Respuesta :
Answer:
1) [tex]31 < p<37[/tex]
For this case the values that satisfy the inequality are: 32,33,34,35,36
And we can analyze one by one the number:
[tex] a=32= 16*2[/tex] so then is a composite number because 2>1 and 16>1
[tex] a=33= 11*3[/tex] so then is a composite number because 3>1 and 11>1
[tex] a=34= 17*2[/tex] so then is a composite number because 2>1 and 17>1
[tex] a=35= 7*5[/tex] so then is a composite number because 7>1 and 5>1
[tex] a=36= 6*6[/tex] so then is a composite number because 6>1 and 6>1
So then part 1 is correct and we can see that the statement is enough or sufficient all the values on 31<P<37 are composite numbers.
2) For this cas this statement is FALSE, since we have a counterexample on this case:
[tex]a=3=1*3[/tex] and 3 is not a composite number since 1 is not >1
And since we have one element that not satisfy the condition that's FALSE.
Step-by-step explanation:
For this question we need to use the following definition "If an integer p can b expressed as the product of two integers, each of which that is greater then 1, then the integer p can be considered as a composite number". And this number is not the same as prime number.
Part 1
[tex]31 < p<37[/tex]
For this case the values that satisfy the inequality are: 32,33,34,35,36
And we can analyze one by one the number:
[tex] a=32= 16*2[/tex] so then is a composite number because 2>1 and 16>1
[tex] a=33= 11*3[/tex] so then is a composite number because 3>1 and 11>1
[tex] a=34= 17*2[/tex] so then is a composite number because 2>1 and 17>1
[tex] a=35= 7*5[/tex] so then is a composite number because 7>1 and 5>1
[tex] a=36= 6*6[/tex] so then is a composite number because 6>1 and 6>1
So then part 1 is correct and we can see that the statement is enough or sufficient all the values on 31<P<37 are composite numbers.
Part 2
For this cas this statement is FALSE, since we have a counterexample on this case:
[tex]a=3=1*3[/tex] and 3 is not a composite number since 1 is not >1
And since we have one element that not satisfy the condition that's FALSE.
