![5 \frac{3}{5} + 2 \frac{1}{10} \\ = \frac{28}{5} + \frac{21}{10} \\ = \frac{56}{10} + \frac{21}{10} \\ = \frac{77}{10} \\ = 7 \frac{7}{10}](https://tex.z-dn.net/?f=5%20%5Cfrac%7B3%7D%7B5%7D%20%20%2B%202%20%5Cfrac%7B1%7D%7B10%7D%20%20%5C%5C%20%20%3D%20%20%5Cfrac%7B28%7D%7B5%7D%20%20%2B%20%20%5Cfrac%7B21%7D%7B10%7D%20%20%5C%5C%20%20%3D%20%20%5Cfrac%7B56%7D%7B10%7D%20%20%2B%20%20%5Cfrac%7B21%7D%7B10%7D%20%20%5C%5C%20%20%3D%20%20%5Cfrac%7B77%7D%7B10%7D%20%20%5C%5C%20%20%3D%207%20%5Cfrac%7B7%7D%7B10%7D%20)
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Answer:
1938
Step-by-step explanation:
Answer:
(a)
The values of X can be 0, 1, 2 , ..., 10 . So, X is a discrete random variable.
(b)
The distribution of X is Binomial distribution with the parameters n = 10 and p = 0.2
(c)
Probability that no one or one person will be injured = P(X = 0) + P(X = 1)
= 10C0 * 0.20 * (1 - 0.2)10-0 + 10C1 * 0.21 * (1 - 0.2)10-1
= 0.810 + 10 * 0.2 * 0.89
= 0.3758096
(d)
Average value of X = np
Average value of X = 10 * 0.2 = 2
(e)
Variance of X = np(1-p)
Variance of X = 10 * 0.2 * (1 - 0.2) = 1.6
(f)
Number of ways in which 2 people gets injured = 10C2 = 10! / ((10-2)! 2!) = (10 * 9) / (2 * 1) = 45
Assume the best player got injures, number of ways in which one people out of remaining 9 people gets injured = 9C1
= 9! / ((9-1)! 1!)
= 9
Probability that the best player got injured = Number of ways in which 1 people gets out of 9 and best person gets injured / Number of ways in which 2 people gets injured
= 9 / 45
= 0.2
Both 17/10 and 2 4/5 should be rounded up in preparation for this estimation.
17/10 is close to 2 and 2 4/5 is close to 3. Thus, your estimated answer should be
2/3, or approx. 0.6666....
Exact answer: divide 17/10 by 14/5. LCD is 10, so convert 14/5 to 28/10.
Now divide 17/10 by 28/10. Answer: 17/28 = approx. = 0.607 approx.
These two results are comparable: 0.6666.... and 0.6071 ....