How to find the domain for problem 42?
2 answers:
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Check the picture below.
so, as I see it, we can simply move from the point P, 360° counter-clockwise, the green one, and we'll be landing on a positive angle on the 1st Quadrant, the angle of 360° + 30° = 390°.
we can also move 360° clockwise, the red one, from P, so we'll be landing at -360° + 30 = -330°, also on the 1st Quadrant, but a negative counterpart.
now, if we move by jumps of 360° on either direction, we'll be landing coterminal with P all the time, so if we use simply multiply 360° by "n", some integer, then that angle added or subtracted from 30°, will be coterminal with P.
( 6 , 30°±360n°) where n ∈ ℤ.
Answer:
The probability, to the nearest 10th of a percent, that both marbles drawn will be red is 28.5%
Step-by-step explanation:
Number of red marbles = 4
Number of blue marbles = 8
Number of Green marbles = 2
Total number of marbles = 4+8+2 =14
We need to find the probability, to the nearest 10th of a percent, that both marbles drawn will be red?
The formula used will be:
In our case, Number of favourable outcomes = 4 (as red marbles are drawn)
Number of possible outcome = 14 (Total number of marbles)
The probability will be:
Now, calculating nearest 10th of a percent
So, the probability, to the nearest 10th of a percent, that both marbles drawn will be red is 28.5%