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azamat
3 years ago
6

Metodo igualacion A) x+ y = 12 B) x - y = 4 ?

Mathematics
1 answer:
Olenka [21]3 years ago
6 0
<span>No sé la forma correcta de hacerlo, pero encontrar conjuntos de dos números que pueden igualar 12, luego encontrar cuál de los encaja en la segunda ecuación correctamente.</span>
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AlladinOne [14]

Answer:

What do you want us to find???

Step-by-step explanation:

4 0
2 years ago
A trough has ends shaped like isosceles triangles, with width 2 m and height 5 m, and the trough is 18 m long. Water is being pu
fenix001 [56]

9514 1404 393

Answer:

   5/9 m/min

Step-by-step explanation:

The depth of the water is 2/5 of the depth of the trough, so the width of the surface will be 2/5 of the width of the trough:

  2/5 × 2 m = 4/5 m

Then the surface area of the water is ...

  A = LW = (18 m)(4/5 m) = 14.4 m²

The rate of change of height multiplied by the area gives the rate of change of volume:

  8 m³/min = (14.4 m²)(h')

  h' = (8 m³/min)/(14.4 m²) = 5/9 m/min

8 0
3 years ago
Five-year-old students at an elementary school were given a 30-yard head start in a race. The graph shows how far the avera
zimovet [89]

Answer:

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Step-by-step explanation:

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4 0
3 years ago
One of the earliest applications of the Poisson distribution was in analyzing incoming calls to a telephone switchboard. Analyst
grandymaker [24]

Answer:

(a) P (X = 0) = 0.0498.

(b) P (X > 5) = 0.084.

(c) P (X = 3) = 0.09.

(d) P (X ≤ 1) = 0.5578

Step-by-step explanation:

Let <em>X</em> = number of telephone calls.

The average number of calls per minute is, <em>λ</em> = 3.0.

The random variable <em>X</em> follows a Poisson distribution with parameter <em>λ</em> = 3.0.

The probability mass function of a Poisson distribution is:

P(X=x)=\frac{e^{-\lambda}\lambda^{x}}{x!};\ x=0,1,2,3...

(a)

Compute the probability of <em>X</em> = 0 as follows:

P(X=0)=\frac{e^{-3}3^{0}}{0!}=\frac{0.0498\times1}{1}=0.0498

Thus, the  probability that there will be no calls during a one-minute interval is 0.0498.

(b)

If the operator is unable to handle the calls in any given minute, then this implies that the operator receives more than 5 calls in a minute.

Compute the probability of <em>X</em> > 5  as follows:

P (X > 5) = 1 - P (X ≤ 5)

              =1-\sum\limits^{5}_{x=0} { \frac{e^{-3}3^{x}}{x!}} \,\\=1-(0.0498+0.1494+0.2240+0.2240+0.1680+0.1008)\\=1-0.9160\\=0.084

Thus, the probability that the operator will be unable to handle the calls in any one-minute period is 0.084.

(c)

The average number of calls in two minutes is, 2 × 3 = 6.

Compute the value of <em>X</em> = 3 as follows:

<em> </em>P(X=3)=\frac{e^{-6}6^{3}}{3!}=\frac{0.0025\times216}{6}=0.09<em />

Thus, the probability that exactly three calls will arrive in a two-minute interval is 0.09.

(d)

The average number of calls in 30 seconds is, 3 ÷ 2 = 1.5.

Compute the probability of <em>X</em> ≤ 1 as follows:

P (X ≤ 1 ) = P (X = 0) + P (X = 1)

             =\frac{e^{-1.5}1.5^{0}}{0!}+\frac{e^{-1.5}1.5^{1}}{1!}\\=0.2231+0.3347\\=0.5578

Thus, the probability that one or fewer calls will arrive in a 30-second interval is 0.5578.

5 0
3 years ago
Twenty percent of a number is 14. find the number
7nadin3 [17]
The answer is:   70  .
___________________________________________
Explanation:
_________________________________

(20/100) x = 14;  solve for "x".

(20/100) = 2/10 = 1/5 ; 

(1/5) x = 14

x/5 = 14 ; 

x = 14 * 5 ; 

x = 70 .  The answer is:  70 .
___________________________________
5 0
3 years ago
Read 2 more answers
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