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slava [35]
3 years ago
7

Evaluate the following line integral. ModifyingBelow Integral from nothing to nothing With Upper C xy font size decreased by 5 d

s​; C is the portion of the unit circle Bold r (s )equals left angle cosine s comma sine s right angle​, for 0 less than or equals s less than or equals StartFraction 7 pi Over 6 EndFraction

Mathematics
1 answer:
Tema [17]3 years ago
6 0

Answer:

1/2

Step-by-step explanation:

Please see attachment .

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Sam bought a new couch on sale at the discount furniture store. The original cost of the couch was $550. The sale price was $449
dolphi86 [110]
30%x550=165
550-165=385
385+99=385.00
4 0
3 years ago
Need help!
omeli [17]
Rational numbers: B C D G
Irrational numbers: A E F
Irrational numbers are numbers that cannot be written as fractions, rational numbers are all other numbers
5 0
4 years ago
Find the interest earned: <br><br> $11,500 for 3.5 years at 7.25% annual simple interest
Otrada [13]

Answer:

Future Value, using...

 Simple Interest: $  

14,418.13

 Annually Compounded Interest: $  

14,692.25

Step-by-step explanation:

8 0
3 years ago
Please do right and show work
Elis [28]

Let's thing

949 ----- 100% of the price

499 ------ x% of the price

Multiply in cross

949.x = 100.499

949x = 49900

x = 49900/949

x = 52,58%

So, rounding it 52,58% ≈ 53%

Alternative B

5 0
3 years ago
Read 2 more answers
The probability of winning on a single toss of the dice is p. A starts, and if he fails, he passes the dice to B, who then attem
alekssr [168]

Answer:

The probability of A winning is \frac{1}{2-p} and the probability of B winning is \frac{1-p}{2-p}.

Step-by-step explanation:

The probability of winning is <em>p</em>.

The game stops when either A or B wins.

The sample space of A winning is as follows:

{S, FFS, FFFFS, FFFFFFS, ...}

The sample space of B winning is as follows:

{FS, FFFS, FFFFFS, FFFFFFFS, ...}

Compute the probability of <em>A</em> winning as follows:

P (A winning) = P (S) + P (FFS) + P (FFFFS) + ...

                      =p+(1-p)(1-p)p+(1-p)(1-p)(1-p)(1-p)p+...\\=p\sum\limits^{\infty}_{i=0} {(1-p)^{2i}}\\=p\times \frac{1}{1-(1-p)^{2}}\\=\frac{p}{1-(1-p)^{2}}\\=\frac{p}{1-1-p^{2}+2p}\\=\frac{1}{2-p}

Compute the probability of <em>B</em> winning as follows:

P (B winning) = P (FS) + P (FFFS) + P (FFFFFS) + ...

    =(1-p)p+(1-p)(1-p)(1-p)p+(1-p)(1-p)(1-p)(1-p)(1-p)p+...\\=(1-p)p\sum\limits^{\infty}_{i=0} {(1-p)^{2i}}\\=(1-p)p\times \frac{1}{1-(1-p)^{2}}\\=\frac{(1-p)p}{1-(1-p)^{2}}\\=\frac{(1-p)p}{1-1-p^{2}+2p}\\=\frac{1-p}{2-p}

Thus, the probability of A winning is \frac{1}{2-p} and the probability of B winning is \frac{1-p}{2-p}.

3 0
3 years ago
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