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Vaselesa [24]
2 years ago
9

There are 50 students in an auditorium, of which 2x are boys and y are girls. After (y - 6) boys leave the auditorium and (2x -

5) girls enter the auditorium, the probability of selecting a girl at random becomes 9/13. Find the value of x and of y?
Mathematics
1 answer:
Dmitrij [34]2 years ago
7 0
Total number of students=50
Number of boys=2x
Number of girls=y
total will be:
2x+y=50
⇒y=50-2x

when (y-6) boys left the auditorium the new number of boys was:
2x-(y-6)
=2x-y+6
but y=50-2x
thus the new number will be:
2x-(50-2x)+6
=4x-44

when (2x-5) girls left the auditorium the remaining number will be:
y-(2x-5)
=y-2x+5
but 
y=50-2x
thus the new number of girls will be:
50-2x-2x+5
=55-4x
new total number of students:
(55-4x)+(4x-44)
=11

probability of selecting a girl at random will be:
(55-4x)/11=9/13
13(55-4x)=9*11
715-52x=99
616=52x
x=12
thus
y=50-12=38
thus
x=12 and y=38


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point M with coordinates (3,4) is the midpoint of the line AB and A has the point (-1,6). what is the point of B?
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_____

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5 0
1 year ago
Suppose that 4% of the 2 million high school students who take the SAT each year receive special accommodations because of docum
dangina [55]

Answer:

a. 0.0122

b. 0.294

c. 0.2818

d. 30.671%

e. 2.01 hours

Step-by-step explanation:

Given

Let X represents the number of students that receive special accommodation

P(X) = 4%

P(X) = 0.04

Let S = Sample Size = 30

Let Y be a selected numbers of Sample Size

Y ≈ Bin (30,0.04)

a. The probability that 1 candidate received special accommodation

P(Y = 1) = (30,1)

= (0.04)¹ * (1 - 0.04)^(30 - 1)

= 0.04 * 0.96^29

= 0.012244068467946074580191760542164986632531806368667873050624

P(Y=1) = 0.0122 --- Approximated

b. The probability that at least 1 received a special accommodation is given by:

This means P(Y≥1)

But P(Y=0) + P(Y≥1) = 1

P(Y≥1) = 1 - P(Y=0)

Calculating P(Y=0)

P(Y=0) = (0.04)° * (1 - 0.04)^(30 - 0)

= 1 * 0.96^36

= 0.293857643230705789924602253011959679180763352848028953214976

= 0.294 --- Approximated

c.

The probability that at least 2 received a special accommodation is given by:

P (Y≥2) = 1 -P(Y=0) - P(Y=1)

= 0.294 - 0.0122

= 0.2818

d. The probability that the number among the 15 who received a special accommodation is within 2 standard deviations of the number you would expect to be accommodated?

First, we calculate the standard deviation

SD = √npq

n = 15

p = 0.04

q = 1 - 0.04 = 0.96

SD = √(15 * 0.04 * 0.96)

SD = 0.758946638440411

SD = 0.759

Mean =np = 15 * 0.04 = 0.6

The interval that is two standard deviations away from .6 is [0, 2.55] which means that we want the probability that either 0, 1 , or 2 students among the 20 students received a special accommodation.

P(Y≤2)

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=.

P(0) + P(1) = 0.0122 + 0.294

Calculating P(2)

P(2) = (0.04)² * (1 - 0.04)^(30 - 2(

P(2) = 0.00051

So,

P(0) +P(1) + P(2). = 0.0122 + 0.294 + 0.00051

= 0.30671

Thus it 30.671% probable that 0, 1, or 2 students received accommodation.

e.

The expected value from d) is .6

The average time is [.6(4.5) + 19.2(3)]/30 = 2.01 hours

8 0
2 years ago
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