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aleksandrvk [35]
3 years ago
15

Please can anyone help answer this question? It would be greatly appreciated!!! :)

Mathematics
2 answers:
Ray Of Light [21]3 years ago
7 0

Answer:

0.667

Step-by-step explanation:

Angle in a circle = 360°

Area of total shaded parts = 60 + 60 = 120°

Area of total unshaded parts = 360-120 = 240°

Probability that a random selected point within the circle falls in the unshaded area

= 240/360

= 2/3

≈ 0.667

I apologize in advance if I made a mistake.

Diano4ka-milaya [45]3 years ago
4 0

Answer:

The probability that point falls in the white area is 2/3 or 0.667

Step-by-step explanation:

Consider the provided figure.

As we know there are 360 degrees in a circle.

The diameter divides the circle in two equal half 180° each.

The angle measure of white part in upper half is 180°-60°-60°=60°

The total angle measure of white part in the provided figure is: 180°+60°=240°

Now we need to find the probability that a random selected point falls in the white area

Probability=\frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}}

Substitute Number of favorable outcomes=240° and Total number of outcomes = 360° in the above formula.

Probability=\frac{240}{360}

Probability=\frac{2}{3}\ \text{or}\ 0.667

Hence, the probability that point falls in the white area is 2/3 or 0.667

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This is a question that needs a bit of common sense to get to the actual answer that is desired. Firstly there is one information given .The other information has to be found from common sense. ! single handshake requires 2 people. Now we can easily determine the number of people doing 66 handshakes.
Then
The number of people required for 1 hand shake = 2
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Then it can be said that there were 132 people in the party. I hope the process of doing this problem is clear to you.
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Step-by-step explanation:

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Area: 9.3 x 2 x 1/2 = 9.3

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Can you please tell me the first 20 digits of pi. I need o memorize them for a test
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3.14159265358979323846

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It is known that diskettes produced by a cer- tain company will be defective with probability .01, independently of each other.
zheka24 [161]

Answer:

1.27%

Step-by-step explanation:

To solve this problem, we may consider a binomial distribution where a customer can either accept or reject (and return) the diskette package.

Lets consider  some aspects:

1. From the formulation of the exercise we know that a package is accepted if it has at most 1 defective diskette. So our event A is defined as:

A = 0 or 1 defective diskette

2. The probability of a diskette being defective is 0.01

3. Each package contains 10 diskettes.

If X is defined as number of defective diskettes in the package, the probability of X is given by a binomial distribution with probability 0.01 and n=10

X ~ Bin(p=0.01, n=10)

Let us remember the calculation of probability for the binomial distribution:

P(X=x)=nCx*p^{x}*(1-p)^{(n-x)} with x = 0, 1, 2, 3,…, n

Where

n: number of independent trials

p: success probability  

x: number of successes in n trials

In our case success means finding a defective diskette, therefore

n=10

p=0.01

And for x we just need 0 or 1 defective diskette to reject the package

Hence,

P(X=x)=10Cx*0.01^{x}*(1-0.01)^{(10-x)} with x = 0, 1

So,

P(A)=P(X=0)+P(X=1)

P(A)=10C0*0.01^{0}*(1-0.01)^{(10-0)} + 10C1*0.01^{1}*(1-0.01)^{(9)}

P(A)=0.99^{10}+10*0.01*0.99^{9}

P(A)=0.9957

Now, because we have 3 packages and we might reject just 1 of them, we can find this probability like this:

3*(1-P(A))*P(A)*P(A) = (1-0.9957)*0.9957*0.9957=0.0127

Finally, we have that the probability of returning exactly one of the three packages is 1.27%

3 0
3 years ago
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