What do I select? Please answer it is very hard for me

To avoid modeling the physical characteristics of wheel encoders such the square wave outputs, it is desirable to generate encod

grandymaker [24]

Answer:

The probability distribution is Normal continuous

Step-by-step explanation:

The basic idea is that velocity values have different noise level and an important thing regarding continuous probability distributions is that the probability of the random variable is equal to a specific outcome is 0

In other words, is practically impossible that one value of velocity could be the same as others.

5 0

3 years ago

Suppose you deposit $2,000 in a savings account that pays interest at an annual rate of %5. If no money is added or withdrawn

Inessa [10]

Answer:

For the answer, go to:

brainly.com/question/3965798

7 0

2 years ago

Please show me this but solved on paper?

kobusy [5.1K]

its about drive its about power we stay hungry we devour pick in the work put in the hours and take whats ours

4 0

2 years ago

Read 2 more answers

1 1/2. What are the 5 principles our government is based upon? Explain what each means.. 1.The first ten amendments are known as

sergij07 [2.7K]

Step-by-step explanation:

supremacy of the law ,rule of law ,franchise,equality before the law

3 0

4 years ago

<img src="https://tex.z-dn.net/?f=%5Csf%20%5Clim_%7Bx%20%5Cto%20%5Cinfty%7D%20%5Ccfrac%7B%5Csqrt%7Bx-1%7D-2x%20%7D%7Bx-7%7D" id=

BARSIC [14]

<h3>Answer: -2</h3>

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Work Shown:

$\displaystyle L = \lim_{x\to\infty} \frac{ \sqrt{x-1}-2x }{ x-7 }\\\\\\\displaystyle L = \lim_{x\to\infty} \frac{ \frac{1}{x}\left(\sqrt{x-1}-2x\right) }{ \frac{1}{x}\left(x-7\right) }\\\\\\\displaystyle L = \lim_{x\to\infty} \frac{ \frac{1}{x}*\sqrt{x-1}-\frac{1}{x}*2x }{ \frac{1}{x}*x-\frac{1}{x}*7 }\\\\\\$

$\displaystyle L = \lim_{x\to\infty} \frac{ \sqrt{\frac{1}{x^2}}*\sqrt{x-1}-2 }{ 1-\frac{7}{x} }\\\\\\\displaystyle L = \lim_{x\to\infty} \frac{ \sqrt{\frac{1}{x^2}*(x-1)}-2 }{ 1-\frac{7}{x} }\\\\\\\displaystyle L = \lim_{x\to\infty} \frac{ \sqrt{\frac{1}{x}-\frac{1}{x^2}}-2 }{ 1-\frac{7}{x} }\\\\\\\displaystyle L = \frac{ \sqrt{0-0}-2 }{ 1-0 }\\\\\\\displaystyle L = \frac{-2}{1}\\\\\\\displaystyle L = -2\\\\\\$

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Explanation:

In the second step, I multiplied top and bottom by 1/x. This divides every term by x. Doing this leaves us with various inner fractions that have the variable in the denominator. Those inner fractions approach 0 as x approaches infinity.

I'm using the rule that

$\displaystyle \lim_{x\to\infty} \frac{1}{x^k} = 0\\\\\\$

where k is some positive real number constant.

Using that rule will simplify the expression greatly to leave us with -2/1 or simply -2 as the answer.

In a sense, the leading terms of the numerator and denominator are -2x and x respectively. They are the largest terms for each, so to speak. As x gets larger, the influence that -2x and x have will greatly diminish the influence of the other terms.

This effectively means,

$\displaystyle L = \lim_{x\to\infty} \frac{ \sqrt{x-1}-2x }{ x-7 } = \lim_{x\to\infty} \frac{ -2x }{ x} = -2\\\\\\$

I recommend making a table of values to see what's going on. Or you can graph the given function to see that it slowly approaches y = -2. Keep in mind that it won't actually reach y = -2 itself.

5 0

3 years ago