I need this working of this question

joyce is thinking of a 3 digit number. her number has the digits 8, 4, and 6. to the nearest hundred, it rounds to 600. what is

Artemon [7]

Answer:

648

Step-by-step explanation:

It will round down (since 4 and down round down) to 600. Therefore, the answer is 648.

<em>Hope that helps!</em>

<em>-Sabrina</em>

3 0

3 years ago

Read 2 more answers

The black graph is the graph of

satela [25.4K]

Answer:

There is no graph attached.

Step-by-step explanation:

Check the question again.

7 0

3 years ago

Select the correct answer.

balu736 [363]

Answer:

d. 11

> Step-by-step explanation:

7 0

4 years ago

Differential Equation

ANEK [815]

1. The given equation is probably supposed to read

y'' - 2y' - 3y = 64x exp(-x)

First consider the homogeneous equation,

y'' - 2y' - 3y = 0

which has characteristic equation

r² - 2r - 3 = (r - 3) (r + 1) = 0

with roots r = 3 and r = -1. Then the characteristic solution is

$y = C_1 e^{3x} + C_2 e^{-x}$

and we let y₁ = exp(3x) and y₂ = exp(-x), our fundamental solutions.

Now we use variation of parameters, which gives a particular solution of the form

$y_p = u_1y_1 + u_2y_2$

where

$\displaystyle u_1 = -\int \frac{64xe^{-x}y_2}{W(y_1,y_2)} \, dx$

$\displaystyle u_2 = \int \frac{64xe^{-x}y_1}{W(y_1,y_2)} \, dx$

and W(y₁, y₂) is the Wronskian determinant of the two fundamental solutions. This is

$W(y_1,y_2) = \begin{vmatrix}e^{3x} & e^{-x} \\ (e^{3x})' & (e^{-x})'\end{vmatrix} = \begin{vmatrix}e^{3x} & e^{-x} \\ 3e^{3x} & -e^{-x}\end{vmatrix} = -e^{2x} - 3e^{2x} = -4e^{2x}$

Then we find

$\displaystyle u_1 = -\int \frac{64xe^{-x} \cdot e^{-x}}{-4e^{2x}} \, dx = 16 \int xe^{-4x} \, dx = -(4x + 1) e^{-4x}$

$\displaystyle u_2 = \int \frac{64xe^{-x} \cdot e^{3x}}{-4e^{2x}} \, dx = -16 \int x \, dx = -8x^2$

so it follows that the particular solution is

$y_p = -(4x+1)e^{-4x} \cdot e^{3x} - 8x^2\cdot e^{-x} = -(8x^2+4x+1)e^{-x}$

and so the general solution is

$\boxed{y(x) = C_1 e^{3x} + C_2e^{-x} - (8x^2+4x+1) e^{-x}}$

2. I'll again assume there's typo in the equation, and that it should read

y''' - 6y'' + 11y' - 6y = 2x exp(-x)

Again, we consider the homogeneous equation,

y''' - 6y'' + 11y' - 6y = 0

and observe that the characteristic polynomial,

r³ - 6r² + 11r - 6

has coefficients that sum to 1 - 6 + 11 - 6 = 0, which immediately tells us that r = 1 is a root. Polynomial division and subsequent factoring yields

r³ - 6r² + 11r - 6 = (r - 1) (r² - 5r + 6) = (r - 1) (r - 2) (r - 3)

and from this we see the characteristic solution is

$y_c = C_1 e^x + C_2 e^{2x} + C_3 e^{3x}$

For the particular solution, I'll use undetermined coefficients. We look for a solution of the form

$y_p = (ax+b)e^{-x}$

whose first three derivatives are

${y_p}' = ae^{-x} - (ax+b)e^{-x} = (-ax+a-b)e^{-x}$

${y_p}'' = -ae^{-x} - (-ax+a-b)e^{-x} = (ax-2a+b)e^{-x}$

${y_p}''' = ae^{-x} - (ax-2a+b)e^{-x} = (-ax+3a-b)e^{-x}$

Substituting these into the equation gives

$(-ax+3a-b)e^{-x} - 6(ax-2a+b)e^{-x} + 11(-ax+a-b)e^{-x} - 6(ax+b)e^{-x} = 2xe^{-x}$

$(-ax+3a-b) - 6(ax-2a+b) + 11(-ax+a-b) - 6(ax+b) = 2x$

$-24ax+26a-24b = 2x$

It follows that -24a = 2 and 26a - 24b = 0, so that a = -1/12 = -12/144 and b = -13/144, so the particular solution is

$y_p = -\dfrac{12x+13}{144}e^{-x}$

and the general solution is

$\boxed{y = C_1 e^x + C_2 e^{2x} + C_3 e^{3x} - \dfrac{12x+13}{144} e^{-x}}$

5 0

3 years ago

When the outliers are removed, how does the mean change?

Natalka [10]

Answer:

Removing the outlier decreases the number of data by one and therefore you must decrease the divisor. For instance, when you find the mean of 0,10,10,12,12, you must divide the sum by 5, but when you remove the outlier of 0,you must then divide by 4.

Step-by-step explanation:

8 0

3 years ago