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lbvjy [14]
4 years ago
13

0.12, 1.2, 0.012, 12 in ascending order

Mathematics
2 answers:
Vinil7 [7]4 years ago
8 0

The more 0's there are in front of the 12, the lower it is (and if theres a decimal point)

So 0.012, then 0.12, 1.2, and finally 12

Gnom [1K]4 years ago
7 0

0.012 is the least of all these numbers

0.12

1.2

12 is the greatest out of all these numbers

I hope that this helped you bunch hun :)

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What is the simplified expression for the expression belov
11111nata11111 [884]

Answer: 9x + 17

Step-by-step explanation: Add 4x and 5x and you get 9x. Next take away 15 from 32 and you get 17 therefore the answer wuld be 9x + 17.

6 0
3 years ago
Scored 72 times out of 150 what percent of his shots did he score?
Sholpan [36]
Whoever scored 72 shots out of 150. Did, In fact, score 48% of their shots.
 
Solution:
*To calculate the percentage represented by a ratio simply divide the numerator by the denominator and multiply by 100. 

72:150 (72/150)

72/150 

= 0.48

0.48(100) = 48

... 48%
6 0
4 years ago
Please help no du mb answers
vladimir1956 [14]

Answer:

A = (-2,5)

B = (5,5)

the distance between the points is 7

Step-by-step explanation:

6 0
4 years ago
What would you earn in total commissions on the following sales of $5100 $4876 $5215 $6225 and $5235 if you if you earn a commis
lisov135 [29]

Answer:

$5100 would be $76.50

$4876 would be $73.14

$5215 would be $78.23

$6225 would be $93.38

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Step-by-step explanation:

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5 0
3 years ago
student randomly receive 1 of 4 versions(A, B, C, D) of a math test. What is the probability that at least 3 of the 5 student te
alexdok [17]

Answer:

1.2%

Step-by-step explanation:

We are given that the students receive different versions of the math namely A, B, C and D.

So, the probability that a student receives version A = \frac{1}{4}.

Thus, the probability that the student does not receive version A = 1-\frac{1}{4} = \frac{3}{4}.

So, the possibilities that at-least 3 out of 5 students receive version A are,

1) 3 receives version A and 2 does not receive version A

2) 4 receives version A and 1 does not receive version A

3) All 5 students receive version A

Then the probability that at-least 3 out of 5 students receive version A is given by,

\frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}\times \frac{3}{4}\times \frac{3}{4}+\frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}\times \frac{3}{4}+\frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}

= (\frac{1}{4})^3\times (\frac{3}{4})^2+(\frac{1}{4})^4\times (\frac{3}{4})+(\frac{1}{4})^5

= (\frac{1}{4})^3\times (\frac{3}{4})[\frac{3}{4}+\frac{1}{4}+(\frac{1}{4})^2]

= (\frac{3}{4^4})[1+\frac{1}{16}]

= (\frac{3}{256})[\frac{17}{16}]

= 0.01171875 × 1.0625

= 0.01245

Thus, the probability that at least 3 out of 5 students receive version A is 0.0124

So, in percent the probability is 0.0124 × 100 = 1.24%

To the nearest tenth, the required probability is 1.2%.

4 0
3 years ago
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