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ozzi
3 years ago
15

A die is loaded in such a way that an even number is twice as likely to occur as an odd number. If E is the event that a number

less than 4 occurs on a single toss of the die, find P(E).
Mathematics
2 answers:
Annette [7]3 years ago
8 0

Answer:

P(E)=4/9

Step-by-step explanation:

We have six different numbers in a die, i.e., 1, 2, 3, 4, 5, 6. The probability of each sample point should be equal or greater than zero. Let's suppose that x is the probability that an odd number occurs, so, the probability that an even number occurs is 2x, because the die is loaded in such a way that an even number is twice as likely to occur as an odd number. Besides, the sum of the probability associated to an odd number with the probability associated to an even number should be equal to 1/3, because the sum of the probabilities of all sample points should be equal to 1 and we have three odd numbers and three even numbers in a die. Therefore x+2x=1/3, i.e., 3x=1/3, x = 1/9, i.e., the probability that an odd number occurs is 1/9 and the probability that an even number occurs is 2/9. E is the event that a number less than 4 occurs on a single toss of the die, so, P(E) = P(1) + P(2) + P(3) = 1/9 + 2/9 + 1/9 = 4/9.

elena55 [62]3 years ago
5 0

Answer:

\frac{4}{9}

Step-by-step explanation:

Let the probability of an even number occurring be p and that of odd be q. From the question, p = 2q.

Now, p and q are mutually exclusive events. Hence,

p + q = 1

2q + q = 1

3q = 1

q = 1/3

p = 2q = 2/3

There are 3 odd numbers and 3 even numbers on a die. The probability of each even number is p/3 = 2/9 and the probability of each odd number is q/3 = 1/9.

The event E is the event of a 3, 2 or 1 appearing. Here, we have two odd numbers and 1 even number.

P(E) = 2\times\frac{1}{9} + 1\times\frac{2}{9} = \frac{4}{9}

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Phoenix [80]

Answer: The intersection region shown in the graph attached is the solution of the system of inequalities.

Step-by-step explanation:

The equation of the line in Slope-Intercept form is:

y=mx+b

Where "m" is the slope and "b" is the y-intercept.

 Given the following system of inequalities:

\left \{ {{x+3y\leq-3 } \atop {x\geq3 }} \right.

We need to graph the lines.

We have the line:

x+3y=-3

Solving for "y", we get:

3y=-x-3\\\\y=\frac{1}{3}x-1

Knowing that m=-\frac{1}{3} and b=-1, we can graph it.

Since the symbol is \leq, the line must be solid.

The other line is:

x=3

Since "x" only takes one value, we can identify that it is a vertical line, so we can graph it.

Since the symbol is \geq, the line must be solid.

We need to substitute a point that is not on the line, into each inequalities. Let's choose the point (0,0).

Then:

0+3(0)\leq-3\\\\0\leq-3

This is false, therefore, we must shade the half that does not contain this point.

0\geq 3

This is false, so we must shade the half that does not contain this point.

The intersection region of the solutions (Observe the graph attached) of the given inequalities is the solution of the system of inequalities.

8 0
3 years ago
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olga2289 [7]
This can be factored into (2t - 9)(t + 7)
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3 years ago
What three points are solutions to 2x-y =4
DochEvi [55]

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Attached file

Step-by-step explanation:

5 0
2 years ago
If a standard 6 sided dice is rolled. (Write your answers as a decimal, rounded to the nearest hundredth.) Find: a. Probability
pshichka [43]

Answer:

see below

Step-by-step explanation:

The possible outcomes are 1,2,3,4,5,6

P( a prime number and a multiple of two)

The only prime number that is a multiple of 2 is 2

P( a prime number and a multiple of two) = number of outcomes / total outcomes

P( a prime number and a multiple of two) = 1/6 = .17

P(3 or a prime number)

The prime numbers are 2,3,5  and 3 is included in this list so we have 3 outcomes

P( 3 or a prime number) = number of outcomes / total outcomes

P( 3 or a prime number) =3/6 =1/2= .5

P(3 U Multiple of 2)

Multiples of 2 are 2,4,6 and add 3 to the list  so there are 4 outcomes

P( 3 U Multiple of 2) = number of outcomes / total outcomes

P( 3 or a prime number) =4/6 =2/3= .67

P(2 U Even number)

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4 0
3 years ago
Three times as many children as adults attended a concert on Saturday. An adult ticket cost $7 and a child’s ticket cost $3. The
Ierofanga [76]

For this case we propose a system of equations:

x: Let the variable representing the number of children in the concert

y: Let the variable representing the number of adults at the concert

According to the assistance we have:

x = 3y

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3x + 7y = 6000

Substituting the first in the second equation we have:

3 (3y) + 7y = 6000\\9y + 7y = 6000\\16y = 6000\\y = \frac {6000} {16}\\y = 375

Thus, the concert was 375 adults.

On the other hand we have:

x = 3 (375) = 1125

Thus, the concert was 1125 children.

In total they were:

375 + 1125 = 1500 people

ANswer:

The concert was 1500 people

3 0
3 years ago
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