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Eduardwww [97]
3 years ago
10

8x=64 how do I solve this

Mathematics
2 answers:
Drupady [299]3 years ago
4 0
Divide both sides by 8
Shalnov [3]3 years ago
4 0

Answer:

x = 8

Step-by-step explanation:

8x = 64. Divide each side by eight.

8x ÷ 8 = 1x (or just 'x' )

64 ÷ 8 = 8

x = 8

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x²

Step-by-step explanation:

x³ - 1 ÷ x + 2

the first term of the quotient is x³ ÷ x = x²

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Label the quadrants IV​
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Quadrant IV is always in the bottom right corner.

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Please help!!!!!! + lots of points
Ilya [14]

Hey!

Hope this helps...

~~~~~~~~~~~~~~~~~~~~~~~~~~~

The correct answers are A, B, D, and E

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3 years ago
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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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