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3 years ago

What is the answer to m^7 · m^−3?

Mathematics

1 answer:

elena55 [62]3 years ago

8 0

Answer:

M^4

Step-by-step explanation:

Here use product law which state- same base power should add

M^7-3

M^4

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Unit 10 assignment i need help

uranmaximum [27]

Answer:

Step-by-step explanation:

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4 years ago

Factor (x^6-64)(x+3)

PIT_PIT [208]

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3 years ago

Read 2 more answers

Answer correctly for brainliest and also get a thanks ! Don't answer if you don't know it please !

const2013 [10]

What I see here is a triangle sitting on top of a rectangle, and the
base of the triangle is equal to the length of the rectangle.
To see this, just draw a line between 'F' and 'S'. We can find the
area of the triangle, hen find the area of the rectangle, and then
add the two areas to get the area of the whole polygon.

The triangle:
. . . The base of the triangle is 9 units long.
. . . The height of the triangle is 6 units (from point 'N' down to the line FS).
. . . The area of a triangle is
   (1/2) · (base · height)

   = (1/2) · (9units · 6units)

   = (1/2) · (54 units²) = 27 units².

The rectangle:
. . . The length of the rectangle = 9 units. (line FS)
. . . The height of the rectangle = 2 units. (line WF or line CS)
. . . The area of a rectangle is

   (length) · (height)

   = (9units · 2units)

   = 18 units²

The whole polygon:
The area of the whole polygon is

   (area of the triangle) + (area of the rectangle)

   = (27 units²) + (18 units²) = 45 units²

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3 years ago

Emy has 13 1/4 kilograms of sweet potatoes.She gave away 5 12 kilogramsof corn to her neighbors.how many kilograms of corn than

Ahat [919]

Answer:

2 1/4 kg

Step-by-step explanation:

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1 year ago

Let Y1 and Y2 denote the proportion of time during which employees I and II actually performed their assigned tasks during a wor

Lemur [1.5K]

Answer:

Step-by-step explanation:

From the information given:

The joint density of $y_1$ and $y_2$ is given by:

$f_{(y_1,y_2)} \left \{ {{y_1+y_2, \ \ 0\ \le \ y_1 \ \le 1 , \ \ 0 \ \ \le y_2 \ \ \le 1} \atop {0, \ \ \ elsewhere \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ } \right.$

a)To find the marginal density of $y_1$ .

$f_{y_1} (y_1) = \int \limits ^{\infty}_{-\infty} f_{y_1,y_2} (y_1 >y_2) \ dy_2$

$=\int \limits ^{1}_{0}(y_1+y_2)\ dy_2$

$=\int \limits ^{1}_{0} \ \ y_1dy_2+ \int \limits ^{1}_{0} \ y_2 dy_2$

$= y_1 \ \int \limits ^{1}_{0} dy_2+ \int \limits ^{1}_{0} \ y_2 dy_2$

$= y_1[y_2]^1_0 + \bigg [ \dfrac{y_2^2}{2}\bigg]^1_0$

$= y_1 [1] + [\dfrac{1}{2}]$

$= y_1 + \dfrac{1}{2}$

i.e.

$f_{(y_1}(y_1)}= \left \{ {{y_1+\dfrac{1}{2}, \ \ 0\ \ \le \ y_1 \ \le , \ 1} \atop {0, \ \ \ elsewhere \ \\ \ \ \ \ \ \ \ \ } \right.$

The marginal density of $y_2$ is:

$f_{y_1} (y_2) = \int \limits ^{\infty}_{-\infty} fy_1y_1(y_1-y_2) dy_1$

$= \int \limits ^1_0 \ y_1 dy_1 + y_2 \int \limits ^1_0 dy_1$

$=\bigg[ \dfrac{y_1^2}{2} \bigg]^1_0 + y_2 [y_1]^1_0$

$= [ \dfrac{1}{2}] + y_2 [1]$

$= y_2 + \dfrac{1}{2}$

i.e.

$f_{(y_1}(y_2)}= \left \{ {{y_2+\dfrac{1}{2}, \ \ 0\ \ \le \ y_1 \ \le , \ 1} \atop {0, \ \ \ elsewhere \ \\ \ \ \ \ \ \ \ \ } \right.$

$P\bigg[y_1 \ge \dfrac{1}{2}\bigg |y_2 \ge \dfrac{1}{2} \bigg] = \dfrac{P\bigg [y_1 \ge \dfrac{1}{2} . y_2 \ge\dfrac{1}{2} \bigg]}{P\bigg[ y_2 \ge \dfrac{1}{2}\bigg]}$

$= \dfrac{\int \limits ^1_{\frac{1}{2}} \int \limits ^1_{\frac{1}{2}} f_{y_1,y_1(y_1-y_2) dy_1dy_2}}{\int \limits ^1_{\frac{1}{2}} fy_1 (y_2) \ dy_2}$

$= \dfrac{\int \limits ^1_{\frac{1}{2}} \int \limits ^1_{\frac{1}{2}} (y_1+y_2) \ dy_1 dy_2}{\int \limits ^1_{\frac{1}{2}} (y_2 + \dfrac{1}{2}) \ dy_2}$