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Westkost [7]
2 years ago
15

If a number is chosen at random from the set {1, 2, 3, 4, . . ., 18}, what is the probability that the number chosen is a factor

of 17?
Mathematics
1 answer:
choli [55]2 years ago
5 0
Looking at the set, we are given 18 elements. 17 is prime; it has only two factors: 1 and 17, since 1•17=17. So, the question is really asking what is the probability the numbers 1 or 17 is chosen. As mentioned earlier, 17 is prime, so there are two possible choices: 1 and 17.

P (probability) = possible outcomes / total outcomes

It is important to note that these events are “or” events, meaning that the probability can only be determined by choosing a 1 or a 17; you can’t randomly chose a 1 and 17 at the same time. So, the formula is:

P(A or B) = P(A) + P(B)

All this is saying is that given two possible outcomes, the probability occurs independent of each event; they don’t occur at the same time.

P(1 or 17) = P(1)/18 + P(1)/18

P(1 or 17) = 2/18

Since 17 is prime, it’s two and only factors are 1 and 17. The probability of randomly choosing a 1 or 17 is 2/18, meaning that there are 2 elements in the set out of a possible 18 elements that can be randomly chosen.

2/18 simplifies to 1/9


So, your answer is 1/9
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kolezko [41]

Answer:

98

Step-by-step explanation:

So we have the two functions:

g(x)=-2x-2\text{ and } r(x)=x^2-2

And we want to find the value of r(g(4)).

To do so, first find the value of g(4):

g(x)=-2x-2\\g(4)=-2(4)-2\\

Multiply:

g(4)=(-8)-2

Subtract:

g(4)=-10

Now, substitute this into r(g(4)):

r(g(4))\\=r(-10)

And substitute this value into r(x):

r(-10)=(-10)^2-2

Square:

r(-10)=100-2

Subtract:

r(-10)=98

Therefore:

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3 years ago
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noname [10]

Using the normal distribution, it is found that Sue will get a letter grade of B.

In a <em>normal distribution </em>with mean \mu and standard deviation \sigma, the z-score of a measure X is given by:

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In this problem, Sue got a grade of 0.85, hence z = 0.85.

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A similar problem is given at brainly.com/question/25745464

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Hope this helps!

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