . . . .(2) Equating equation (1) and (2) gives: $a(x-1)(x+17)=Read more

    Since the roots are at x=1 and x=-17, the equation is of the form y=a(x-1)(x+17) . . . . . . . .(1)

    Also, since the vertex of the graph is at the point (-8, -81), the vertex form of the quadratic equation is y=a(x+8)^2-81 . . . . . . . .(2)

    Equating equation (1) and (2) gives:

    a(x-1)(x+17)=a(x+8)^2-81

    \Rightarrow a(x^2+16x-17)=a(x^2+16x+64)-81

    \Rightarrow ax^2+16ax-17a=ax^2+16ax+64a-81

    \Rightarrow -17a=64a-81

    \Rightarrow 64a+17a=81

    \Rightarrow 81a=81

    \Rightarrow a=1

    Substituting a=1 into equation (1) gives:

    y=(x-1)(x+17)

    \Rightarrow y=x^2+16x-17

    Therefore, a=1, b=16, and c=-17.

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  9. Asked: January 22, 2021In: ,

    Write the equation of a line if the gradient is 6 and the y-intercept is -\dfrac{1}{3}

    acedstud
    Best Answer
    acedstud
    This answer was edited.

    The equation of a line is given by $y=mx+c$; where $m$ is the slope or gradient, and $c$ is the y-intercept. Given: $m=6$ and $c=-\dfrac{1}{3}$ Therefore, the required equation is $y=6x-\dfrac{1}{3}$.

    The equation of a line is given by y=mx+c; where m is the slope or gradient, and c is the y-intercept.

    Given: m=6 and c=-\dfrac{1}{3}

    Therefore, the required equation is y=6x-\dfrac{1}{3}.

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  10. Asked: November 7, 2020

    At what rate of simple interest will $500 accumulate to $615 in 2.5 years?

    acedstud
    acedstud
    This answer was edited.

    This question deal with simple interest. The total amount (A) accumulated in simple interest after t years is given by $A=P(1+rt)$, where P is the principal. Given: $P=\$500$, $A=\$615$, $t=2.5$ years. Substituting for $P$, $A$, and $t$, into the formula $A=P(1+rt)$, we have: $615=500(1+2.5r)$ DividRead more

    This question deal with simple interest.

    The total amount (A) accumulated in simple interest after t years is given by A=P(1+rt), where P is the principal.

    Given: P=$500, A=$615, t=2.5 years.

    Substituting for P, A, and t, into the formula A=P(1+rt), we have:

    615=500(1+2.5r)

    Dividing both sides by 500 we have:

    \frac{615}{500}=1+2.5r
    \Rightarrow1.23=1+2.5r

    Subtracting 1 from both sides, we have:

    0.23=2.5r

    Dividing both sides by 2.5, we have:

    \frac{0.23}{2.5}=r

    \Rightarrow r=0.092

    Therefore, the rate is 0.092\times100\%=9.2\%.

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