Answer:
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Step-by-step explanation:
Well to find the prime factor we make the prime factorization tree.
Look at the image below↓
<em>Thus,</em>
<em>the prime factorization of 4275 is </em>
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<em>Hope this helps :)</em>
Answer:
See answer below
Step-by-step explanation:
The statement ‘x is an element of Y \X’ means, by definition of set difference, that "x is and element of Y and x is not an element of X", WIth the propositions given, we can rewrite this as "p∧¬q". Let us prove the identities given using the definitions of intersection, union, difference and complement. We will prove them by showing that the sets in both sides of the equation have the same elements.
i) x∈AnB if and only (if and only if means that both implications hold) x∈A and x∈B if and only if x∈A and x∉B^c (because B^c is the set of all elements that do not belong to X) if and only if x∈A\B^c. Then, if x∈AnB then x∈A\B^c, and if x∈A\B^c then x∈AnB. Thus both sets are equal.
ii) (I will abbreviate "if and only if" as "iff")
x∈A∪(B\A) iff x∈A or x∈B\A iff x∈A or x∈B and x∉A iff x∈A or x∈B (this is because if x∈B and x∈A then x∈A, so no elements are lost when we forget about the condition x∉A) iff x∈A∪B.
iii) x∈A\(B U C) iff x∈A and x∉B∪C iff x∈A and x∉B and x∉C (if x∈B or x∈C then x∈B∪C thus we cannot have any of those two options). iff x∈A and x∉B and x∈A and x∉C iff x∈(A\B) and x∈(A\B) iff x∈ (A\B) n (A\C).
iv) x∈A\(B ∩ C) iff x∈A and x∉B∩C iff x∈A and x∉B or x∉C (if x∈B and x∈C then x∈B∩C thus one of these two must be false) iff x∈A and x∉B or x∈A and x∉C iff x∈(A\B) or x∈(A\B) iff x∈ (A\B) ∪ (A\C).
Answer:
The correct option is (b).
Step-by-step explanation:
The (1 - <em>α</em>)% confidence interval for population mean (<em>μ</em>) is:

The confidence interval for population mean can be computed using either the <em>z</em>-interval or <em>t</em>-interval.
The <em>t</em>-interval is used if the following conditions are satisfied:
- The population standard deviation is not known
- The sample size is large enough
- The population from which the sample is selected is normally distributed.
For computing a (1 - <em>α</em>)% confidence interval for population mean , it is necessary for the population to normally distributed if the sample selected is small, i.e.<em>n</em> < 30, because only then the sampling distribution of sample mean will be approximated by the normal distribution.
In this case the sample size is, <em>n</em> = 28 < 30.
Also it is provided that the systolic blood pressure is known to have a skewed distribution.
Since the sample is small and the population is not normally distributed, the sampling distribution of sample mean will not be approximated by the normal distribution.
Thus, no conclusion can be drawn from the 90% confidence interval for the mean systolic blood pressure.
The correct option is (b).
(9-243)/(4-1)=-78
D)−78 contestants per round, and it represents the average rate at which the number of contestants changed from the first round to the fourth roundReport · <span>14/12/2015</span>
Answer:
Papi
Step-by-step explanation: