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3 years ago

Simplify jk · 6 · 3.

Mathematics

2 answers:

Anna [14]3 years ago

8 0

Jk · 18 Is the answer because you would do 3 times 6

Law Incorporation [45]3 years ago

4 0

The simplified form is jk * 18.

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Evaluate the surface integral S F · dS for the given vector field F and the oriented surface S. In other words, find the flux of

Tomtit [17]

Apparently my answer was unclear the first time?

The flux of F across S is given by the surface integral,

$\displaystyle\iint_S\mathbf F\cdot\mathrm d\mathbf S$

Parameterize S by the vector-valued function r(u, v) defined by

$\mathbf r(u,v)=7\cos u\sin v\,\mathbf i+7\sin u\sin v\,\mathbf j+7\cos v\,\mathbf k$

with 0 ≤ u ≤ π/2 and 0 ≤ v ≤ π/2. Then the surface element is

dS = n • dS

where n is the normal vector to the surface. Take it to be

$\mathbf n=\dfrac{\frac{\partial\mathbf r}{\partial v}\times\frac{\partial\mathbf r}{\partial u}}{\left\|\frac{\partial\mathbf r}{\partial v}\times\frac{\partial\mathbf r}{\partial u}\right\|}$

The surface element reduces to

$\mathrm d\mathbf S=\mathbf n\,\mathrm dS=\mathbf n\left\|\dfrac{\partial\mathbf r}{\partial u}\times\dfrac{\partial\mathbf r}{\partial v}\right\|\,\mathrm du\,\mathrm dv$

$\implies\mathbf n\,\mathrm dS=-49(\cos u\sin^2v\,\mathbf i+\sin u\sin^2v\,\mathbf j+\cos v\sin v\,\mathbf k)\,\mathrm du\,\mathrm dv$

so that it points toward the origin at any point on S.

Then the integral with respect to u and v is

$\displaystyle\iint_S\mathbf F\cdot\mathrm d\mathbf S=\int_0^{\pi/2}\int_0^{\pi/2}\mathbf F(x(u,v),y(u,v),z(u,v))\cdot\mathbf n\,\mathrm dS$

$=\displaystyle-49\int_0^{\pi/2}\int_0^{\pi/2}(7\cos u\sin v\,\mathbf i-7\cos v\,\mathbf j+7\sin u\sin v\,\mathbf )\cdot\mathbf n\,\mathrm dS$

$=-343\displaystyle\int_0^{\pi/2}\int_0^{\pi/2}\cos^2u\sin^3v\,\mathrm du\,\mathrm dv=\boxed{-\frac{343\pi}6}$

4 0

3 years ago

State whether the data described below are discrete or continuous, and explain why. The numbers of languages spoken by differen

Tems11 [23]

Answer:

Option C) The data are discrete because the data can take on specific value.

Step-by-step explanation:

We are given the following in the question:

The numbers of languages spoken by different people.

The data would include points like English, Hindi, Marathi, and many more.

Thus, the data is discrete because the number of languages will always be expressed in whole numbers and thus it cannot take any value within an interval.

The value is calculated and not measured.

Thus, the correct answer is

Option C) The data are discrete because the data can take on specific value.

8 0

3 years ago

Find 2 numbers who product is 15 and sum is 16.

Marizza181 [45]

Answer:

15 and 1

Step-by-step explanation:

x and y are two numbers.

Two equations:

x · y = 15

x + y = 16

Rearrange one of the equations (I'll rearrange the sum equation):

x + y = 16

x = 16 - y

Substitute that to the other equation and solve for y:

x · y = 15

(16 - y) · y = 15

16 - y · y = 15

16 - y² = 15

-y² = 15 - 16

-y² = -1

y² = 1

y = √1

y = 1

Now substitute that to any of the equation and solve for x (in here, I'll choose the multiplication one):

x · y = 15

x · 1 = 15

x = 15

Now verify:

15 · 1 = 15

15 + 1 = 16

This is correct

6 0

3 years ago

What’s the answer?!!

Grace [21]

Answer:

6x+10mm=p is the answer

brainlest plz

6 0

3 years ago

Let X represent the amount of energy a city uses (in megawatt-hours) in the Kanto region. Let Y represent the amount of mismanag

Setler [38]

Answer:

Part 2: The probability of X≤2 or X≥4 is 0.5.

Part 3: The value of marginal probability of y is $f_y(y)=\frac{10-y}{32}$ for $2\leq y\leq 10$

Part 4:The value of E(y) is 4.6667.

Part 5:The value of $f_{xy}(x)$ is $\frac{2}{10-y}$ for $2\leq y\leq 2x\leq 10$

Part 6:The value of $M_{x,y}(y)$ is $\frac{y+10}{4}$

Part 7:The value of E(x) is 3.6667.

Part 8:The value of E(x,y) is 36.

Part 9:The value of Cov(x,y) is 18.8886.

