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slega [8]
3 years ago
13

Quiere cercar con alambre un terreno rectangular que mide 180 m (metros) de largo por 85 m de ancho.¿cuantos metros de alambre n

ecesita? ¿Con qué concepto matemático relacionas esta situación?
You want to wire a rectangular piece of land that is 180 m (meters) long by 85 m wide. How many meters of wire do you need? With what mathematical concept do you relate this situation?
Mathematics
1 answer:
blagie [28]3 years ago
5 0

Answer:

The wire required to fence is 530 m.

Step-by-step explanation:

Length, L = 180 m

Width, W = 85 m

The perimeter of the wire is

P = 2 (L + W)

P = 2 (180 + 85)

P =  530 m

So, the wire required to fence is 530 m.

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5). 30 is 200% of what number?
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Which of the following tables represents a function?
Contact [7]

Answer:

x 2-2 3-3

y 5 5  7 7

Step-by-step explanation:

Each domain, or x value, can only be matched with one y value, so given this, the third one is the answer since every domain is only matched with one range, or in other words, there are no repeating domains in on the x row of the table.

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3 years ago
Eighteen telephones have just been received at an authorized service center. Six of these telephones are cellular, six are cordl
LenaWriter [7]

Answer:

a) 0.0498

b) 0.1489

c) 0.1818

Step-by-step explanation:

Given:

Number of telephones = 6+6+6= 18

6 cellular, 6 cordless, and 6 corded.

a) Probability that all the cordless phones are among the first twelve to be serviced:

12 are selected from 18 telephones, possible number of ways of selection = ¹⁸C₁₂

Then 6 cordless telephones are serviced, the remaining telephones are: 12 - 6 = 6.

The possible ways of selecting thr remaining 6 telephones = ¹²C₆

Probability of servicing all cordless phones among the first twelve:

= (⁶C₆) (⁶C₁₂) / (¹⁸C₁₂)

= \frac{1 * 924}{18564}

= 0.0498

b) Probability that after servicing twelve of these phones, phones of only two of the three types remain to be serviced:

Here,

One type must be serviced first

The 6 remaining to be serviced can be a combination of the remaining two types.

Since there a 3 ways to select one type to be serviced, the probability will be:

= 3 [(⁶C₁)(⁶C₅) + (⁶C₂)(⁶C₄) + (⁶C₃)(⁶C₃) + (⁶C₄)(⁶C₂) + (⁶C₅)(⁶C₁)] / ¹⁸C₁₂

= \frac{3 * [(6)(6) + (15)(15) + (20)(20) + (15)(15) + (6)(6)]}{18564}

= \frac{2766}{18564}

= 0.1489

c) probability that two phones of each type are among the first six:

(⁶C₂)³/¹⁸C₆

\frac{3375}{18564}

=0.1818

5 0
4 years ago
Simplify this expression: g-3(f-g)-2f
vodomira [7]

Answer:

4g-5f

Step-by-step explanation:

g-3(f-g)-2f=g-3f+3g-2f=4g-5f

5 0
4 years ago
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