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oksano4ka [1.4K]

3 years ago

14

A bag contains 3 green marbles and 8 white marbles. Suppose a marble is randomly selected. What are the odds in favor of picking

a white marble?

2 answers:

Juli2301 [7.4K]3 years ago

5 0

8/11 or 72.7272727272%

KatRina [158]3 years ago

4 0

It would be 7 out of 11 because there are 8 marbles and you only took out one white marble

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nlexa [21]

wish i did, i'm bored as well

3 0

2 years ago

Read 2 more answers

A movie theater has a seating capacity of 249. The theater charges $5.00 for children, $7.00 for students, and $12.00 of adults.

Sonbull [250]

9514 1404 393

Answer:

126 children
63 adults
60 students

Step-by-step explanation:

Let a represent the number of adults that attended. Then 2a is the number of children, and (249 -a -2a) is the number of students. The total revenue is ...

5(2a) +7(249 -3a) +12a = 1806

a +1743 = 1806 . . . . simplify; next, subtract 1743

a = 63 . . . . . . . . adults

2a = 126 . . . . . . children

249 -3a = 60 . . students

126 children, 63 adults, and 60 students attended.

4 0

3 years ago

find the value of x show your work in the diagram below C is the midpoint of segment a b, a c equals 5x + 2 + CB equals 3x + 12

Dmitrij [34]

Answer:

Initial equation: 5x + 2 = 3x + 12

x = 5

Step-by-step explanation:

Point C is the midpoint between A and B.

This means that:

AC = CB

We have that:

AC = 5x + 2

CB = 3x + 12

Since AC = CB, the initial equation is:

5x + 2 = 3x + 12

Now we can solve

5x + 2 = 3x + 12

5x - 3x = 12 - 2

2x = 10

x = 10/2

x = 5

8 0

10 months ago

A, B and C are events such that P(A) = 1/3, P(B) = ¼, and P(C) = 1/5. Find P(AUBUC) under each of the following assumptions: a)

marta [7]

(a) If $A,B,C$ are mutually exclusive, then

$P(A\cap B)=P(A\cap C)=P(B\cap C)=P(A\cap B\cap C)=0$

so we have

$P(A\cup B\cup C)=P(A)+P(B)+P(C)=\dfrac{47}{60}$

(b) If $A,B,C$ are mutually independent, then

$P(A\cap B)=P(A)P(B),$

$P(A\cap C)=P(A)P(C),$

$P(B\cap C)=P(B)P(C),$

$P(A\cap B\cap C)=P(A)P(B)P(C)$

so that

$P(A\cup B\cup C)=P(A)+P(B)+P(C)-(P(A\cap B)+P(A\cap B)+P(B\cap C))+P(A\cap B\cap C)$

$P(A\cup B\cup C)=\dfrac{47}{60}-\left(\dfrac1{12}+\dfrac1{15}+\dfrac1{20}\right)+\dfrac1{60}$

$P(A\cup B\cup C)=\dfrac35$

5 0

3 years ago

What is 4/5 of 500? Round it to the nearest hundredth

vaieri [72.5K]

To be honest i have no idea what the answer is but do you know a sentence for the word scientific theory

4 0

2 years ago