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marusya05 [52]
3 years ago
13

Can a rational number be a irrational number

Mathematics
1 answer:
jekas [21]3 years ago
4 0

Answer:

No. Rational numbers are different from irrational numbers. Rational numbers are numbers that can be written as a fraction. Rational numbers can be a terminating decimal (a number with digits that end/do not repeat) or a repeating decimal (a number with digits that repeat continuously in a pattern). On the other hand, an irrational number is a number that is continuous, and its digits are random.

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Which is the approximate solution for the system of equations x + 5= 10 and 3x + y = 1?
zheka24 [161]

Answer: The approximate solution is (-0.36, 2.08).

Given system of equations : x+5y=10 and 3x+y=1

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3 years ago
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(6t^4-t+1) - (6t^3-4t)
goldenfox [79]

Answer:

= 6 {t}^{4}   - 6 {t}^{3}  + 3t + 1

Step-by-step explanation:

6 {t}^{4}  - t + 1 - 6 {t}^{3}  + 4t \\  = 6 {t}^{4}   - 6 {t}^{3}  + 3t + 1

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3 years ago
How can I simplify 84/144 to the lowest term.
lions [1.4K]
You can divide 84 and 144 and convert the answer to a fraction.

Correct me if I am wrong.
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3 years ago
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Which graph shows the solution to the inequality -x+2> 1
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I need the graphs to help you, sorry ;_;
5 0
3 years ago
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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