All categories

2 years ago

if you place these marbles in a bag, close your eyes, and choose a marble, what is the probability that it will be

Mathematics

1 answer:

oksian1 [2.3K]2 years ago

8 0

Answer:

P(blue) = 3/7

So '7' belongs in the box

Step-by-step explanation:

There are 14 total marbles, 6 of which are blue. The probability of pulling a blue marble out of the bag is

P(blue) = 6/14, which reduces to 3/7

You might be interested in

Which of the following is an example of the atmosphere interacting with the biosphere?

Vilka [71]

I’d say option 4

However, one may think option 3 might be possible as well. But that it is not interacting, it’s affecting.

So I’d stick with option 4.

Good luck!

6 0

3 years ago

Read 2 more answers

How do I do this <br> X+y=-4<br> 3x-6y=15

avanturin [10]

Answer:

x=-1, y=-3. (-1, -3).

Step-by-step explanation:

x+y=-4

3x-6y=15

---------------

x=-4-y

3(-4-y)-6y=15

-12-3y-6y=15

-12-9y=15

9y=-12-15

9y=-27

y=-27/9

y=-3

x+(-3)=-4

x-3=-4

x=-4+3

x=-1

8 0

3 years ago

An equation and the steps to solve it are shown below:

lisov135 [29]

Answer:

message me and i will tell u

Step-by-step explanation:

5 0

2 years ago

Notebook paper is on sale for 50% off a package regularly sells for $0.58 a customer made by only two packages at the sale price

blagie [28]

Richard had to pay 0.87

7 0

2 years ago

looks out from the crown of the Statue of Liberty, approximately 250 ft above ground. The tourist sees a ship coming into the ha

Zigmanuir [339]

Answer:

The distance from the base of the statue to the ship is 771.60 ft

Step-by-step explanation:

Refer the attached figure

Height of statue AB= 250 feet

The tourist sees a ship coming into the harbor and measures the angle of depression as 18°.

So, ∠ACB = 18°

We are supposed to find the distance from the base of the statue to the ship i.e. BC

In ΔABC

$Tan \theta = \frac{Perpendicular}{Base}$

$Tan 18^{\circ}=\frac{AB}{BC}$

$Tan 18^{\circ}=\frac{250}{BC}$

$BC=\frac{250}{Tan 18^{\circ}}$

$BC=\frac{250}{0.324}$

BC=771.60 ft

Hence the distance from the base of the statue to the ship is 771.60 ft

3 0

3 years ago