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olchik [2.2K]
3 years ago
7

Find three consecutive odd numbers whose sum is 303

Mathematics
2 answers:
CaHeK987 [17]3 years ago
8 0
n;\ n+2;\ n+4-three\ consecutive\ odd\ numbers\\\\n+(n+2)+(n+4)=303\\\\3n+6=303\\\\3n=303-6\\\\3n=297\ \ \ \ /:3\\\\n=99\\\\Answer:99;\ 101;\ 103.
Roman55 [17]3 years ago
5 0
99 + 101 + 103 = 303
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Help please I don't understand!!!
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A- 2b = 2  one of the 2 "A" most be negative so they can cancer each other 
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3 years ago
Boxes of raisins are labled as containing 22 ounces. Following are the weights, in the ounces, of a sample of 12 boxes. It is re
kati45 [8]

Answer:

The 99 % confidence interval on the basis of mean is ( 21.7393  ;   22.0257)

Step-by-step explanation:

Mean = Sum of observations / Number of observations

Mean = 21.88 +21.76 +22.14 +21.63+ 21.81 +22.12+ 21.97+ 21.57+ 21.75+ 21.96 +22.20 +21.80/ 12

Mean =x`= 262.59/12=  21.8825

Standard Deviation = s= ∑x²/n - ( ∑x/n)²

∑x²/n= 478.7344 +473.4976 + 490.1796+467.8569+ 475.6761 + 489.2944+ 482.6809+ 465.2649+ 473.0625+ 482.2416 +492.84 + 475.24/ 12

∑x²/n= 5746.5689/12= 478.8807 = 478.881

Standard Deviation = s= ∑x²/n - ( ∑x/n)²

s= 478.881- (21.8825)²= 478.881-478.843= 0.037

The confidence limit 99% for the mean will be determined by

x` ± α(100-1) √s/n

Putting the values in the above equation

= 21.8825 ± 2.58 √0.037/12

Solving the square root

= 21.8825 ± 2.58 (0.05549)

Multiplying the square root with 2.58

=21.8825 ± 0.1432

Adding and subtracting would give

21.7393  ;   22.0257,

Hence the 99 % confidence interval on the basis of mean is ( 21.7393  ;   22.0257)

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