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xenn [34]
3 years ago
6

En la primera fase del proceso selectivo para Al cuerpo de profesores eliminan a cuatro novenos De los aspirantes mientras que e

n la segunda fase eliminan A 5/7 de los que superaron la primera fase Si se presentaron 315 aspirantes cuál es la fracción De los aspirantes que han superado las dos fases Del proceso selectivo?
Mathematics
1 answer:
irina1246 [14]3 years ago
8 0

Answer:

The fraction of the applicants who have passed the two phases of the selection process is \frac{10}{63}.

Step-by-step explanation:

The question is:

In the first phase of the selection process for the body of teachers they eliminate four-ninth of the applicants while in the second phase they eliminate 5/7 of those who passed the first phase. If there were 315 applicants what is the fraction of the applicants who have passed the two phases of the selection process?

The total number of applicants is

<em>N</em> = 315.

It is provided that 4/9th of the applicants were eliminated in the first phase.

Then the fraction of candidates who passed the first phase is:

Fraction who passed phase I = 1-\frac{4}{9}=\frac{5}{9}

Now, it is also provided that 5/7 of those who passed the first phase were eliminated in the second phase.

Then the fraction of candidates who passed the second phase is:

Fraction who passed phase II = 1-\frac{5}{7}=\frac{2}{7}

That is, 2/7 of the remaining applicants passed the second phase.

The fraction of the applicants who have passed the two phases is:

\frac{5}{9}\times \frac{2}{7}=\frac{10}{63}

Thus, the fraction of the applicants who have passed the two phases of the selection process is \frac{10}{63}.

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3 years ago
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schepotkina [342]

Answer:

Step-by-step explanation:

(n - p) -5

Substitute in -4 and 7

(-4 - 7) - 5

Combine in parenthesis

-11 - 5

Combine

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6 0
4 years ago
300=x^2+15x what is x?
Vlada [557]

Answer:

x = -26.3746, 11.3746

Step-by-step explanation:

Easiest and fastest way to solve for <em>x</em> is to graph x² + 15x - 300 and analyze where the graph crosses the x-axis.

Alternatively, you can use the Quadratic Formula to help solve for <em>x</em>.

6 0
3 years ago
A ball is dropped from a height of 10 meters. Each time it bounces, it reaches 50 percent of its previous height. The total vert
velikii [3]
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5/2=2.5
2.5/2=1.25
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3 0
3 years ago
You need two bottles of fertilizer to treat the flower garden shown. How many bottles do you need to treat a similar garden with
Scilla [17]

Answer:

26.25 bottles

Step-by-step explanation:

A flower garden in the form of trapezoid is shown. The length of longer base is labeled 18 feet, shorter base is 15 feet, height is 4 feet, and one leg is 5 feet.

Formula for perimeter of a trapezoid = Side a + side b + length a + length b

For the trapezium shown, its perimeter step 1,

we have to find the missing length

Length a

We are given height = 4 feet

Length of the longer base = 18 feet

Length of the shorter base = 15 feet

Different would give us the base of the right angle triangle

= 18 - 15 = 3 feet

Using Pythagoras Theorem we can find the side a = c

a² + b² = c²

3² + 4² = c²

9 + 16 = c²

c = √25

c = 5 feet

Hence side a = 5 feet

Perimeter of the Trapezoid shown =

18 feet + 15 feet + 5 feet + 5 feet

= 38 feet.

From the question,

You need two bottles of fertilizer to treat the flower garden shown. How many bottles do you need to treat a similar garden with a perimeter of 105 feet

Hence:

38 feet = 2 bottles

105 feet = x bottles

Cross y

= 38 × x = 105 × 2

= 38x = 105 × 2

x = 105 × 2/38

x = 26.25 bottles

26.25 bottles of fertilizer is needed to treat the garden with a perimeter of 105

3 0
3 years ago
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