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4 years ago

6

Your driver license will be _____ if you race another driver on a public road, commit a felony using a motor vehicle, or are fou

nd guilty of reckless driving three times in one year.
A. suspended
B. denied
C. canceled
D. revoked

2 answers:

Georgia [21]4 years ago

6 0

Hello there,

In the problems given in the question, the driver's license is confiscated and suspended.

So our answer is: A)

Achievements.

3 years ago

that is wrong

olganol [36]4 years ago

4 0

Answer:

It is actually D revoked

Explanation:

You might be interested in

Three tool materials (high-speed steel, cemented carbide, and ceramic) are to be compared for the same turning operation on a ba

Tpy6a [65]

Answer:

Among all three tools, the ceramic tool is taking the least time for the production of a batch, however, machining from the HSS tool is taking the highest time.

Explanation:

The optimum cutting speed for the minimum cost

$V_{opt}= \frac{C}{\left[\left(T_c+\frac{C_e}{C_m}\right)\left(\frac{1}{n}-1\right)\right]^n}\;\cdots(i)$

Where,

C,n = Taylor equation parameters

$T_h$ =Tool changing time in minutes

$C_e$ =Cost per grinding per edge

$C_m$ = Machine and operator cost per minute

On comparing with the Taylor equation $VT^n=C$ ,

Tool life,

$T= \left[ \left(T_t+\frac{C_e}{C_m}\right)\left(\frac{1}{n}-1\right)\right]}\;\cdots(ii)$

Given that,

Cost of operator and machine time $=\$40/hr=\$0.667/min$

Batch setting time = 2 hr

Part handling time: $T_h=2.5$ min

Part diameter: $D=73 mm$ $=73\times 10^{-3} m$

Part length: $l=250 mm=250\times 10^{-3} m$

Feed: $f=0.30 mm/rev= 0.3\times 10^{-3} m/rev$

Depth of cut: $d=3.5 mm$

For the HSS tool:

Tool cost is $20 and it can be ground and reground 15 times and the grinding= $2/grind.

So, $C_e=$ $\$20/15+2=\$3.33/edge$

Tool changing time, $T_t=3$ min.

$C= 80 m/min$

$n=0.130$

(a) From equation (i), cutting speed for the minimum cost:

$V_{opt}= \frac {80}{\left[ \left(3+\frac{3.33}{0.667}\right)\left(\frac{1}{0.13}-1\right)\right]^{0.13}}$

$\Rightarrow 47.7$ m/min

(b) From equation (ii), the tool life,

$T=\left(3+\frac{3.33}{0.667}\right)\left(\frac{1}{0.13}-1\right)\right]}$

$\Rightarrow T=53.4$ min

(c) Cycle time: $T_c=T_h+T_m+\frac{T_t}{n_p}$

where,

$T_m=$ Machining time for one part

$n_p=$ Number of pieces cut in one tool life

$T_m= \frac{l}{fN}$ min, where $N=\frac{V_{opt}}{\pi D}$ is the rpm of the spindle.

$\Rightarrow T_m= \frac{\pi D l}{fV_{opt}}$

$\Rightarrow T_m=\frac{\pi \times 73 \times 250\times 10^{-6}}{0.3\times 10^{-3}\times 47.7}=4.01 min/pc$

So, the number of parts produced in one tool life

$n_p=\frac {T}{T_m}$

$\Rightarrow n_p=\frac {53.4}{4.01}=13.3$

Round it to the lower integer

$\Rightarrow n_p=13$

So, the cycle time

$T_c=2.5+4.01+\frac{3}{13}=6.74$ min/pc

(d) Cost per production unit:

$C_c= C_mT_c+\frac{C_e}{n_p}$

$\Rightarrow C_c=0.667\times6.74+\frac{3.33}{13}=\$4.75/pc$

(e) Total time to complete the batch= Sum of setup time and production time for one batch

$=2\times60+ {50\times 6.74}{50}=457 min=7.62 hr$ .

