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scoray [572]
3 years ago
12

An after school music program has 15 out 50 students practicing. Write 15/50 (15 over 50) as a decimal and as a percent.

Mathematics
2 answers:
ValentinkaMS [17]3 years ago
7 0
15/50 * 2/2 = 30/100
Now you can easily see decimal and percent form

Decimal: 0.30
Percent: 30 percent
Elodia [21]3 years ago
5 0

Answer:

Percent- 30

decimal-0.3

Step-by-step explanation:

Hope this helps and have a wonderful day!!!

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After multiplying each side of the equation by the LCD and simplify , the resulting equation is
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After multiplying each side of the equation by the LCD, the result will be
2(x² - x - 6) = x(x - 6) + 2x(2x + 12)

Then simplify and move all expression the the left of equal sign
2(x² - x - 6) = x(x - 6) + 2x(2x + 12)
2x² - 2x - 12 = x² - 6x + 4x² + 24x
2x² - x² - 4x² -2x + 6x - 24x - 12 = 0
-3x² -20x - 12 = 0

Multiply each side by -1
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The answer is third option
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4 years ago
sophia used 1/10 of a liter of water to fill 1/4 of a fish aquarium. How much water, in liter is required to fill the aquarium?​
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4/10 of a liter

Step-by-step explanation:

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3 years ago
The diameter of a circle is 2 kilometers. What is the radius?
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An Article a Florida newspaper reported on the topics that teenagers most want to discuss with their parents. The findings, the
Sergio039 [100]

Answer:

The estimate is  P__{hat}} \pm E  = 0.37 \pm 0.0348

Step-by-step explanation:

From the question we are told that  

    The sample size is  n =  522

    The sample proportion of students  would like to talk about school is  \r p__{hat}} =  0.37

  Given that the confidence level is  90 % then the level of significance can be mathematically evaluated as

                  \alpha  =  100 - 90

                  \alpha  =  10\%

                  \alpha  =  0.10

Next we obtain the critical value of  \frac{\alpha }{2} from the normal distribution table, the value is  

               Z_{\frac{\alpha }{2} } =Z_{\frac{0.10}{2} } =  1.645

Generally the margin of error can be mathematically represented as

               E =  Z_{\frac{\alpha }{2} } *  \sqrt{\frac{\r P_{hat}(1- \r P_{hat} )}{n } }

=>            E = 1.645 *  \sqrt{\frac{0.37 (1- 0.37  )}{522 } }

=>             E = 0.0348

Generally the estimate the proportion of all teenagers who want more family discussions about school at 90% confidence level is  

                       P__{hat}} \pm  E

substituting values

                     0.37 \pm 0.0348

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