1 3/9 but if reduce to lowest term than 1 1/3
Answer:
c = 32/3
Step-by-step explanation:
Eliminate the fraction by mult. all terms by 8: 96 + 3c = 64.
Subtract 64 from both sides, obtaining: 3c = 32. Solving for c, c = 32/3.
Answer:
D
Step-by-step explanation:
Say the short leg is x and the hypotenuse is y. Then, y = 3x (because "The length of the hypotenuse of a right triangle is three times the length of the shorter leg.")
Use the Pythagorean Theorem: a^2 + b^2 = c^2 (where a and b are the legs and c is the hypotenuse). Here, we can say a = x and b = 12 and c = y. So:
x^2 + 12^2 = y^2
x^2 + 144 = (3x)^2
x^2 + 144 = 9x^2
8x^2 = 144
x^2 = 18
x = 
The answer is D.
Hope this helps!
There are 9 marbles in the bag. We pick 2 without replacement and get a probability of 1/6.
Each draw of a marble has a probability associated with it. Multiplying these gives 1/6 so let us assume the probabilities are (1/3) and (1/2).
In order for the first draw to have a probability of 1/3 we need to draw a color that has (1/3)(9)=3 marbles. So let's say there are 3 red marbles. The P(a red marble is drawn) = 1/3.
Now that a marble has been drawn there are 8 marbles left. In order for the second draw to have a probability of 1/2 we must draw a color that has (1/2)(8) = 4 marbles. So let's say there are 4 blue marbles out of the 8.
Since there are 9 marbles to start and we have 3 red marbles and 4 blue marbles, the remaining 2 marbles must be a different color. Let us say they are green.
The problem is: There are 3 red marbles, 4 blue marbles and 2 green marbles in a jar. A marble is picked at random, it's color is noted and the marble is not replaced. A second marble is drawn at random and its color noted. What is the probability that the first marble is red and the second blue?