(f) The proportion of time spent actually cutting metal

$=\frac{50\times4.01}{457}=0.4387=43.87\%$

Now, for the cemented carbide tool:

Cost per edge,

$C_e=$ $\$8/6=\$1.33/edge$

Tool changing time, $T_t=1min$

$C= 650 m/min$

$n=0.30$

(a) Cutting speed for the minimum cost:

$V_{opt}= \frac {650}{\left[ \left(1+\frac{1.33}{0.667}\right)\left(\frac{1}{0.3}-1\right)\right]^{0.3}}=363m/min$ [from(i)]

(b) Tool life,

$T=\left[ \left(1+\frac{1.33}{0.667}\right)\left(\frac{1}{0.3}-1\right)\right]=7min$ [from(ii)]

(c) Cycle time:

$T_c=T_h+T_m+\frac{T_t}{n_p}$

$T_m= \frac{\pi D l}{fV_{opt}}$

$\Rightarrow T_m=\frac{\pi \times 73 \times 250\times 10^{-6}}{0.3\times 10^{-3}\times 363}=0.53min/pc$

$n_p=\frac {7}{0.53}=13.2$

$\Rightarrow n_p=13$ [ nearest lower integer]

So, the cycle time

$T_c=2.5+0.53+\frac{1}{13}=3.11 min/pc$

(d) Cost per production unit:

$C_c= C_mT_c+\frac{C_e}{n_p}$

$\Rightarrow C_c=0.667\times3.11+\frac{1.33}{13}=\$2.18/pc$

(e) Total time to complete the batch $=2\times60+ {50\times 3.11}{50}=275.5 min=4.59 hr$ .

(f) The proportion of time spent actually cutting metal

$=\frac{50\times0.53}{275.5}=0.0962=9.62\%$

Similarly, for the ceramic tool:

$C_e=$ $\$10/6=\$1.67/edge$

$T_t-1min$

$C= 3500 m/min$

$n=0.6$

(a) Cutting speed:

$V_{opt}= \frac {3500}{\left[ \left(1+\frac{1.67}{0.667}\right)\left(\frac{1}{0.6}-1\right)\right]^{0.6}}$

$\Rightarrow V_{opt}=2105 m/min$

(b) Tool life,

$T=\left[ \left(1+\frac{1.67}{0.667}\right)\left(\frac{1}{0.6}-1\right)\right]=2.33 min$

(c) Cycle time:

$T_c=T_h+T_m+\frac{T_t}{n_p}$

$\Rightarrow T_m=\frac{\pi \times 73 \times 250\times 10^{-6}}{0.3\times 10^{-3}\times 2105}=0.091 min/pc$

$n_p=\frac {2.33}{0.091}=25.6$

$\Rightarrow n_p=25 pc/tool\; life$

So,

$T_c=2.5+0.091+\frac{1}{25}=2.63 min/pc$

(d) Cost per production unit:

$C_c= C_mT_c+\frac{C_e}{n_p}$

$\Rightarrow C_c=0.667\times2.63+\frac{1.67}{25}=$1.82/pc$

(e) Total time to complete the batch

$=2\times60+ {50\times 2.63}=251.5 min=4.19 hr$ .

(f) The proportion of time spent actually cutting metal

$=\frac{50\times0.091}{251.5}=0.0181=1.81\%$

3 0

4 years ago

Biologists use a sequence of letters A, C, T, and G to model a genome. A gene isa substring of a genome that starts after a trip

kogti [31]

Answer:

You did not mention the programming language for implementation so i am writing a JAVA code.

import java.util.Scanner; // to get input from user

public class Genome{

public static void main(String[] args) { //start of main() function body

Scanner input = new Scanner(System.in); //creates Scanner object

System.out.print("Enter a genome string: ");

//prompts user to enter a genome string

String genome = input.nextLine();

//reads the input genome string and stores it into genome variable

boolean gene_found = false;

//variable gene_found of boolean type that has two value true or false

int startGene = 0; // stores starting of the gene string

for (int i = 0; i < genome.length() - 2; i++) {

//loop moves through genome string until the third last gene character

String triplet = genome.substring(i, i + 3);

