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Sphinxa [80]
3 years ago
13

What is .5 as a percent?

Mathematics
1 answer:
Luden [163]3 years ago
7 0
Percent means parts out of 100
x%=x/100
0.50=0.50/1=50/100=50%
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I need help please answer correctly
Bas_tet [7]

Answer:

AC

Step-by-step explanation:

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2 years ago
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Rohan did a survey in four bird parks and wrote the following observations:
stiks02 [169]

Answer:  Park P

Step-by-step explanation:

Since, According to the question,

Number of yellow bird in park P = 33

Total number of the bird in park P = 53

Thus, the percentage of yellow bird in park P = \frac{33}{53} \times 100= 62.264150943% ≈ 62 %

Number of yellow bird in park Q = 20

Total number of the bird in park Q = 48

Thus, the percentage of yellow bird in park P = \frac{20}{48} \times 100=41.666666666% ≈ 42%

Number of yellow bird in park R = 54

Total number of the bird in park R = 90

Thus, the percentage of yellow bird in park R = \frac{54}{90} \times 100= 60 %

Number of yellow bird in park S = 44

Total number of the bird in park S = 83

Thus, the percentage of yellow bird in park P = \frac{44}{83} \times 100=53.012048192 % ≈ 53 %

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Therefore, highest percentage of the yellow bird is 62 %

Which is present in park P.

8 0
2 years ago
Can I get help with this?
kiruha [24]

Answer:

F'(-3,2)

G'(-1,0)

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7 0
3 years ago
Which is the best estimate for StartFraction (8.9 times 10 Superscript 8 Baseline) Over (3.3 times 10 Superscript 4 Baseline) En
slavikrds [6]

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C. 3 x 10 ^4

Step-by-step explanation:

i promise this is right i got it right on the edge test

3 0
3 years ago
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Q1 How many subsets of a set with 100 elements have morethan one element ?
Damm [24]
A set of n elements has 2^n possible subsets, where two classes of those sets are the empty set (1) and all the possible singleton sets (n). So a set of n elements has 2^n-1-n possible subsets with more than one elements. For Q1 take n=100.

Q2a. Assuming not containing the same digits twice also includes not numbers with three or four of the same digit, and assuming digits are chosen from the usual 0-9, there are

4!\dbinom{10}4=\dfrac{4!10!}{4!(10-4)!}=10\times9\times8\times7=5040

possible strings.

Q2b. The first three digits can be chosen freely from 0-9, while the last digit has to be one of 0, 2, 4, 6, or 8. This means you have

10^3\times5=5000

possible strings

Q2c. Any such string will take the form 999X, 99X9, 9X99, or X999, where X has 9 possible choices (0-9 excluding 9, since we want exactly three 9s in any such string). So there are

4\times9=36

possible strings.
4 0
2 years ago
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