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sergey [27]
3 years ago
10

If a probability distribution is 5/12, 1/12, 1/4, x, what is the value of x?

Mathematics
1 answer:
babunello [35]3 years ago
3 0

Answer:

Correct option: A

Step-by-step explanation:

The properties of a probability distribution are:

  1. 0 < P (x = x) < 1.
  2. ∑ P (X = x) = 1.

The second property implies that the sum of all the probabilities of a random variable is 1.

Use this property to determine the value of <em>x</em> as follows:

\sum P(X=k)=\frac{5}{12}+\frac{1}{12}+\frac{1}{4}+x

                 1=\frac{5+1+3+12x}{12}

                 1=\frac{9+12x}{12}

               12=9+12x\\

             12x=12-9

             12x=3

                 x=\frac{3}{12}

                 x=\frac{1}{4}

Thus, the missing probability value is \frac{1}{4}.

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if x is a positive integer, for how many dofferent values of x is square root 48 over x a whole number
Mazyrski [523]

Answer:

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5 0
3 years ago
Write an<br> explicit formula for<br> then<br> th<br> an.<br> term of the sequence 45, 15,5. ....
7nadin3 [17]

Answer:

The sequence above is a geometric sequence

For an nth in a geometric sequence

u(n) = a {r}^{n - 1}

where n is the number of terms

a is the first term

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From the question

a = 45

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Hope this helps you

7 0
3 years ago
If f(1) = 0, what are all the roots of the function f(x)=x^3+3x^2-x-3? Use the Remainder Theorem.
Sophie [7]
There's no if about it, 

f(x)=x^3+3x^2-x-3&#10;

has a zero f(1)=0 so x-1 is a factor.   That's the special case of the Remainder Theorem; since f(1)=0 we'll get a remainder of zero when we divide f(x) by x-1.

At this point we can just divide or we can try more little numbers in the function.  It doesn't take too long to discover f(-1)=0 too, so  x+1 is a factor too by the remainder theorem.  I can find the third zero as well; but let's say that's out of range for most folks.

So far we have 

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where r is the zero we haven't guessed yet.  Again we could divide f(x) by (x-1)(x+1)=x^2-1 but just looking at the constant term we must have

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x^3+3x^2-x-3 = (x-1)(x+1)(x+3)

We check f(-3)=(-3)^3+3(-3)^2 -(-3)-3 = 0 \quad\checkmark

We usually talk about the zeros of a function and the roots of an equation; here we have a function f(x) whose zeros are

x=1, x=-1, x=-3

8 0
3 years ago
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What’s the correct name for that line?
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GM, because when naming a line, you go from left to right just like how we read.

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An experiment consists of rolling two fair dice and adding the dots on the two sides facing up. What is the probability that the
notka56 [123]
C. 1/12
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