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ICE Princess25 [194]
3 years ago
15

Based on the line of best fit, which of these statements is true?​

Mathematics
1 answer:
lara31 [8.8K]3 years ago
7 0

There’s no pic of the answer sorry

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Pls hurry by Friday! Whoever answers first will get points
leonid [27]

Answer:

14 inches for every side

Step-by-step explanation:

44/4 = 22

circumfrence of half circle is 3.14r = 22

22/3.14 is about 7

the radius is half the side length so multiply 7 by 2 to get 14

mark brainliest please

3 0
2 years ago
When steve woke up. his temerature was 102. two hours later it was 3 lower what is his temerature now?
valentina_108 [34]
Steve's original temperature - 102

2 hours later it dropped 3 degrees

102 - 3 = 99

Steve's current temperature is 99.
8 0
3 years ago
A.)12 <br> B.) 25<br> C.)51<br> D.) 77
givi [52]

Answer:

4x+3=2x+27

4x-2x=27-3

2x=24

x=24/2

x=12

3 0
2 years ago
Read 2 more answers
PLEASE HELP HURRY WITH THE QUESTIONS BELOW!!!!!
AnnZ [28]

From left to right, the answers are:  152, D(line e), C(obtuse), 44 degrees.

Now I feel guilty and sleezy.  Giving answers alone without explanations
is something I almost never do, because it doesn't help anybody do
anything but cheat.  But it's all I have time for right now.  If a mod comes
along and decides to delete this because of no explanations, then let him
delete it.  It'll serve us right, you and me both.


8 0
3 years ago
Using the bijection rule to count binary strings with even parity.
AleksandrR [38]

Answer:

Lets denote c the concatenation of strings. For a binary string <em>a</em> in B9, we define the element f(a) in E10 this way:

  • f(a) = a c {1} if a has an odd number of 1's
  • f(a) = a c {0} if a has an even number of 1's

Step-by-step explanation:

To show that the function f defined above is a bijective function, we need to prove that f is well defined, injective and surjective.

f   is well defined:

To see this, we need to show that f sends elements fromo b9 to elements of E10. first note that f(a) has 1 more binary integer than a, thus, it has 10. if a has an even number of 1's, then f(a) also has an even number because a 0 was added. On the other hand, if a has an odd number of 1's, then f(a) has one more 1, as a consecuence it will have an even number of 1's. This shows that, independently of the case, f(a) is an element of E10. Thus, f is well defined.

f is injective (or one on one):

If a and b are 2 different binary strings, then f(a) and f(b) will also be different because the first 9 elements of f(a) form a and the first elements of f(b) form b, thus f(a) is different from f(b). This proves that f in injective.

f is surjective:

Let y be an element of E10, Let x be the first 9 elements of y, then f(x) = y:

  • If x has an even number of 1's, then the last digit of y has to be 0, and f(x) = x c {0} = y
  • If x has an odd number of 1's, then the last digit of y has to be a 1, otherwise it wont be an element of E10, and f(x) = x c {1} = y

This shows that f is well defined from B9 to E10, injective, and surjective, thus it is a bijection.

3 0
3 years ago
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