Answer:
40
Step-by-step explanation:
The correct question is:
Suppose x = c1e^(-t) + c2e^(3t) a solution to x''- 2x - 3x = 0 by substituting it into the differential equation. (Enter the terms in the order given. Enter c1 as c1 and c2 as c2.)
Answer:
x = c1e^(-t) + c2e^(3t)
is a solution to the differential equation
x''- 2x' - 3x = 0
Step-by-step explanation:
We need to verify that
x = c1e^(-t) + c2e^(3t)
is a solution to the differential equation
x''- 2x' - 3x = 0
We differentiate
x = c1e^(-t) + c2e^(3t)
twice in succession, and substitute the values of x, x', and x'' into the differential equation
x''- 2x' - 3x = 0
and see if it is satisfied.
Let us do that.
x = c1e^(-t) + c2e^(3t)
x' = -c1e^(-t) + 3c2e^(3t)
x'' = c1e^(-t) + 9c2e^(3t)
Now,
x''- 2x' - 3x = [c1e^(-t) + 9c2e^(3t)] - 2[-c1e^(-t) + 3c2e^(3t)] - 3[c1e^(-t) + c2e^(3t)]
= (1 + 2 - 3)c1e^(-t) + (9 - 6 - 3)c2e^(3t)
= 0
Therefore, the differential equation is satisfied, and hence, x is a solution.
Answer:
d a m i dont even know this
Step-by-step explanation:
Answer:
c) Is not a property (hence (d) is not either)
Step-by-step explanation:
Remember that the chi square distribution with k degrees of freedom has this formula

Where N₁ , N₂m ....
are independent random variables with standard normal distribution. Since it is a sum of squares, then the chi square distribution cant take negative values, thus (c) is not true as property. Therefore, (d) cant be true either.
Since the chi square is a sum of squares of a symmetrical random variable, it is skewed to the right (values with big absolute value, either positive or negative, will represent a big weight for the graph that is not compensated with values near 0). This shows that (a) is true
The more degrees of freedom the chi square has, the less skewed to the right it is, up to the point of being almost symmetrical for high values of k. In fact, the Central Limit Theorem states that a chi sqare with n degrees of freedom, with n big, will have a distribution approximate to a Normal distribution, therefore, it is not very skewed for high values of n. As a conclusion, the shape of the distribution changes when the degrees of freedom increase, because the distribution is more symmetrical the higher the degrees of freedom are. Thus, (b) is true.