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2 years ago

a plastic bag contains 5 marbles. two of the marbles are red and 3 of the marbles are blue. without looking, what is the probabi

lity of pulling out a blue marble and then another blue marble?(Assume that you don't put the first blue marble back in the bag) write your answer as a fraction.

Mathematics

1 answer:

Hitman42 [59]2 years ago

4 0

$|\Omega|=5\cdot4=20\\|A|=3\cdot2=6\\\\P(A)=\dfrac{6}{20}=\dfrac{3}{10}$

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A house has increased in value by 34% since it was purchased the current value is 469,000 what is the value when it was purchase

Rashid [163]

Answer:

$ 710606.606

Step-by-step explanation:

Let original price be x

x - 34% of x = 469000

=> X - 34x/100 = 469000

=> 100x -34x/100 = 469000

=> 66x/100 = 469000

=> 66x = 46900000

=> X = 46900000/66

=> X = 710606.606

8 0

3 years ago

Beatrix ate 21 raisins. This represents 30% of the box of raisins. How many raisins were originally in the box?

leonid [27]

21/0.3= 70
there were 70 raisins in the box

8 0

3 years ago

What is 18% of 700? Enter your answer in the box.

ella [17]

18% of 700 is 126 .18x700=126

4 0

3 years ago

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I need help matching these

Ber [7]

Answer:

1. 4

2. 64

3. 16

4. 1

5. 2

6 12

Step-by-step explanation:

4 0

2 years ago

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Given: ΔABC ≅ ΔEDF

V125BC [204]

we know that

the triangle ABC and triangle EDF are congruent triangles-----> given problem

therefore

$AC=EF\\AB=ED\\BC=DF$

we know that

the distance between two points is equal to

$d=\sqrt{(y2-y1)^{2}+(x2-x1)^{2}}$

Step $1$

Find the distance AB

$A(0,1)\\B(0,2)$

substitute in the formula of the distance

$d=\sqrt{(2-1)^{2}+(0-0)^{2}}$

$d=\sqrt{(1)^{2}+(0)^{2}}$

$dAB=1\ unit$

Step $2$

Find the distance BC

$B(0,2)\\C(3,2)$

substitute in the formula of the distance

$d=\sqrt{(2-2)^{2}+(3-0)^{2}}$

$d=\sqrt{(0)^{2}+(3)^{2}}$

$dBC=3\ units$

Step $3$

Find the distance AC

we know that

the triangle ABC is a right triangle

Applying the Pythagorean Theorem

$AC^{2}=AB^{2} +BC^{2}$

substitute the values in the formula

$AC^{2}=1^{2} +3^{2}$

$AC=\sqrt{10}\ units=3.16\ units$

round to the nearest tenth

$AC=3.2\ units$

therefore

$EF=3.2\ units$

$ED=1\ unit$

$DF=3\ units$

the answer is

The length of the hypotenuse is equal to $3.2\ units$

4 0

3 years ago

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