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arlik [135]
3 years ago
5

Five people each write a positive integer on a piece of paper. If the average of these five integers is 8, what is the greatest

possible value of one of these integers?
A. 36
B. 38
C. 39
D. 40
Mathematics
2 answers:
frozen [14]3 years ago
8 0

Answer: 36

Step-by-step explanation:

erik [133]3 years ago
6 0

Answer:

Option D. 40

Step-by-step explanation:

Let

x------>the greatest possible value  

we know that

If four people write the number zero, then the average will be

8=(0+0+0+0+x)/5\\ \\40=0+x\\ \\x=40

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A game wheel has 38 slots numbered 1-38, respectively. Matt places a bet on the odd numbers and Julie places a bet on numbers 10
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Odd numbers are of the from a(n)=a+d(n-1) where a=1 and d=2 so

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3 years ago
A group of students estimated the length of one minute without reference to a watch or​ clock, and the times​ (seconds) are list
alexandr402 [8]

Complete Question

The complete question is shown on the first uploaded image  

Answer:

Yes the students are reasonably good at estimating one​ minute

a

  Option A is  correct

b

  The test statistics is  t  =  0.354

Step-by-step explanation:

From the question we are told that

    The  set of data is  

               68, 82 ,  38 ,  62 , 41, 25 , 57 ,  64, 67, 47, 61, 71, 91, 87, 64

     The  population mean is  \mu  = 60

The  level of significance is given as\alpha =  0.01

    The  critical value for this level of significance obtained from the normal distribution table is  

         Z_{\alpha } =  2.33

   The  null hypothesis is  

           H_o  :  \mu  =  60 \ seconds

   The alternative hypothesis is  

          Ha :  \mu \ne 60 \ seconds      

Generally the sample mean is mathematically represented as

       \= x  =  \frac{\sum x_i }{n  }

where  n = 15

 So  

      \= x  =  \frac{ 68+ 82 +  38 +  62 + 41+ 25 + 57 + 64+67+ 47+ 61+ 71+ 91+ 87+ 64}{15}

      \= x  =  61.67

The standard deviation is mathematically represented as

     \sigma  =\sqrt{  \frac{ \sum  (x_i -  \= x )^2}{n} }

substituting values

      \sigma  =\sqrt{  \frac{ \sum  (68-61.67)^2 + ( 82-61.67 )^2  +   (38 -61.67)^2 +  62-61.67)^2 + (41-61.67)^2 +(25-61.67)^2 +( 57+61.67)^2 }{15} }           \sqrt{  \frac{  ( \cdot  \cdot  + (  64-61.67)^2 + (67-61.67)^2 +(47-61.67)^2 + (61-61.67)^2+  (71-61.67)^2 + (91 -61.67)^2+( 87-61.67)^2 +  (64-61.67)^2}{15} }=>   \sigma  =  18.23

The  test statistics is evaluated as  

      t  =  \frac{\= x  - \mu  }{ \frac{\sigma }{\sqrt{n} } }

substituting values

       t  =  \frac{61.67  -60  }{ \frac{18.23 }{\sqrt{ 15} } }

      t  =  0.354

Now comparing the statistics and the critical value of the level of significance we see that the the test statistics is less than the critical value

Hence the we fail to reject the null hypothesis which mean that these times are from a population with a mean equal to 60 seconds

So we can state that yes the students are reasonably good at estimating one minute given that the sample mean is not far from the population mean

4 0
3 years ago
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