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DedPeter [7]
2 years ago
6

What is a fraction that is repeating between 0.5 and 1?

Mathematics
1 answer:
ExtremeBDS [4]2 years ago
6 0

Answer:

2/3

Step-by-step explanation:

2/3=.66666666......

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Write the equation for the area (A) in terms of length (l) of a playground if the width (w) is one half as long as the length (l
Dvinal [7]
Let, the length = l
Width = l/2

Area = l * w = l * l/2 = l²/2

So, Your Final Answer would be  A = l²/2

Hope this helps!
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HELP PLZ There are 14 parking spaces in each row of a parking lot and there are 15 rows in the lot. How many parking spaces are
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There are 14x15 parking spaces=210
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What point is a solution to the inequality shown in this graph? (-3,-6) & (3,2)
NARA [144]

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(0,5)

Step-by-step explanation:

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3 years ago
Your cell phone plan allows you 500 minutes to
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Answer:

500 ≤ 260 + 6x

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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
2 years ago
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