The bag contains,
Red (R) marbles is 9, Green (G) marbles is 7 and Blue (B) marbles is 4,
Total marbles (possible outcome) is,
Let P(R) represent the probablity of picking a red marble,
P(G) represent the probability of picking a green marble and,
P(B) represent the probability of picking a blue marble.
Probability , P, is,
Probablity of drawing a Red marble (R) and then a blue marble (B) without being replaced,
That means once a marble is drawn, the total marbles (possible outcome) reduces as well,
Hence, the best option is G.
Answer:
less than/equal to -2
Or, greater than/equal to 2.5
Step-by-step explanation:
2x² - x - 10 ≥ 0
2x² - 5x + 4x - 10 ≥ 0
x(2x - 5) + 2(2x - 5) ≥ 0
(x + 2)(2x - 5) ≥ 0
Roots: -2, 5/2
x《 -2 or x》5/2
Answer:
as the cycle continues, you will always get 50 as the answer, since you are dividing both the numerator and the denominator with 10 in each expression. In other words, the 10:00 you divide the numerator which cancels out with the 10 you divide the denominator with. So the answer will always be 50 in each next expression.
Step-by-step explanation:
(I) 45÷90= 4500/90 = 50
(ii) 450÷9= 450/9 =50
(iii) 45÷0.9 = 45/0.9=50
(iv) 4.5 ÷ 0.09 = 45/0.09
starting with (I) 4500÷90 = 4500/90 =50
you divide the numerator and the denominator by 10 to get ; (ii) 450÷9 = 450/9 = 50
as the cycle continues you will always get 50 as the answer since you are dividing both the numerator and the denominator with 10 in each expression. In other words, the 10:00 you divide the numerator with cancels out with the 10 you divide the denominator with. So the answer will always be 50.
Answer:
(5,3)
I'm not soooo sure I'm %50 sure
9514 1404 393
Answer:
59. t ≥ -1
63. x ≤ -1/2
67. (x ≤ 0) ∪ (x ≥ 6)
71. x > 1/2
Step-by-step explanation:
The domain is the set of values of x where the function is defined. In rational functions it excludes any values of x that make the denominator zero. When even-index roots are involved, it excludes values of x that make the argument of the root function negative.
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59. t+1 ≥ 0 ⇒ t ≥ -1
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63. 1 -2x ≥ 0 ⇒ x ≤ -1/2
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67. x² -6x ≥ 0 ⇒ x(x -6) ≥ 0 ⇒ (x ≤ 0) ∪ (x ≥ 6)
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71. 2x -1 > 0 ⇒ x > 1/2 . . . . note we left off the =0 case for this one
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The attachment is a plot of the g(x) in 67. You can see that it is undefined for 0 < x < 6.
As you can see, a graphing calculator can be helpful in identifying the domain. Of course, any vertical asymptotes are excluded, as are any "holes" in a function.
