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skelet666 [1.2K]
3 years ago
15

Please help me with this I really need my grade up

Mathematics
1 answer:
Katarina [22]3 years ago
6 0

Answer:15 but including the beginning and the end 17

Step-by-step explanation:

I would do 17

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In deciding the best medium to tell a story, which step is first?(1 point)
zavuch27 [327]

Answer:

identifying the audience

Step-by-step explanation:

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3 years ago
7. Which of these polynomial functions gives the number of lights in a hexagonal rig (six sides) with n rows? (Hint: Look for a
morpeh [17]
Thank you for posting your question here at brainly. I hope the answer will help you. Feel free to ask more questions.
Below are the answers:

<span>7. C. H(n) = ½n^2 + n + 1
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3 0
3 years ago
Let $s$ be a subset of $\{1, 2, 3, \dots, 100\}$, containing $50$ elements. how many such sets have the property that every pair
Tamiku [17]

Let A be the set {1, 2, 3, 4, 5, ...., 99, 100}.

The set of Odd numbers O = {1, 3, 5, 7, ...97, 99}, among these the odd primes are :

P={3, 5, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97}

we can count that n(O)=50 and n(P)=24.

 

 

Any prime number has a common factor >1 with only multiples of itself.

For example 41 has a common multiple >1 with 41*2=82, 41*3=123, which is out of the list and so on...

For example consider the prime 13, it has common multiples >1 with 26, 39, 52, 65, 78, 91, and 104... which is out of the list.

Similarly, for the smallest odd prime, 3, we see that we are soon out of the list:

3, 3*2=6, 3*3=9, ......3*33=99, 3*34=102.. 

we cannot include any non-multiple of 3 in a list containing 3. We cannot include for example 5, as the greatest common factor of 3 and 5 is 1.

This means that none of the odd numbers can be contained in the described subsets.

 

 

Now consider the remaining 26 odd numbers:

{1, 9, 15, 21, 25, 27, 33, 35, 39, 45, 49, 51, 55, 57, 63, 65, 69, 75, 77, 81, 85, 87, 91, 93, 95, 99}

which can be written in terms of their prime factors as:

{1, 3*3, 3*5, 3*7, 5*5,3*3*3, 3*11,5*7, 3*13, 2*2*3*3, 7*7, 3*17, 5*11 , 3*19,3*21, 5*13, 3*23,3*5*5, 7*11, 3*3*3*3, 5*17, 3*29, 7*13, 3*31, 5*19, 3*3*11}

 

1 certainly cannot be in the sets, as its common factor with any of the other numbers is 1.

3*3 has 3 as its least factor (except 1), so numbers with common factors greater than 1, must be multiples of 3. We already tried and found out that there cannot be produced enough such numbers within the set { 1, 2, 3, ...}

 

3*5: numbers with common factors >1, with 3*5 must be 

either multiples of 3: 3, 3*2, 3*3, ...3*33 (32 of them)

either multiples of 5: 5, 5*2, ...5*20 (19 of them)

or of both : 15, 15*2, 15*3, 15*4, 15*5, 15*6 (6 of them)

 

we may ask "why not add the multiples of 3 and of 5", we have 32+19=51, which seems to work.

The reason is that some of these 32 and 19 are common, so we do not have 51, and more important, some of these numbers do not have a common factor >1:

for example: 3*33 and 5*20

so the largest number we can get is to count the multiples of the smallest factor, which is 3 in our case.

 

By this reasoning, it is clear that we cannot construct a set of 50 elements from {1, 2, 3, ....}  containing any of the above odd numbers, such that the common factor of any 2 elements of this set is >1.

 

What is left, is the very first (and only) obvious set: {2, 4, 6, 8, ...., 48, 50}

 

<span>Answer: only 1: the set {2, 4, 6, …100}</span>

8 0
3 years ago
PLEASE HELP I HAVE 10 MORE MINUTES TO FINISH THIS!!!
evablogger [386]

Answer:

b

Step-by-step explanation:

add 5+3 then add 1/2 plus 3/4

7 0
3 years ago
A veterinarian wishes to compare the number of times race horses are reported to be lame during a 12‑month period, as compared w
cestrela7 [59]

Answer:

Step-by-step explanation:

Hello!

The objective of the veterinarian is to check if racing contributes to lameness in racing horses. To conduct a test the veterinarian should select at random two groups of horses, group one will be of racing horses and group two should be of not racing horses. Other than racing, all the other differences between the horses that may cause lame or resistance to lame should be controlled( confounding variables) such as age, breed, the number of weakly hours spent exercising, diet, obstacle-free environment, etc...

This type of study is an experimental design.

The experimental design is an analytic essay characterized by the fact that the investigator controls the study variable and the randomization of the samples. In which all the variables are controlled except the ones that you wish to study.

I hope it helps!

5 0
3 years ago
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