//stores the triplet of genome substring

if (triplet.equals("ATG")) {

//if value in triplet is equal to ATG

startGene = i + 3;

//3 is added to i-th position of the genome string

}

else if (((triplet.equals("TAG")) || (triplet.equals("TAA")) || (triplet.equals("TGA"))) &&(startGene != 0))

//checks if the genome ends with one the given triplets TAG TAA and TGA

{ String gene = genome.substring(startGene, i);

gene stores substring of genome string from startGene to the position i

if (gene.length() % 3 == 0)

//if the the mod of gene length is 0 then the gene is found

{gene_found = true;

System.out.println(gene); //returns the found gene

startGene = 0;} } }

if (!gene_found) //if gene is not found returns the message below

System.out.println("no gene is found"); } }

Explanation:

This program first asks user to enter a genome string.

The loop starts from the first character of the entered string and this loop continues to execute until the value of i is 2 less than the genome input string length.

triplet variable stores first 3 characters of the genome string in first iteration and then moves through the string taking 3 characters each. This is done by dividing genome string to substring of 3 characters.

If condition checks if the 3 characters of genome string matches ATG using equals() function. If it is true this means start of genome is reached and these triplets are stored in startGene.

Else condition checks the end of the genome as the genome ends before one of TAG, TAA or TGA triplets. So this is checked here.

gene variable holds the triplet value stored in startGene and the value stored in index position i which means it holds the start of the genome till the end of the genome sequence. The end which is pointed by i variable is 2 less than the genome length and it is stored in gene variable.

After the loop ends the substring stored in gene variable is checked for a valid genome sequence by mod operator. If the length of the value stored in gene variable mod 0 is equal to 0 this means genome sequence is found and this string sequence stored in gene is displayed else the no gene is found message is displayed on output screen.

7 0

3 years ago

The team needs to choose a primary view for the part drawing. Three team members make suggestions:

dexar [7]

Answer:

<u>Option 1</u>

Explanation:

As the team has already submitted the plans for the part drawing, the best way to proceed would be how it was given in the plans. Hence, the option to be selected :

<u>Team member 1 suggests an orthographic top view because that is how the plans for the part were submitted.</u>

6 0

2 years ago

Who is the worst clown in rouge linage

Lerok [7]

All of them
explanation:
you don’t need one

4 0

3 years ago

University administrators have developed a Markov model to simulate graduation rates at their school. Students might drop out, r

abruzzese [7]

Solution :

The percentage of the students who have a chance of repeating their current year = 3%

The drop out students for the first year and the sophomores = 6%

Drop out rate of first year and the seniors = 4%

Now for the state space :

S = { first year(1), sophomores(2), juniors(3), seniors(4), graduates(G), Dropouts(D) }

Therefore

the first year students are indicated as '1'

Sophomores are indicated as '2'

Juniors are indicated as '3'

Seniors are indicated as '4

Graduates are indicated as 'G'

Dropouts are indicated as 'D'

The transition diagram is attached below.

The probability of the students who have the chance of repeating their current year = 3/100 = 0.03

Probability of first year dropouts and sophomores = 6/100 = 0.06

Probability of dropout rate of juniors and seniors = 4/100 = 0.04

Therefore, the probability matrix can be made as :

1 2 3 4 G D

$\begin{matrix}1\\ 2\\ 3\\ 4\\ G\\ D\end{matrix}$ $\begin{bmatrix}0.03 & 0.91 & 0 & 0 & 0 & 0.06\\ 0& 0.03 & 0.91 & 0 & 0 & 0.06\\ 0& 0 & 0.03 & 0.93 & 0 & 0.04\\ 0& 0 & 0 & 0.03 & 0.93 & 0.04\\ 0& 0 & 0 & 0 & 1 & 0\\ 0& 0 & 0 & 0 & 0 & 1\end{bmatrix}$

Here, G represents 'graduates' and D represents 'Dropouts.'

5 0

3 years ago