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Mrac [35]
3 years ago
12

There are 1250 students at Cypress Middle School. 42% of the students are enrolled in PE. How many students are enrolled in PE?

Mathematics
2 answers:
yuradex [85]3 years ago
8 0

Answer:

The answer is 525 students.

Step-by-step explanation:

1250 total students

42% = PE

42% of 1250 = 525

Thus, the answer is 525.

slamgirl [31]3 years ago
4 0

Answer: 525 students are enrolled in PE.

Step-by-step explanation:

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Which three values are greater than 1/2? <br> 1/3<br> 45%<br> 0.62<br> 59%<br> 0.315<br> 3/4
Sloan [31]
0.62, 59%, and 3/4
these are all more than 50% which is what 1/2 is.
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2 years ago
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The amount of time required for an oil and filter change on an automobile is normally distributed with a mean of 45 minutes and
Firdavs [7]

Answer:

1a

  P(39 <  X < 48  ) = 0.8767

1b

    95% of all sample means will fall between 40.1  <  \mu < 49.9

1c

    \= x = 41. 795

2

   n =  25

Step-by-step explanation:

From the question we are told that

   The mean is n   =  45

   The population standard deviation is  \sigma =  10

   The sample size is n  =  16

Generally the standard error of the mean is mathematically represented as

       \sigma_{x} =  \frac{ \sigma}{\sqrt{n} }

=>    \sigma_{x} =  \frac{ 10 }{\sqrt{16 } }

=>    \sigma_{x} = 2.5

Generally the probability that the sample mean will be between 39 and 48 minutes is

    P(39 <  X < 48  ) =  P( \frac{ 39 - 45}{ 2.5} <  \frac{X - \mu }{\sigma } <  \frac{ 48 - 45}{ 2.5} )

=> P(39 <  X < 48  ) =  P(-2.4 < Z< 1.2 )

=> P(39 <  X < 48  ) =  P( Z< 1.2 ) - P(Z <  -2.4)

From the z table  the area under the normal curve to the left corresponding to  1.2  and  -2.4  is

=> P( Z< 1.2 ) = 0.88493

and  

    P( Z< - 2.4 ) = 0.0081975

So

   P(39 <  X < 48  ) = 0.88493 -0.0081975

=> P(39 <  X < 48  ) = 0.8767

From the question we are told the confidence level is  95% , hence the level of significance is    

      \alpha = (100 - 95 ) \%

=>   \alpha = 0.05

Generally from the normal distribution table the critical value  of   is  

   Z_{\frac{\alpha }{2} } =  1.96

Generally the margin of error is mathematically represented as  

      E = Z_{\frac{\alpha }{2} } *  \frac{\sigma }{\sqrt{n} }

=>   E = 1.96 * 2.5  

=>   E =4.9  

Generally the  95% of all sample means will fall between

      \mu  -E <  and   \mu   +E

=>   45  -4.9\   and \  45  + 4.9

Generally the value which  90% of sample means is  greater than is mathematically represented

      P( \= X >  \= x  ) = 0.90

=>   P( \= X >  \= x  ) =  P( \frac{\= X  - \mu }{ \sigma_x} >  \frac{\= x  -45 }{ 2.5}  ) = 0.90

=>  P( \= X >  \= x  ) =  P( Z >  z  ) = 0.90

Generally from the z-table  the critical  value  of  0.90  is  

      z = -1.282

      \frac{\= x  -45 }{ 2.5}  = -1.282

=>   \= x = 41. 795

Considering question 2

 Generally we are told that the standard deviation of the mean to be one fifth of the population standard deviation, this is mathematically represented as

         s = \frac{1}{5} \sigma

  Generally the standard deviation of the sample mean is mathematically  represented as

          s = \frac{\sigma }{ \sqrt{n} }

=>       \frac{1}{5} \sigma = \frac{\sigma }{ \sqrt{n} }

=>       n =  5^2

=>       n =  25

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2 years ago
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Please help me I’m stuck on this.
Sidana [21]

Answer:

The cost of the game is $7 .

Step-by-step explanation:

By looking at the graph, the line at x = 4 is equals to y = 7

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Jack scored a mean of 15 points per game in his first 3 basketball games. In his 4th game, he scored 27 points. What is his mean
timurjin [86]

His mean score for the first 4 games will be 18 points.

Explanation

In the first 3 games, the mean of the scores is 15 points . Mean is the simple average of some numbers.

So, the total score in first 3 games = (15*3)= 45 points

Now, in the 4th game, he scored 27 points. So, the total score in the first 4 games = 45+27= 72 points

Thus, the average of scores in first 4 games = \frac{72}{4} = 18 points

So, his mean score for the first 4 games will be 18 points.

7 0
3 years ago
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