maybe bc of how many times you can add the number itself
Answer:
0
Step-by-step explanation:
any log with a base of one and it becomes logv5 (1) after logv5(logV3 (3) because log3(3) equal one so then logv5 (1) is 0
<h3>Answer: y = (3/2)x + 0</h3>
This is the same as y = (3/2)x
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Work Shown:
Find the slope of the line through (x1,y1) = (-2,-3) and (x2,y2) = (2,3)
m = (y2 - y1)/(x2 - x1)
m = (3 - (-3))/(2 - (-2))
m = (3 + 3)/(2 + 2)
m = 6/4
m = 3/2
The slope is the fraction 3/2. This is going to be in front of the x, or to the left of the x.
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Plug m = 3/2 and (x1,y1) = (-2,-3) into the point slope formula. Solve for y.
y - y1 = m(x - x1)
y - (-3) = (3/2)(x - (-2))
y + 3 = (3/2)(x + 2)
y + 3 = (3/2)x + (3/2)(2)
y + 3 = (3/2)*x + 3
y + 3 - 3 = (3/2)*x + 3 - 3
y = (3/2)x + 0
The y intercept is zero. This matches up with the fact the graph crosses the y axis at y = 0.
Answer:
16.7%.
Step-by-step explanation:
There are initially
pencils in the bag.
Take a pencil out of this bag of 47 pencils. 15 out of the 47 pencils blue. Let
represent the event of getting a blue pencil on the first pick. The probability of getting a blue pencil is:
.
There are now
pencils left in the bag. However, given that the first pencil removed from the bag is blue, the number of red pencils in the bag will still be 24. Take another pencil out of this bag of 46 pencils. Let
represent the event of getting a red pencil on the second pick. The possibility that the second pencil is red given that the first pencil is blue will be:
.
The question is asking for the possibility that the first pencil is blue and the second pencil is red. That is:
.