=3 and 1/2 X 3 and 1/2

PtichkaEL [24]

Add 3 on both sides

8y - 3 + 3 = 13 + 3

8y = 13 + 3

8y = 16

Divide 8 on both sides

8y/8 = 16/8

y = 16/8

y = 2

y = 2

8 0

3 years ago

Read 2 more answers

What is the solution to this system of leaner equation <br> 2x + y = 1 <br> 3x - y = -6

scZoUnD [109]

Answer:

Solution :

{x,y} = {3,7}

Step-by-step explanation:

5 0

3 years ago

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If f(x, y, z) = x sin(yz), (a) find the gradient of f and (b) find the directional derivative of f at (2, 4, 0) in the direction

valentina_108 [34]

Answer:

a) $\nabla f(x,y,z) = \sin{yz}\mathbf{i} + xz\cos{yz}\mathbf{j} + xy \cos{yz}\mathbf{k}$ .

b) $Du_{f}(2,4,0) = -\frac{8}{\sqrt{11}}$

Step-by-step explanation:

Given a function $f(x,y,z)$ , this function has the following gradient:

$\nabla f(x,y,z) = f_{x}(x,y,z)\mathbf{i} + f_{y}(x,y,z)\mathbf{j} + f_{z}(x,y,z)\mathbf{k}$ .

(a) find the gradient of f

We have that $f(x,y,z) = x\sin{yz}$ . So

$f_{x}(x,y,z) = \sin{yz}$

$f_{y}(x,y,z) = xz\cos{yz}$

$f_{z}(x,y,z) = xy \cos{yz}$ .

$\nabla f(x,y,z) = f_{x}(x,y,z)\mathbf{i} + f_{y}(x,y,z)\mathbf{j} + f_{z}(x,y,z)\mathbf{k}$ .

$\nabla f(x,y,z) = \sin{yz}\mathbf{i} + xz\cos{yz}\mathbf{j} + xy \cos{yz}\mathbf{k}$

(b) find the directional derivative of f at (2, 4, 0) in the direction of v = i + 3j − k.

The directional derivate is the scalar product between the gradient at (2,4,0) and the unit vector of v.

We have that:

$\nabla f(x,y,z) = \sin{yz}\mathbf{i} + xz\cos{yz}\mathbf{j} + xy \cos{yz}\mathbf{k}$

$\nabla f(2,4,0) = \sin{0}\mathbf{i} + 0\cos{0}\mathbf{j} + 8 \cos{0}\mathbf{k}$ .

$\nabla f(2,4,0) = 0i+0j+8k=(0,0,8)$

The vector is $v = i + 3j - k = (1,3,-1)$

To use v as an unitary vector, we divide each component of v by the norm of v.

$|v| = \sqrt{1^{2} + 3^{2} + (-1)^{2}} = \sqrt{11}$

So

$v_{u} = (\frac{1}{\sqrt{11}}, \frac{3}{\sqrt{11}}, \frac{-1}{\sqrt{11}})$

Now, we can calculate the scalar product that is the directional derivative.

$Du_{f}(2,4,0) = (0,0,8).(\frac{1}{\sqrt{11}}, \frac{3}{\sqrt{11}}, \frac{-1}{\sqrt{11}}) = -\frac{8}{\sqrt{11}}$

6 0

3 years ago

Motorola used the normal distribution to determine the probability of defects and the number

vovangra [49]

Answer:

a.P<x<9.85 orx>10.15)=0.3174, Total defects=317.4

b.p=0.0026,total defects=2.6

c.Less of the items produced will be classified as defects.

Step-by-step explanation:

a.The standard score,z, is esentially x reduced by process mean then divided my process standard deviation.

$\mu=10,\sigma=0.15\\Therefore:-\\z=\frac{x-\mu}{\sigma}=\frac{9.85-10}{0.15}\approx-1.0\\z=\frac{x-\mu}{\sigma}=\frac{10.15-10}{0.15}\approx+1.0\\P(x10.15)=P(Z+1.0)\\=2P(Z$

Total defects=Production*Probability

=0.3174*1000

$=317.4$

b. $\mu=10,\sigma=0.05$

therefore:-

$z=\frac{x-\mu}{\sigma}=\frac{9.85-10}{0.05}\approx-3.0\\z=\frac{x-\mu}{\sigma}=\frac{10.15-10}{0.05}\approx+3.0\\\\=P(x10.15)=P(Z+3.0)\\=2P(Z$

Defects=Probability*Production

=0.0026*1000

=2.6

c.Reducing process variation results in a significant reduction in the number of unit defects.

3 0

3 years ago

find the common difference of the arithmetic sequence. 9, 13, 17, 21, . . . a. nine thirteenths b. 4 c. one and four ninths d. 2

shusha [124]

The answer is letter b.4

3 0

3 years ago