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patriot [66]
2 years ago
13

Which type of option helps the importer to hedge risks of exchange rate? A. Put option B. All names are not correct c. Call opti

on d. Call option and put option
Mathematics
1 answer:
dalvyx [7]2 years ago
5 0

Answer:

Call option and put option ( D )

Step-by-step explanation:

During hedging in stock/financial markets both the Call and put option can be used to hedge the trading position of the trader against the change in exchange. This is because the call or put option is used depending on the initial position of the trader.

<em>Call option is used when the trader is currently holding a short position</em>

<em>Put option is used when the trader is currently holding a long position</em>

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GIVING BRAINLIEST AND 20 points!
mestny [16]

Answer:

The answer is 27

Step-by-step explanation:

Step 1: Simplify both sides of the equation.

2(3y+6+3)=196−16

(2)(3y)+(2)(6)+(2)(3)=196+−16(Distribute)

6y+12+6=196+−16

(6y)+(12+6)=(196+−16)(Combine Like Terms)

6y+18=180

6y+18=180

Step 2: Subtract 18 from both sides.

6y+18−18=180−18

6y=162

Step 3: Divide both sides by 6.

6y

/6

=

162

/6

y=27

3 0
2 years ago
Subtract. 3- (-5) - (-2) Plot the solution on the number line.​
Whitepunk [10]

Answer:

10. put a dot on the number 10

Step-by-step explanation:

3-(-5)-(-2)

when subtracting negatives, you actually end up adding

so

3+5+2=10

3 0
2 years ago
Read 2 more answers
Find the derivative of ln(cosx²).​
RoseWind [281]

Answer:

\displaystyle \frac{dy}{dx} = -2x \tan (x^2)

General Formulas and Concepts:

<u>Calculus</u>

Differentiation

  • Derivatives
  • Derivative Notation

Basic Power Rule:

  1. f(x) = cxⁿ
  2. f’(x) = c·nxⁿ⁻¹

Derivative Rule [Chain Rule]:                                                                                 \displaystyle \frac{d}{dx}[f(g(x))] =f'(g(x)) \cdot g'(x)

Step-by-step explanation:

<u>Step 1: Define</u>

<em>Identify</em>

\displaystyle y = \ln (\cos x^2)

<u>Step 2: Differentiate</u>

  1. Logarithmic Differentiation [Derivative Rule - Chain Rule]:                       \displaystyle y' = \frac{(\cos x^2)'}{\cos x^2}
  2. Trigonometric Differentiation [Derivative Rule - Chain Rule]:                   \displaystyle y' = \frac{-\sin x^2 (x^2)'}{\cos x^2}
  3. Basic Power Rule:                                                                                         \displaystyle y' = \frac{-2x \sin x^2}{\cos x^2}
  4. Rewrite [Trigonometric Identities]:                                                              \displaystyle y' = -2x \tan (x^2)

Topic: AP Calculus AB/BC (Calculus I/I + II)

Unit: Differentiation

8 0
3 years ago
Find the coefficient:
guapka [62]

Answer:

1/3

Step-by-step explanation:

Ask me in the comments of you need an explanation.

:)

3 0
3 years ago
Read 2 more answers
According to the article "Characterizing the Severity and Risk of Drought in the Poudre River, Colorado" (J. of Water Res. Plann
mihalych1998 [28]

Answer:

(a) P (Y = 3) = 0.0844, P (Y ≤ 3) = 0.8780

(b) The probability that the length of a drought exceeds its mean value by at least one standard deviation is 0.2064.

Step-by-step explanation:

The random variable <em>Y</em> is defined as the number of consecutive time intervals in which the water supply remains below a critical value <em>y₀</em>.

The random variable <em>Y</em> follows a Geometric distribution with parameter <em>p</em> = 0.409<em>.</em>

The probability mass function of a Geometric distribution is:

P(Y=y)=(1-p)^{y}p;\ y=0,12...

(a)

Compute the probability that a drought lasts exactly 3 intervals as follows:

P(Y=3)=(1-0.409)^{3}\times 0.409=0.0844279\approx0.0844

Thus, the probability that a drought lasts exactly 3 intervals is 0.0844.

Compute the probability that a drought lasts at most 3 intervals as follows:

P (Y ≤ 3) =  P (Y = 0) + P (Y = 1) + P (Y = 2) + P (Y = 3)

              =(1-0.409)^{0}\times 0.409+(1-0.409)^{1}\times 0.409+(1-0.409)^{2}\times 0.409\\+(1-0.409)^{3}\times 0.409\\=0.409+0.2417+0.1429+0.0844\\=0.8780

Thus, the probability that a drought lasts at most 3 intervals is 0.8780.

(b)

Compute the mean of the random variable <em>Y</em> as follows:

\mu=\frac{1-p}{p}=\frac{1-0.409}{0.409}=1.445

Compute the standard deviation of the random variable <em>Y</em> as follows:

\sigma=\sqrt{\frac{1-p}{p^{2}}}=\sqrt{\frac{1-0.409}{(0.409)^{2}}}=1.88

The probability that the length of a drought exceeds its mean value by at least one standard deviation is:

P (Y ≥ μ + σ) = P (Y ≥ 1.445 + 1.88)

                    = P (Y ≥ 3.325)

                    = P (Y ≥ 3)

                    = 1 - P (Y < 3)

                    = 1 - P (X = 0) - P (X = 1) - P (X = 2)

                    =1-[(1-0.409)^{0}\times 0.409+(1-0.409)^{1}\times 0.409\\+(1-0.409)^{2}\times 0.409]\\=1-[0.409+0.2417+0.1429]\\=0.2064

Thus, the probability that the length of a drought exceeds its mean value by at least one standard deviation is 0.2064.

6 0
3 years ago
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