LCD (1/7, 14/7, 12/13, 5/6)
LCM = (7, 7, 13, 6)
= 2 * 3 * 7 * 13
= 546
1/7 = 78/546
14/7 = 1092/546
12/13 = 504/546
5/6 = 455/546
Calculation:
1/7 + 14/7
= 1 + 14/7
= 15/7
The common denominator you can calculate as the least common multiple of the both denominators: LCM (7, 7) = 7
Add:
15/7 + 12/13
= 15 . 13/7. 13 + 12 . 7/13 . 7
= 195/91 + 84/91
= 195 + 84/91
= 279/91
The common denominator you can calculate as the least common multiple of the both denominators: LCM (7, 13) = 91
Add:
279/91 + 5/6
= 279 . 6/91 . 6 + 5 . 91/6. 91
= 1674/546 + 455/546
= 1674 + 455/546
= 2129/546
The common denominator you can calculate as the least common multiple of the both denominators: LCM (91, 6) = 546
Hence, 546 is the LCM/LCD of (1/7, 14/17, 13/13, 5/6).
Hope that helps!!!!!!
Answer:
Answer = d. Chi-Square Goodness of Fit
Step-by-step explanation:
A decision maker may need to understand whether an actual sample distribution matches with a known theoretical probability distribution such as Normal distribution and so on. The Goodness-of-fit Test is a type of Chi-Square test that can be used to determine if a data set follows a Normal distribution and how well it fits the distribution. The Chi-Square test for Goodness-of-fit enables us to determine the extent to which theoretical probability distributions coincide with empirical sample distribution. To apply the test, a particular theoretical distribution is first hypothesized for a given population and then the test is carried out to determine whether or not the sample data could have come from the population of interest with hypothesized theoretical distribution. The observed frequencies or values come from the sample and the expected frequencies or values come from the theoretical hypothesized probability distribution. The Goodness-of-fit now focuses on the differences between the observed values and the expected values. Large differences between the two distributions throw doubt on the assumption that the hypothesized theoretical distribution is correct and small differences between the two distributions may be assumed to be resulting from sampling error.
Answer:
Solution for A combination lock uses 3 numbers, each of which can be 0 to 33. If there are no restrictions on the numbers, how many possible ...
Step-by-step explanation:
