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muminat
3 years ago
9

A company knows that 30% of Customers who come to the store will check out the merchandise and then order it online because it i

s cheaper. The company wants to know the probability that it will take at least three customers to find one who shops on line. How could the company find out this information?
Mathematics
1 answer:
valentinak56 [21]3 years ago
7 0
Let X be a discrete random variable with geometric distribution.
 Let x be the number of tests and p the probability of success in each trial, then the probability distribution is:
 P (X = x) = p * (1-p) ^ (x-1). With x = (1, 2, 3 ... n).
 This function measures the probability P of obtaining the first success at the x attempt.
 We need to know the probability of obtaining the first success at the third trial.
  Where a success is defined as a customer buying online.
 The probability of success in each trial is p = 0.3.
 So:
 P (X = 3) = 0.3 * (1-0.3) ^ (3-1)
 P (X = 3) = 0.147
 The probability of obtaining the first success at the third trial is 14.7%
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Help with 30 please. thanks.​
Svet_ta [14]

Answer:

See Below.

Step-by-step explanation:

We have the equation:

\displaystyle  y = \left(3e^{2x}-4x+1\right)^{{}^1\! / \! {}_2}

And we want to show that:

\displaystyle y \frac{d^2y }{dx^2} + \left(\frac{dy}{dx}\right) ^2 = 6e^{2x}

Instead of differentiating directly, we can first square both sides:

\displaystyle y^2 = 3e^{2x} -4x + 1

We can find the first derivative through implicit differentiation:

\displaystyle 2y \frac{dy}{dx}  = 6e^{2x} -4

Hence:

\displaystyle \frac{dy}{dx} = \frac{3e^{2x} -2}{y}

And we can find the second derivative by using the quotient rule:

\displaystyle \begin{aligned}\frac{d^2y}{dx^2} & = \frac{(3e^{2x}-2)'(y)-(3e^{2x}-2)(y)'}{(y)^2}\\ \\ &= \frac{6ye^{2x}-\left(3e^{2x}-2\right)\left(\dfrac{dy}{dx}\right)}{y^2} \\ \\ &=\frac{6ye^{2x} -\left(3e^{2x} -2\right)\left(\dfrac{3e^{2x}-2}{y}\right)}{y^2}\\ \\ &=\frac{6y^2e^{2x}-\left(3e^{2x}-2\right)^2}{y^3}\end{aligned}

Substitute:

\displaystyle y\left(\frac{6y^2e^{2x}-\left(3e^{2x}-2\right)^2}{y^3}\right) + \left(\frac{3e^{2x}-2}{y}\right)^2 =6e^{2x}

Simplify:

\displaystyle \frac{6y^2e^{2x}- \left(3e^{2x} -2\right)^2}{y^2} + \frac{\left(3e^{2x}-2\right)^2}{y^2}= 6e^{2x}

Combine fractions:

\displaystyle \frac{\left(6y^2e^{2x}-\left(3e^{2x} - 2\right)^2\right) +\left(\left(3e^{2x}-2\right)^2\right)}{y^2} = 6e^{2x}

Simplify:

\displaystyle \frac{6y^2e^{2x}}{y^2} = 6e^{2x}

Simplify:

6e^{2x} \stackrel{\checkmark}{=} 6e^{2x}

Q.E.D.

6 0
3 years ago
Can someone solve this
zysi [14]
You know how Soh Cah Toa applies to right triangles; you can assign the adjacent side = 4 and hypotenuse = 7

CosФ = (adjacent) / (hypotenuse)

Then with the Pythagorean theorem
4² + b² = 7²
16 + b² = 49
b² = 49 - 16
b² = 33
b = √(33)

So "b" is the opposite side.

SinФ = (opposite) / (hypotenuse) = √(33)/7
TanФ = (opposite) / (adjacent) = √(33)/4
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