This is a linear differential equation of first order. Solve this by integrating the coefficient of the y term and then raising e to the integrated coefficient to find the integrating factor, i.e. the integrating factor for this problem is e^(6x).
<span>Multiplying both sides of the equation by the integrating factor: </span>
<span>(y')e^(6x) + 6ye^(6x) = e^(12x) </span>
<span>The left side is the derivative of ye^(6x), hence </span>
<span>d/dx[ye^(6x)] = e^(12x) </span>
<span>Integrating </span>
<span>ye^(6x) = (1/12)e^(12x) + c where c is a constant </span>
<span>y = (1/12)e^(6x) + ce^(-6x) </span>
<span>Use the initial condition y(0)=-8 to find c: </span>
<span>-8 = (1/12) + c </span>
<span>c=-97/12 </span>
<span>Hence </span>
<span>y = (1/12)e^(6x) - (97/12)e^(-6x)</span>
Answer: The probability in (b) has higher probability than the probability in (a).
Explanation:
Since we're computing for the probability of the sample mean, we consider the z-score and the standard deviation of the sampling distribution. Recall that the standard deviation of the sampling distribution approximately the quotient of the population standard deviation and the square root of the sample size.
So, if the sample size higher, the standard deviation of the sampling distribution is lower. Since the sample size in (b) is higher, the standard deviation of the sampling distribution in (b) is lower.
Moreover, since the mean of the sampling distribution is the same as the population mean, the lower the standard deviation, the wider the range of z-scores. Because the standard deviation in (b) is lower, it has a wider range of z-scores.
Note that in a normal distribution, if the probability has wider range of z-scores, it has a higher probability. Therefore, the probability in (b) has higher probability than the probability in (a) because it has wider range of z-scores than the probability in (a).
Answer:
0+5+20= 25 the answer is 25
The value of g would be 75 degrees.
