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densk [106]
3 years ago
5

A bowl contains candies of the same size in three flavors: orange, strawberry, and pineapple. If the probability of randomly pul

ling out an orange candy is 1 over 9, and the probability of randomly pulling out a strawberry candy is 3 over 9, what is the probability of randomly pulling out a pineapple candy?
Mathematics
2 answers:
adelina 88 [10]3 years ago
7 0
To find this add 3 and 1 together.  This gives you 4.  Then, you subtract 4 from 9.  The answer is 5.  Therefore, the probability to pull out a pineapple candy would be 5/9.
lapo4ka [179]3 years ago
5 0
<h2>Answer:</h2>

The probability of randomly pulling out a pineapple candy is:

                  \dfrac{5}{9}

<h2>Step-by-step explanation:</h2>

There are three flavors in a ice cream namely:

orange, strawberry, and pineapple.

We know that for any three events in a single experiment the sum of the probability of three events is equal to 1.

i.e. if A, B and C are three events and P denotes the probability of an event then:

         P(A)+P(B)+P(C)=1

Here let A= flavor is orange

B= flavor is strawberry

and  C= flavor is pineapple.

and we have:

P(A)=\dfrac{1}{9},\ P(B)=\dfrac{3}{9}

Hence, we have:

\dfrac{1}{9}+\dfrac{3}{9}+P(C)=1\\\\\\i.e.\\\\\\\dfrac{4}{9}+P(C)=1\\\\\\i.e.\\\\\\P(C)=1-\dfrac{4}{9}\\\\\\i.e.\\\\\\P(C)=\dfrac{9-4}{9}\\\\i.e.\\\\P(C)=\dfrac{5}{9}

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3 years ago
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Use Stokes' Theorem to evaluate C F · dr where C is oriented counterclockwise as viewed from above. F(x, y, z) = yzi + 4xzj + ex
natima [27]

Answer:

The result of the integral is 81π

Step-by-step explanation:

We can use Stoke's Theorem to evaluate the given integral, thus we can write first the theorem:

\displaystyle \int\limits_C \vec F \cdot d\vec r = \int \int_S curl \vec F \cdot d\vec S

Finding the curl of F.

Given F(x,y,z) = < yz, 4xz, e^{xy} > we have:

curl \vec F =\left|\begin{array}{ccc} \hat i &\hat j&\hat k\\ \cfrac{\partial}{\partial x}& \cfrac{\partial}{\partial y}&\cfrac{\partial}{\partial z}\\yz&4xz&e^{xy}\end{array}\right|

Working with the determinant we get

curl \vec F = \left( \cfrac{\partial}{\partial y}e^{xy}-\cfrac{\partial}{\partial z}4xz\right) \hat i -\left(\cfrac{\partial}{\partial x}e^{xy}-\cfrac{\partial}{\partial z}yz \right) \hat j + \left(\cfrac{\partial}{\partial x} 4xz-\cfrac{\partial}{\partial y}yz \right) \hat k

Working with the partial derivatives

curl \vec F = \left(xe^{xy}-4x\right) \hat i -\left(ye^{xy}-y\right) \hat j + \left(4z-z\right) \hat k\\curl \vec F = \left(xe^{xy}-4x\right) \hat i -\left(ye^{xy}-y\right) \hat j + \left(3z\right) \hat k

Integrating using Stokes' Theorem

Now that we have the curl we can proceed integrating

\displaystyle \int\limits_C \vec F \cdot d\vec r = \int \int_S curl \vec F \cdot d\vec S

\displaystyle \int\limits_C \vec F \cdot d\vec r = \int \int_S curl \vec F \cdot \hat n dS

where the normal to the circle is just \hat n= \hat k since the normal is perpendicular to it, so we get

\displaystyle \int\limits_C \vec F \cdot d\vec r = \int \int_S \left(\left(xe^{xy}-4x\right) \hat i -\left(ye^{xy}-y\right) \hat j + \left(3z\right) \hat k\right) \cdot \hat k dS

Only the z-component will not be 0 after that dot product we get

\displaystyle \int\limits_C \vec F \cdot d\vec r = \int \int_S 3z dS

Since the circle is at z = 3 we can just write

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Thus the integral represents the area of a circle, the given circle x^2+y^2 = 9 has a radius r = 3, so its area is A = \pi r^2 = 9\pi, so we get

\displaystyle \int\limits_C \vec F \cdot d\vec r = 9(9\pi)\\\displaystyle \int\limits_C \vec F \cdot d\vec r = 81 \pi

Thus the result of the integral is 81π

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spayn [35]

Answer:

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2. 4 remaining in the bag

3. 2 marbles remain in the bag

4. 2/4=1/2=50%

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