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3 years ago

Plz answer it fast I will give you ten point

Mathematics

2 answers:

notka56 [123]3 years ago

8 0

i think that true answer is b

vovangra [49]3 years ago

5 0

Answer:C

Step-by-step explanation: Add 2.76 +0.40

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Score: 0 of 1 pt

cupoosta [38]

Answer:

Step-by-step explanation:

21/.3

5 0

3 years ago

A company purchased a delivery van for $28,000 with a salvage value of $3,000 on September 1, Year 1. It has an estimated useful

Tems11 [23]

Answer:

Depreciation till December 31, Year 1 will be equal to 1,250$

Step-by-step explanation:

Purchasing Cost = 28,000$

Salvage Value = 3,000$

Total Depreciation:

Total Depreciation over 5 years (60 Months) = Purchasing Cost - Salvage Value

Total Depreciation over 5 years (60 Months) = 28,000 - 3,000

Total Depreciation over 5 years (60 Months) = 25,000$

Monthly Depreciation:

Using the unity method we have monthly depreciation by dividing the total depreciation by the total no. of months as below:

Total Depreciation over a single month =25,000/60

Total Depreciation over a single month = 416.67$ (Monthly Depreciation)

Depreciation till December 31, Year 1

As from September 1, Year 1 to December 31, Year 1, its been 3 months therefore total depreciation will be = 3 * Monthly Depreciation

Depreciation till December 31, Year 1 = 3 * 416.67

Depreciation till December 31, Year 1 = 1,250$

4 0

3 years ago

Suppose that are Yi.....Yn are i.i.d random variables with a Nâ(âÎ¼Y, Ï^2Y â) distribution. How would the probability density of

svp [43]

Answer:

b. As the sample size â increases, the variance of decreases. â So, the distribution of becomes highly concentrated around.

Step-by-step explanation:

Let : Yi,.... Yn are = i.i.d are random variables. The probability density of the distribution varies along with the sample size. When the sample size changes, the probability density of $E^3$ also changes.

The probability distribution may be defined as the statistical expression which defines the likelihood of any outcome for the discrete random variable and it can be opposed to the continuous random variable.

In the context, when the size of the sample of the distribution size increases, it causes a decrease in the variance and so the distribution becomes highly concentrated around.

5 0

3 years ago

Find the absolute maximum and absolute minimum values of the function f(x, y) = x 2 + y 2 − x 2 y + 7 on the set d = {(x, y) : |

dsp73

Looks like $f(x,y)=x^2+y^2-x^2y+7$ .

$f_x=2x-2xy=0\implies2x(1-y)=0\implies x=0\text{ or }y=1$

$f_y=2y-x^2=0\implies2y=x^2$

If $x=0$ , then $y=0$ - critical point at (0, 0).
If $y=1$ , then $x=\pm\sqrt2$ - two critical points at $(-\sqrt2,1)$ and $(\sqrt2,1)$

The latter two critical points occur outside of $D$ since $|\pm\sqrt2|>1$ so we ignore those points.

The Hessian matrix for this function is

$H(x,y)=\begin{bmatrix}f_{xx}&f_{xy}\\f_{yx}&f_{yy}\end{bmatrix}=\begin{bmatrix}2-2y&-2x\\-2x&2\end{bmatrix}$

The value of its determinant at (0, 0) is $\det H(0,0)=4>0$ , which means a minimum occurs at the point, and we have $f(0,0)=7$ .

Now consider each boundary:

If $x=1$ , then

$f(1,y)=8-y+y^2=\left(y-\dfrac12\right)^2+\dfrac{31}4$

which has 3 extreme values over the interval $-1\le y\le1$ of 31/4 = 7.75 at the point (1, 1/2); 8 at (1, 1); and 10 at (1, -1).

If $x=-1$ , then

$f(-1,y)=8-y+y^2$

and we get the same extrema as in the previous case: 8 at (-1, 1), and 10 at (-1, -1).

If $y=1$ , then

$f(x,1)=8$

which doesn't tell us about anything we don't already know (namely that 8 is an extreme value).

If $y=-1$ , then

$f(x,-1)=2x^2+8$

which has 3 extreme values, but the previous cases already include them.

Hence $f(x,y)$ has absolute maxima of 10 at the points (1, -1) and (-1, -1) and an absolute minimum of 0 at (0, 0).

3 0

3 years ago

Look at the picture so you know what the problem is.

My name is Ann [436]

Excuse me, is this I ready because you know you can get reported right... just try your best, it's not a real grade! :)

& possibly delete this question because a bot can track and report this to your school. People at my school did that and got in 3 days detention...

7 0

3 years ago