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:
Step-by-step explanation:
cos 43=cos (90-47)=sin 47
Step-by-step explanation:
- Let
be the smallest integer
- Let
be the larger integer
As the sum of the reciprocals of the two positive integers is frac 3/16
So,



Simplify






So,

<h2>Verification:</h2>






Therefore,

Keywords: word problem
, integer
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Example of Roots of Complex Number
(64)1/6={2cis(60k)∘} for k=0,1,2,3,4,5.
Answer:
$189.75
Step-by-step explanation:
Whenever you want to find the percent of something, you need to turn the percent into a decimal. In this case, we need to find 115% of the wholesale price because there is a markup, so the cost will be higher.
115% as a decimal is 1.15. Here is how the equation will be set up:
$165 • 1.15 = $189.75
No rounding is necessary for this problem because it is already to the nearest cent.