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Mathematics 17 Online
OpenStudy (anonymous):

A jar contains 14 nickels, 10 dimes, 6 quarters, and 22 pennies. A coin is chosen at random from the jar. What is the probability that the coin chosen is a dime?

OpenStudy (anonymous):

10/52

OpenStudy (strawberry17):

Add all of them together (14+10+6+22=52). There are 10 dimes, so it would be a probability of 10/52.

OpenStudy (anonymous):

but am been told that it could be 5/21,1/21,5/26,1/26

OpenStudy (strawberry17):

Your right, I forgot, you also have to simplify 10/52.

OpenStudy (strawberry17):

You would do that, by dividing the numerator and the denominator both by 2. That would give you 5/26. (10/2=5, 52/2=26)

OpenStudy (strawberry17):

Do you get it?

OpenStudy (anonymous):

would it be 5 over 26

OpenStudy (strawberry17):

yes.. 5/26 means the same thing as 5 over 26. :)

OpenStudy (anonymous):

thank u been working on that for 20min.

OpenStudy (strawberry17):

your welcome

OpenStudy (anonymous):

i got one more if u can help

OpenStudy (anonymous):

A triangle is graphed in the coordinate plane. The vertices of the triangle have coordinates (3, 1), (1, 1), and (1, 2). What is the perimeter of the triangle?

OpenStudy (strawberry17):

do you what the distance formula is?

OpenStudy (strawberry17):

It is: d = √(x2 - x1)^2 + (y2 - y1)^2

OpenStudy (strawberry17):

You would use this formula to find the perimeter of each side of the triangle.

OpenStudy (strawberry17):

Let name the sides: A: (3, 1) B: (1, 1) C: (1, 2) So first you would find side AB. You would put in the values into the distance formula.

OpenStudy (strawberry17):

The coordinates of A are x1 and y1 and the coordinates of B are x2 and y2. So: x1 = 3, x2 = 1, y1 = 1, y2 = 1 d = √(x2 - x1)^2 + (y2 - y1)^2 d = √(1 - 3)^2 + (1 - 1)^2 d = √(-2)^2 + (0)^2 d = √(4) + (0) d = √4 d = 2 So the distance of AB is 2 units.

OpenStudy (strawberry17):

the distance equals the perimeter

OpenStudy (strawberry17):

Next, you would find the distance of the next side, which is BC. Do you understand so far...?

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