Part 10:X and Y are not independent variables as $f_{xy}(x,y)\neq f_x(x).f_y(y)\\$

Step-by-step explanation:

As the complete question is here, however some of the values are not readable, thus the question is found online and is attached herewith.

From the given data, the joint distribution is given as

$f(x,y)=\frac{1}{16}$ for $2\leq y\leq 2x\leq 10$

Now the distribution of x is given as

$f_x(x)=\int\limits^{\infty}_{-\infty} {f(x,y)} \, dy$

Here the limits for y are $2\leq y\leq 2x$ So the equation becomes

$f_x(x)=\int\limits^{\infty}_{-\infty} {f(x,y)} \, dy\\f_x(x)=\int\limits^{2x}_{2} \frac{1}{16} \, dy\\f_x(x)=\frac{1}{16} (2x-2)\\f_x(x)=\frac{x-1}{8} \,\,\,\,\,\,\,\,\,\,\,\, for \,\,\,\,\,\,\,\,\,\ 1\leq x\leq 5$

Part 2:

The probability is given as

$P(X\leq 2 U X\geq 4)=\int\limits^2_1 {f_x(x)} \, dx +\int\limits^5_4 {f_x(x)} \, dx\\P(X\leq 2 U X\geq 4)=\int\limits^2_1 {\frac{x-1}{8}} \, dx +\int\limits^5_4 {\frac{x-1}{8}} \, dx\\P(X\leq 2 U X\geq 4)=\frac{1}{16}+\frac{7}{16}\\P(X\leq 2 U X\geq 4)=0.5$

So the probability of X≤2 or X≥4 is 0.5.

Part 3:

The distribution of y is given as

$f_y(y)=\int\limits^{\infty}_{-\infty} {f(x,y)} \, dx$

Here the limits for x are $y/2\leq x\leq 5$ So the equation becomes

$f_y(y)=\int\limits^{\infty}_{-\infty} {f(x,y)} \, dx\\f_y(y)=\int\limits^{5}_{y/2} \frac{1}{16} \, dx\\f_y(y)=\frac{1}{16} (5-\frac{y}{2})\\f_y(y)=\frac{10-y}{32} \,\,\,\,\,\,\,\,\,\,\,\, for \,\,\,\,\,\,\,\,\,\ 2\leq y\leq 10$

So the value of marginal probability of y is $f_y(y)=\frac{10-y}{32}$ for $2\leq y\leq 10$

Part 4

The value is given as

$E(y)=\int\limits^{10}_2 {yf_y(y)} \, dy\\E(y)=\int\limits^{10}_2 {y\frac{10-y}{32}} \, dy\\E(y)=\frac{1}{32}\int\limits^{10}_2 {10y-y^2} \, dy\\E(y)=4.6667$

So the value of E(y) is 4.6667.

Part 5

This is given as

$f_{xy}(x)=\frac{f_{xy}(x,y)}{f_y(y)}\\f_{xy}(x)=\frac{\frac{1}{16}}{\frac{10-y}{32}}\\f_{xy}(x)=\frac{2}{10-y}$

So the value of $f_{xy}(x)$ is $\frac{2}{10-y}$ for $2\leq y\leq 2x\leq 10$

Part 6

The value is given as

$\geq M_{x,y}(y)=E(f_{xy}(x))=\int\limits^5_{y/2} {x f_{xy}(x)} \, dx \\M_{x,y}(y)=\int\limits^5_{y/2} {x \frac{2}{10-y}} \, dx \\M_{x,y}(y)=\frac{2}{10-y}\left[\frac{x^2}{2}\right]^5_{\frac{y}{2}}\\M_{x,y}(y)=\frac{2}{10-y}\left(\frac{25}{2}-\frac{y^2}{8}\right)\\M_{x,y}(y)=\frac{y+10}{4}$

So the value of $M_{x,y}(y)$ is $\frac{y+10}{4}$

Part 7

The value is given as

$E(x)=\int\limits^{5}_1 {xf_x(x)} \, dx\\E(x)=\int\limits^{5}_1 {x\frac{x-1}{8}} \, dx\\E(x)=\frac{1}{8}\left(\frac{124}{3}-12\right)\\E(x)=\frac{11}{3} =3.6667$

So the value of E(x) is 3.6667.

Part 8

The value is given as

$E(x,y)=\int\limits^{5}_1 \int\limits^{10}_2 {xyf_{x,y}(x,y)} \,dy\, dx\\E(x,y)=\int\limits^{5}_1 \int\limits^{10}_2 {xy\frac{1}{16}} \,dy\, dx\\E(x,y)=\int\limits^{5}_1 \frac{x}{16}\left[\frac{y^2}{2}\right]^{10}_2\, dx\\E(x,y)=\int\limits^{5}_1 3x\, dx\\\\E(x,y)=3\left[\frac{x^2}{2}\right]^5_1\\E(x,y)=36$