Answer:
sectores industriales, comerciales o públicos, o para el uso doméstico.
Explanation:
Answer:
more than 10 joules jjsjdnjdfjnfjfnjfnnfndnndjnfjgngjntjfjmrmdkdmjd
Answer:
16 seconds
Explanation:
Given:
C = 60
L = 4 seconds each = 4*4 =16
In this problem, the first 3 timing stages are given as:
200, 187, and 210 veh/h.
We are to find the estimated effective green time of the fourth timing stage. The formula for the estimated effective green time is:
Let's first find the fourth stage critical lane group ratio
, using the formula:
![C = \frac{1.5L +5}{1 - ( \frac{200}{1800} + \frac{187}{1800} + \frac{210}{1800}) + ( \frac{v}{s})}](https://tex.z-dn.net/?f=%20C%20%3D%20%5Cfrac%7B1.5L%20%2B5%7D%7B1%20-%20%28%20%5Cfrac%7B200%7D%7B1800%7D%20%2B%20%5Cfrac%7B187%7D%7B1800%7D%20%2B%20%5Cfrac%7B210%7D%7B1800%7D%29%20%2B%20%28%20%5Cfrac%7Bv%7D%7Bs%7D%29%7D%20)
![60 = \frac{1.5*16 + 5}{1 - ( \frac{200}{1800} + \frac{187}{1800} + \frac{210}{1800}) + ( \frac{v}{s})}](https://tex.z-dn.net/?f=%2060%20%3D%20%5Cfrac%7B1.5%2A16%20%2B%205%7D%7B1%20-%20%28%20%5Cfrac%7B200%7D%7B1800%7D%20%2B%20%5Cfrac%7B187%7D%7B1800%7D%20%2B%20%5Cfrac%7B210%7D%7B1800%7D%29%20%2B%20%28%20%5Cfrac%7Bv%7D%7Bs%7D%29%7D%20)
Solving for
, we have:
Let's also calculate the volume capacity ratio X,
![X = (\frac{200}{1800} + \frac{187}{1800} + \frac{210}{1800} + 0.185)(\frac{60}{60-16}](https://tex.z-dn.net/?f=X%20%3D%20%28%5Cfrac%7B200%7D%7B1800%7D%20%2B%20%5Cfrac%7B187%7D%7B1800%7D%20%2B%20%5Cfrac%7B210%7D%7B1800%7D%20%2B%200.185%29%28%5Cfrac%7B60%7D%7B60-16%7D%20)
X = 0.704
For the the estimated effective green time of the fourth timing stage, we have:
Substituting figures in the equation, we now have:
15.78 ≈ 16 seconds
The estimated effective green time of the fourth timing stage is 16 seconds
Answer:
E
Explanation:
I have a big brain and I just took the test and got it correct.
Answer:
potato<-100
print(potato)
sqrt(potato)
potato<-potato*2
print(potato)
Explanation:
The complete question is as follows
Create a variable called potato whose value corresponds to the number of potatoes you’ve eaten in the last week. Or something equally ridiculous. Print out the value of potato.
Calculate the square root of potato using the sqrt() function. Print out the value of potato again to verify that the value of potato hasn’t changed.
Reassign the value of potato to potato * 2.
Print out the new value of potato to verify that it has changed
The question was answered using R programming language.
At line 1, I assumed that I ate 100 potatoes in the previous week.
So, potato = 100
At line 2, the value of potato is printed as 100.
At line 3, the square root of potato is calculated using sqrt function: Square for of 100 = 10
At line 4,the initial value of potato is doubled and stored in potato variable. 100 * 2 = 200
At line 5, the new value of potato is printed: 200.