( sorry for commenting, I need points to ask questions I dearly apologize!)
All you have to do is change the whole number to a fraction. Say you have 4 divided by 1/2. All you have to do to make a whole number a fraction is put it over 1. So now you would have 4/1 (4 being the numerator) divided by 1/2. When you divide fractions always remember this; Keep, Switch, Flip. Keep 4/1, Flip the division sign to multiplication, and Flip 1/2 to make it 2/1. Then you multiply the numerators and the denominators. So 4/1 * 2/1 = 8/1. Hope this helped.
Answer:
(-∞,∞)
Step-by-step explanation:
Using Chebyshev's Theorem, it is found that:
At least 75% of the exam scores fall between 76 and 92.
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- Chebyshev's Theorem states that the percentage of measures within k standard deviations of the mean is of at least
.
- With k = 2, that is, within 2 standard deviations of the mean:
![P = 100(1 - \frac{1}{2^{2}})](https://tex.z-dn.net/?f=P%20%3D%20100%281%20-%20%5Cfrac%7B1%7D%7B2%5E%7B2%7D%7D%29)
![P = 100(1 - \frac{1}{4})](https://tex.z-dn.net/?f=P%20%3D%20100%281%20-%20%5Cfrac%7B1%7D%7B4%7D%29)
![P = 100(\frac{3}{4})](https://tex.z-dn.net/?f=P%20%3D%20100%28%5Cfrac%7B3%7D%7B4%7D%29)
![P = 75](https://tex.z-dn.net/?f=P%20%3D%2075)
- At least 75% of the measures are within 2 standard deviations of the mean.
84 - 2(4) = 76
84 + 2(4) = 92
- Thus, at least 75% of scores between 76 and 92.
A similar problem is given at brainly.com/question/23612895