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ycow [4]
3 years ago
13

A bag contains 7 red marbles, 3 blue marbles and 2 green. If two marbles are drawn out of the bag, what is the probability, to t

he nearest 1000th, that both marbles drawn will be blue?
Mathematics
1 answer:
Setler [38]3 years ago
5 0

Answer:

0.0455

Step-by-step explanation:

<u>The formula for calculator probability is </u>

# of favorable outcomes / # of possible outcomes

<u>The total number of marbles in the bag: </u>

7+3+2=12

<u>Probability of the first marble being blue:</u>

3/12

<u>Probability of the second marble being blue:</u>

2/11 -- 2 because only 2 blue marbles left after drawing the first one out, 11 because after taking the first marble out, we have 11 marbles left in the bag

<u>Formula for both events happening:</u>

p(A and B) = p(A) * p(B)

In this problem,

A = first one blue

B = second one blue

<u>Probability that if two marbles are drawn out of the bag, both are blue:</u>

3/12 x 2/11 = 0.045454

<u>Rounded answer:</u>

0.0455 (or 4.55%)

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3 less than g is greater than 17
lawyer [7]

Step-by-step explanation:

what do you need to create ?

g - 3 > 17

g > 20

valid for all g in the interval (20, +infinity) with both ends excluded.

that is all I can see for your very limited problem description.

6 0
2 years ago
Two friends sold many pieces of furniture and made ​$1550 during their garage sale. They had fifteen more​ $10 bills than​ $50 b
Novosadov [1.4K]

Answer:

The number of $ 10 bills is 37, the number of $ 20 is 6 and the number of $ 50 bills is 22.

Step-by-step explanation:

Let n₁ = number of $ 10 bills,

Let n₂ = number of $ 20 bills and

Let n₃ = number of $ 50 bills

Given that 10n₁ + 20n₂ + 50n₃ = 1550    (1)

Also, since we have fifteen more​ $10 bills than​ $50 bills, n₁ = n₃ + 15  (2)

and we have 4 more than three times as many​ $20 bills as​ $50 bills. n₃ = 3n₂ + 4. (3)

substituting equations (2) and (3) into (1), we have

10(n₃ + 15) + 20n₂ + 50n₃ = 1550

expanding the bracket, we have

10n₃ + 150 + 20n₂ + 50n₃ = 1550

collecting like terms, we have

60n₃ + 150 + 20n₂ = 1550

inserting equation (3), we have

60(3n₂ + 4) + 150 + 20n₂ = 1550

expanding the bracket, we have

180n₂ + 240 + 150 + 20n₂ = 1550

collecting like terms, we have

200n₂ + 390 = 1550

subtracting 390 from both sides, we have

200n₂ = 1550 - 390

200n₂ = 1160

dividing both sides by 200, we have

n₂ = 1160/200

n₂ = 5.8

n₂ ≅ 6 since it cannot be a fraction.

Substituting this into (3), we have

n₃ = 3n₂ + 4 = 3(6) + 4 = 18 + 4 = 22

substituting n₃ into (2), we have

n₁ = n₃ + 15 = 22 + 15 = 37

So, the number of $ 10 bills is 37, the number of $ 20 is 6 and the number of $ 50 bills is 22.

3 0
3 years ago
Solve for m.<br> 4+ 7m=5
Simora [160]
First you would isolate (7m) by subtracting 4 from both sides:
4 + 7m - 4 = 5 - 4
This would leave you with:
7m = 1
Then you would isolate m by dividing both sides by 7:
7m/7 = 1/7
Your final answer:
m = 1/7
3 0
3 years ago
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F(x)=1/3x+7 find inverse
jeka94

Answer:

Step-by-step explanation:

8 0
3 years ago
Simplify and write the trigonometric expression in terms of sine and cosine:
asambeis [7]

Answer:

we have the expression as;

1/sin u cos u

Step-by-step explanation:

tan u = sin u/cos u

cot u = cos u/sin u

Thus;

sin u/cos u + cos u/sin u

The lcm is sin u cos u

Thus, we have that;

(sin^2 u + cos^2 u)/sin u cos u

But ; sin^2 u + cos^2 u = 1

so we have ;

1/sin u cos u

4 0
3 years ago
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