Solving'' an inequality<span> means </span>finding<span> all of its solutions. A "solution'' of an </span>inequality<span> is a number which when substituted for the </span>variable<span> makes the </span>inequality<span> a true statement. ... </span>Rule<span> 3b. Multiplying/dividing by the same NEGATIVE number on both sides AND changing the ... All real </span>numbers<span> less than 1 solve the </span>inequality<span>.</span>
1/3 of 60 to 1/5 of 100 is 20 to 20.
1/3 x 60/1 =20
1/5 x 100/1=20
Reminder: You're able to cross-multiply.
36....we need to break this down...12 * 3 = 36
12 * 3....we can break this down because 12 is not prime...4 * 3 = 12
4 * 3 * 3...we can break the 4 down because it is not prime
2 * 2 * 3 * 3 <==== prime factorization
Answer:
NO
Step-by-step explanation:
The changeability of a sampling distribution is measured by its variance or its standard deviation. The changeability of a sampling distribution depends on three factors:
- N: The number of observations in the population.
- n: The number of observations in the sample.
- The way that the random sample is chosen.
We know the following about the sampling distribution of the mean. The mean of the sampling distribution (μ_x) is equal to the mean of the population (μ). And the standard error of the sampling distribution (σ_x) is determined by the standard deviation of the population (σ), the population size (N), and the sample size (n). That is
μ_x=p
σ_x== [ σ / sqrt(n) ] * sqrt[ (N - n ) / (N - 1) ]
In the standard error formula, the factor sqrt[ (N - n ) / (N - 1) ] is called the finite population correction. When the population size is very large relative to the sample size, the finite population correction is approximately equal to one; and the standard error formula can be approximated by:
σ_x = σ / sqrt(n).
The problem here takes a brilliant mind to answer this. This problem can easily be answered using programming because we can not then and there push all the possibilities using paper and pen.
The answer is 3816547290.
Trying all the possibilities starting form 1000000080 (we are sure that the last number should be 0). Then traversing that number until 9999999990. Each traverse, check the number if its divisible to n, so on and so forth.
