The spinner below is spun. Find the probability that it lands on an odd number, given that it lands on a white space.
2 answers:
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Answer:
Since a/2⁽ⁿ ⁺ ¹⁾b < a/2ⁿb, we cannot find a smallest positive rational number because there would always be a number smaller than that number if it were divided by half.
Step-by-step explanation:
Let a/b be the rational number in its simplest form. If we divide a/b by 2, we get another rational number a/2b. a/2b < a/b. If we divide a/2b we have a/2b ÷ 2 = a/4b = a/2²b. So, for a given rational number a/b divided by 2, n times, we have our new number c = a/2ⁿb where n ≥ 1
Since = a/(2^∞)b = a/b × 1/∞ = a/b × 0 = 0, the sequence converges.
Now for each successive division by 2, a/2⁽ⁿ ⁺ ¹⁾b < a/2ⁿb and
a/2⁽ⁿ ⁺ ¹⁾b/a/2ⁿb = 1/2, so the next number is always half the previous number.
So, we cannot find a smallest positive rational number because there would always be a number smaller than that number if it were divided by half.
Answer:
c) skewed to the right.
Step-by-step explanation:
We need to remember that is a distribution is skewed to the right then we have the following condition satisfied:
And if is skewed to the left then we have:
If the distribution is symmetric we need to satisfy:
For this case since we have most of the values between 200000 and 500000 when we put atypical values like 15000000 that would affect the sample mean and on this case the sample mean would larger than the sample median because the median is a robust measure of central tendency not affected by outliers.
So for this special case we can say that the . And probably the median would be higher than the mode so then we can conclude that the best answer for this case would be:
c) skewed to the right.