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nikdorinn [45]
3 years ago
12

What is the center of the circle shown below

Mathematics
2 answers:
ivanzaharov [21]3 years ago
7 0

J

That's the only point within the circle, and it seems like the center. In these questions, just assuming it's perfectly in the center typically works.

Vinvika [58]3 years ago
6 0

the correct answer to your question is : J

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Solve for x. x +1/2 = 3/4
Alexeev081 [22]

x + 1/2 = 3/4

set denominators equal:

x + 2/4 = 3/4

-2/4 for both sides:

x = 1/4

there you go! hope this helps!

4 0
3 years ago
Read 2 more answers
Find the sum of 0.2126 and 0.7145 correct to 3 significant figures​
Afina-wow [57]

Answer:

0.927

Step-by-step explanation:

add then round the last digit (it doesnt matter if it is a decimal number or not).

3 0
2 years ago
Each letter of the word "supercalifragilisticexpialidocious" is placed into a bag and drawn at 3 times, replacing the letter aft
Makovka662 [10]

Answer:

P(X ≥ 1) = 0.50

Step-by-step explanation:

Given that:

The word "supercalifragilisticexpialidocious" has 34 letters in which 'i' appears 7 times in the word.

Then; the probability of success = 7/34 = 0.20588

Using Binomial distribution to determine the probability; we have:

P(X = x)  = ^nC_x  \ \beta^x   \  (1 - \beta)^{n-x}

where;

x = 0,1,2,...n    and    0  <  β   <   1

and x represents the  number of successes.

However; since the letter is drawn thrice; the probability that the letter "i" is drawn at least once can be computed as:

P(X ≥ 1) = 1 - P(X< 1)

P(X ≥ 1) = 1 - P(X =0)

P(X \ge 1) =  1 - \bigg [ {^3C__0} (0.21)^0 (1-0.21)^{3-0} \bigg]

P(X \ge 1) =  1 - \bigg [ 1 \times 1 (0.79)^{3} \bigg]

P(X ≥ 1) = 1 - 0.50

P(X ≥ 1) = 0.50

4 0
3 years ago
un problema de matemáticas y es así felpe y luis se encontraron en el pedromo para practicar velocidad en bicicleta felpe dijo a
dsp73
I do not know ur language I think it is spanish sorry I couldn't help
3 0
3 years ago
¡Hello!
KIM [24]

Answer:

Step-by-step explanation:

number 2= 2 (rounded)

same thing for 1

4 0
3 years ago
Read 2 more answers
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