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sweet [91]
3 years ago
12

En la ciudad de Huaraz se instalaron tuberias para distribuir gas natural el 2018.Si sabemos que durante la instalacion se hizo

la primera revicionde las conexiones y que las reviciones se llevan a cabo de 3 años ¿que numero de revision se realizara el 2048?
Mathematics
1 answer:
nekit [7.7K]3 years ago
7 0

Answer:

11th revision

Step-by-step explanation:

The first revision is in 2018, and for each 3 years there is a new revision.

From 2018 to 2048, there are 30 years.

So to find how many revisions there are, we can divide the 30 years by the period of 3 years:

Number of revisions = total period / period per revision = 30 / 3 = 10 revisions.

As the year 2018 also has a revision, we have 10 + 1 = 11 total revisions.

So in 2048 there will be the 11th revision.

You might be interested in
12/5 rename the fractions as mixed numbers
Alex_Xolod [135]

Answer:

2 2/5

Step-by-step explanation:

12 can go into 5 2 times with a left over of two, so it would be 2 which is 10 and a left over of two. The denominator stays the same. So it would be 2 2/5.

5 0
2 years ago
When x pounds of force is applied to one end of a lever that is L feet long, the resulting force y on the other end is
nadezda [96]
The question is worded poorly, but it looks like you have a lever in equilibrium, with a force x at a distance d from the fulcrum, and a force y at a distance L - d from the fulcrum. You already have the equilibrium formula for this situation:

xd = y(L - d)

If you know x, y, and d, you can solve for the length L.
3 0
2 years ago
Please solve the system of equations below thank you<br> explain if u can
ivanzaharov [21]
First let’s try to cancel out the x
5x + -5x = 0
Add the y and the numbers together
-3y + -2y = -5y
26 + -16 = 10
-5y = 10
y = -2
Use y=-2 in one of the equations
-3(-2) + 5x = 26
6 + 5x = 26
5x = 20
X = 4

So
Y= -2
X= 4

https://mathsmadeeasy.co.uk/gcse-maths-revision/simultaneous-equations-gcse-maths-revision/

Check this link out for more help!
7 0
2 years ago
Suppose a certain computer virus can enter a system through an email or through a webpage. There is a 40% chance of receiving th
DedPeter [7]

Answer:

P = 0.42

Step-by-step explanation:

This probability problem can be solved by building a Venn like diagram for each probability.

I say that we have two sets:

-Set A, that is the probability of receiving this virus through the email.

-Set B, that is the probability of receiving it through the webpage.

The most important information in these kind of problems is the intersection. That is, that he virus enters the system simultaneously by both email and webpage with a probability of 0.17. It means that A \cap B = 0.17.

By email only

The problem states that there is a 40 chance of receiving it through the email. It means that we have the following equation:

A + (A \cap B) = 0.40

A + 0.17 = 0.40

A = 0.23

where A is the probability that the system receives the virus just through the email.

The problem states that there is a 40% chance of receiving it through the email. 23% just through email and 17% by both the email and the webpage.

By webpage only

There is a 35% chance of receiving it through the webpage. With this information, we have the following equation:

B + (A \cap B) = 0.35

B + 0.17 = 0.35

B = 0.18

where B is the probability that the system receives the virus just through the webpage.

The problem states that there is a 35% chance of receiving it through the webpage. 18% just through the webpage and 17% by both the email and the webpage.

What is the probability that the virus does not enter the system at all?

So, we have the following probabilities.

- The virus does not enter the system: P

- The virus enters the system just by email: 23% = 0.23

- The virus enters the system just by webpage: 18% = 0.18

- The virus enters the system both by email and by the webpage: 17% = 0.17.

The sum of the probabilities is 100% = 1. So:

P + 0.23 + 0.18 + 0.17 = 1

P = 1 - 0.58

P = 0.42

There is a probability of 42% that the virus does not enter the system at all.

5 0
3 years ago
A manager in a call center needs to determine how many agents to schedule next week.
lukranit [14]

Answer:

Number of calls expected in next week by manager = 7940

Average Number of calls that call center agent will attend in an hour =7 calls

It is also given that, Call center remain open for 10 hours 5 days a week.

Also, it is given that, full time agents work 40 hours a week but are only on call for 35 hours per week  ,Part time agents work 20 hours a week but are only on calls 17 hours per week .

⇒Number of hours worked by full time agents × Number of calls attended in an hour × Number of full time agents + Number of hours worked by Part time agents × Number of calls attended in an hour × Number of Part time agents ≤ 7940

⇒35 × 7×Number of full time agents +17 × 7 ×Number of Part time agents ≤ 7940

Option A

⇒35×15×7+17×7×15

= 3675+1785

= 5460

Option B

⇒35 ×7×20+17×7×7

=4900 +833

= 5733

Option C

⇒35×20×7 +17×20×7

=4900+2380

=7280

Option D

⇒25 × 35×7+17×7×5

=6125 +595

=6720

Option E

⇒28×35×7+17×7×10

=6860+1190

=8050

Option E, ⇒ 28 full time agents and 10 part time agents , is best to meet the scheduling needs is most appropriate, that is nearer to 7940 calls.

7 0
2 years ago
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