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3 years ago

ANSWER ASAP , NO LINKS

Mathematics

1 answer:

gayaneshka [121]3 years ago

4 0

……………………………………the answer is 1/6.

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Use crayons to color to shows at least two of it's parts<br><br><br>

Luden [163]

Answer:

im not exactly sure what you mean by that, could you please rephrase and expand on the pregunta

Step-by-step explanation:

5 0

2 years ago

Suppose z equals f (x comma y ), where x (u comma v )space equals space 2 u plus space v squared, y (u comma v )space equals spa

barxatty [35]

$z=f(x(u,v),y(u,v)),\begin{cases}x(u,v)=2u+v^2\\y(u,v)=3u-v\end{cases}$

We're given that $f_x(6,1)=3$ and $f_y(6,1)=-1$ , and want to find $\frac{\partial z}{\partial v}(1,2)$ .

By the chain rule, we have

$\dfrac{\partial z}{\partial v}=\dfrac{\partial z}{\partial x}\dfrac{\partial x}{\partial v}+\dfrac{\partial z}{\partial y}\dfrac{\partial y}{\partial v}$

and

$\dfrac{\partial x}{\partial v}=2v$

$\dfrac{\partial y}{\partial v}=-1$

Then

$\dfrac{\partial z}{\partial v}(1,2)=\dfrac{\partial z}{\partial x}(6,1)\dfrac{\partial x}{\partial v}(1,2)+\dfrac{\partial z}{\partial y}(6,1)\dfrac{\partial y}{\partial v}(1,2)$

(because the point $(x,y)=(6,1)$ corresponds to $(u,v)=(1,2)$ )

$\implies\dfrac{\partial z}{\partial v}(1,2)=3\cdot2\cdot2+(-1)\cdot(-1)=\boxed{13}$

4 0

3 years ago

How to solve this? I'm new to it

Elodia [21]

Answer:

2 = 65

1= 115

Step-by-step explanation:

angles 2 and 65° are equal

so input that in for 2

then subtract it by 180°

6 0

3 years ago

How do i graph y=1/2+4

galben [10]

y=1/2+4

y=2+4

y=6

y=1/2+4

y=1/6

y=6

Those homeworks are so easy lmaøoo

5 0

3 years ago

Read 2 more answers

An optical inspection system is used to distinguish among different part types. The probability of a correct classification of a

dsp73

Answer:

P(X=0)=0.000008

P(X=1)=0.00176

P(X=2)=0.057624

P(X=3)=0.941192

Step-by-step explanation:

Probability of correct classification = p = 0.98

Probability of incorrect classification = q = 1 - p = 0.02

The probability of success and failure is the same for all the trials. The trials are independent of each other and the number of trials is fixed i.e. n = 3.

This satisfies all the conditions of a Binomial Experiment. So we can use Binomial experiment to model the probability mass function.

The general formula of a binomial probability is:

$P(X=x)=^{n}C_{x}(p)^{x}q^{n-x}$

Here x denote the number of successes, which can be {0, 1, 2, 3}. So we need to evaluate the above equation for each value of x to determine the probability Mass function of X, as shown below:

$P(X=0)=^{3}C_{0}(0.98)^{0}(0.02)^{3-0}=0.000008\\\\ P(X=1)=^{3}C_{1}(0.98)^{1}(0.02)^{3-1}=0.001176\\\\ P(X=2)=^{3}C_{2}(0.98)^{2}(0.02)^{3-2}=0.057624\\\\ P(X=3)=^{3}C_{3}(0.98)^{3}(0.02)^{3-3}=0.941192$

7 0

3 years ago