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2 years ago

PLEASE FIX THIS LUA SCRIPT

Engineering

2 answers:

yKpoI14uk [10]2 years ago

6 0

Answer: not sure

Explanation: brainly please

zavuch27 [327]2 years ago

5 0

Answer:

When declaring variables, you said “workspace.MeanEnemy”. What you should have done was “game.Workspace.MeanEnemy”. Same for the variable right under it. Other than that, seems to be all good!

Hope This Helped!

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Using only the sequential operations described in Section 2.2.2, write an algorithm that gets two values: the price for item A a

suter [353]

Answer:

total_cost = cost + tax

Explanation:

Step1) Let the 2 variable for take input from user e.g price and quantity

var price ;

var quantity ;

var cost ;

var tax ;

var total_cost ;

Step2) take input from user quantity of item 'A'

step3) cost = price * quantity

Step4) tax = 0.08 * cost

Step5) total_cost = cost + tax

Step6) print the total_cost

7 0

3 years ago

The number of weaving errors in a twenty-foot by ten-foot roll of carpet has a mean of 0.8 What is the probability of observing

Viktor [21]

Answer:

0.14% probability of observing more than 4 errors in the carpet

Explanation:

When we only have the mean, we use the Poisson distribution.

In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by the following formula:

$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}$

In which

x is the number of sucesses

e = 2.71828 is the Euler number

$\mu$ is the mean in the given interval.

The number of weaving errors in a twenty-foot by ten-foot roll of carpet has a mean of 0.8.

This means that $\mu = 0.8$

What is the probability of observing more than 4 errors in the carpet

Either we observe 4 or less errors, or we observe more than 4. The sum of the probabilities of these outcomes is 1. So

$P(X \leq 4) + P(X > 4) = 1$

We want P(X > 4). Then

$P(X > 4) = 1 - P(X \leq 4)$

In which

$P(X \leq 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4)$

$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}$

$P(X = 0) = \frac{e^{-0.8}*(0.8)^{0}}{(0)!} = 0.4493$

$P(X = 1) = \frac{e^{-0.8}*(0.8)^{1}}{(1)!} = 0.3595$

$P(X = 2) = \frac{e^{-0.8}*(0.8)^{2}}{(2)!} = 0.1438$

$P(X = 3) = \frac{e^{-0.8}*(0.8)^{3}}{(3)!} = 0.0383$

$P(X = 4) = \frac{e^{-0.8}*(0.8)^{4}}{(4)!} = 0.0077$

$P(X \leq 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) = 0.4493 + 0.3595 + 0.1438 + 0.0383 + 0.0077 = 0.9986$

$P(X > 4) = 1 - P(X \leq 4) = 1 - 0.9986 = 0.0014$

0.14% probability of observing more than 4 errors in the carpet

5 0

3 years ago

This question allows you to practice proving a language is non-regular via the Pumping Lemma. Using the Pumping Lemma (Theorem 1

Ulleksa [173]

Answer:

L is not a regular language with formal proofs

Explanation:

(a) To prove that L is not a regular language, we will use a proof by contradiction. the assumption entails that L is a regular language. Then by the Pumping Lemma for Regular Languages, 

there exists a pumping length p for L such that for any string s ∈ L where |s| ≥ p, 

s = xyz subject to the following conditions: 

(a) |y| > 0 

(b) |xy| ≤ p, and 

(c) ∀i > 0, xyi 

z ∈ L

(b) To determine that L is not a regular language, we mke use of proof by contradiction. lets assume, that L is regular. Then by the Pumping Lemma for Regular Languages, it states also,

The pumping length, p for L such that for any string s ∈ L where |s| ≥ p, s = xyz subject to the condtions as follows : 

(a) |y| > 0 

(b) |xy| ≤ p, and 

(c) ∀i > 0, xyi 

z ∈ L. 

Choose s = 0p10p 

. Clearly, |s| ≥ p and s ∈ L. By condition (b) above, it follows is shown. by the first condition x and y are zeros.

for some k > 0. Per (c), we can take i = 0 and the resulting string will still be in L. Thus, xy0 

z should be in L. xy0 

z = xz = 0(p−k)10p 

It is shown that is is not in L. This is a contraption with the pumping lemma. our assumption that L is regular is incorrect, and L is not a regular language

6 0

3 years ago

The following median grain size data were obtained during isothermal liquid phase sintering of an 82W-8Mo-8Ni-2Fe alloy. What is

Morgarella [4.7K]

Answer:

The probable grain-coarsening mechanism is : Ideal grain growth mechanism

( $d^{2}$ - $d_{0} ^{2}$ = kt )