$16,744.00
Link: http://www.moneychimp.com/features/simple_interest_calculator.htm
Answer:
10
Step-by-step explanation:
j=6
k=4
Rewrite the equation. So let's do that-
2.5(6*4)/6
6 and the other 6 cancel out so we are left with 2.5*4
Which is then equal to 10.
By the way, whenever you see the same number in a division problem on the numerator and denominator, just cancel them out because if you still did 2.5(6*4)/6, you would still get 10. I was just simplifying it!
Answer:
Step-by-step explanation:
The question says,
A roulette wheel has 38 slots, of which 18 are black, 18 are red,and 2 are green. When the wheel is spun, the ball is equally likely to come to rest in any of the slots. One of the simplest wagers chooses red or black. A bet of $1 on red returns $2 if the ball lands in a red slot. Otherwise, the player loses his dollar. When gamblers bet on red or black, the two green slots belong to the house. Because the probability of winning $2 is 18/38, the mean payoff from a $1 bet is twice 18/38, or 94.7 cents. Explain what the law of large numbers tells us about what will happen if a gambler makes very many betson red.
The law of large numbers tells us that as the gambler makes many bets, they will have an average payoff of which is equivalent to 0.947.
Therefore, if the gambler makes n bets of $1, and as the n grows/increase large, they will have only $0.947*n out of the original $n.
That is as n increases the gamblers will get $0.947 in n places
More generally, as the gambler makes a large number of bets on red, they will lose money.
Answer:
I believe its the second one.
Step-by-step explanation:
Answer:
titutex=cos\alp,\alp∈[0:;π]
\displaystyle Then\; |x+\sqrt{1-x^2}|=\sqrt{2}(2x^2-1)\Leftright |cos\alp +sin\alp |=\sqrt{2}(2cos^2\alp -1)Then∣x+
1−x
2
∣=
2
(2x
2
−1)\Leftright∣cos\alp+sin\alp∣=
2
(2cos
2
\alp−1)
\displaystyle |\N {\sqrt{2}}cos(\alp-\frac{\pi}{4})|=\N {\sqrt{2}}cos(2\alp )\Right \alp\in[0\: ;\: \frac{\pi}{4}]\cup [\frac{3\pi}{4}\: ;\: \pi]∣N
2
cos(\alp−
4
π
)∣=N
2
cos(2\alp)\Right\alp∈[0;
4
π
]∪[
4
3π
;π]
1) \displaystyle \alp \in [0\: ;\: \frac{\pi}{4}]\alp∈[0;
4
π
]
\displaystyle cos(\alp -\frac{\pi}{4})=cos(2\alp )\dotscos(\alp−
4
π
)=cos(2\alp)…
2. \displaystyle \alp\in [\frac{3\pi}{4}\: ;\: \pi]\alp∈[
4
3π
;π]
\displaystyle -cos(\alp -\frac{\pi}{4})=cos(2\alp )\dots−cos(\alp−
4
π
)=cos(2\alp)…
1
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