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Ronch [10]
3 years ago
8

En un curso hay 11 mujeres y 13 hombres. Si se eligen dos alumnos para representarlos en el campeonato de debate, ¿cuál es la pr

obabilidad de que ambos sean hombres?
Mathematics
1 answer:
Tresset [83]3 years ago
8 0

Answer:

106.34%  

Step-by-step explanation:

hay 24 personas en total (11 mujeres+13 hombres)

la probabilidad para que uno de los alumnos es un hombre es 13/24 (porque hay 13 hombres en la clase)

Entonces, si uno hombre está elegido, quedan 12 hombres, así que, la probabilidad si es un hombre es 12/23 (porque una persona ya está elegido y quedan 23 personas)

la probabilidad es 13/24+12/23=587/552=106.34%  (Redondeado a la décima más cercana)

(lo siento mi español no es perfecto)

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(URGENT) Need this for today!​
Goryan [66]

Answer:

5.7in

Step-by-step explanation:

using Pythagorean theorem

c^2=a^2+b^2

c=9in ,a=7in,b=xin

9^2=7^2+x^2

81=49+x^2

x^2=81-49

x^2=32

x=√32

x=5.7in(to the nearest tenth)

6 0
3 years ago
to the nearest tenth, what is the length of the hypotenuse of an isosceles right triangle with a leg of 7 square root of 3 inche
hoa [83]

since its isosceles its a 45,45,90 right triangle which means the sides are in the proportion of x,x,x root 2

since the leg is 7 root 3 that means the hypotenuse is 7 root 6

8 0
3 years ago
What does "PEMDAS" stand for?
Fed [463]
P= parentheses or ()
E= exponents such as x^{2}
M= multiplying
D= division
A= Addition
S= Subtraction

Another popular acronym is BEDMAS, which is pretty much the same except for division and multiplication are switched and it uses the word Brackets to describe these: (). 
8 0
3 years ago
Read 2 more answers
How many subsets of {1, 2, 3, 4, 6, 8, 10, 15} are there for which the sum of the elements is 15?
stepladder [879]

Answer:

512

Step-by-step explanation:

Suppose we ask how many subsets of {1,2,3,4,5} add up to a number ≥8. The crucial idea is that we partition the set into two parts; these two parts are called complements of each other. Obviously, the sum of the two parts must add up to 15. Exactly one of those parts is therefore ≥8. There must be at least one such part, because of the pigeonhole principle (specifically, two 7's are sufficient only to add up to 14). And if one part has sum ≥8, the other part—its complement—must have sum ≤15−8=7

.

For instance, if I divide the set into parts {1,2,4}

and {3,5}, the first part adds up to 7, and its complement adds up to 8

.

Once one makes that observation, the rest of the proof is straightforward. There are 25=32

different subsets of this set (including itself and the empty set). For each one, either its sum, or its complement's sum (but not both), must be ≥8. Since exactly half of the subsets have sum ≥8, the number of such subsets is 32/2, or 16.

8 0
4 years ago
Please help me "solve the system"&lt;br /&gt;&lt;br /&gt;<img src="https://tex.z-dn.net/?f=x%20%5E%7B2%7D%20%20%2B%20y%20%3D%207
SVEN [57.7K]

Equation 1: x² + y = 7

Equation 2: 3x + 7 = 9

Use Equation 2 to solve for "x" and plug in "x" into Equation 1 to solve for "y".

3x + 7 = 9

3x = 2

x = \frac{2}{3}

*********************************

x² + y = 7

(\frac{2}{3})² + y = 7

\frac{4}{9} + y = 7

y = 7 - \frac{4}{9}

y = \frac{63}{9} - \frac{4}{9}

y = \frac{59}{9}

Answer: (\frac{2}{3}, \frac{59}{9})




3 0
4 years ago
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