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Lena [83]
3 years ago
14

Plz help

Mathematics
1 answer:
Aleonysh [2.5K]3 years ago
8 0
 1)Rewrite the table:
70,  49,  34.3,  24.01,  11.807

2) write the quotient of each number by the number before & notice the value:

49/70= 0.7
34.3/49 = 0.7
24.01/34.3 =0.7
16.0807/24.01 = 0.67 ≈0.7
You notice this is a geometric progression with r 0.7
The last term in a GP =ar^(n-1)
a=70; r= 0.7 n-1= number of terms (days -1)
For last term = 1 ==> then 1=70(0.7)^(n-1)
1/70 = (0.7)^(n-1)
log(1/70) =log[(0.7)^(n-1) ==> log1 - log 70 = (n-1) log(0.7) & you will find that it need 11.92 days to equal on & to be less than it will necessitate 12 Days

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Solution :

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$n_1=39, \ \  n_2=37$

M.E. , Margin of error,

$=t_{n_1+n_2-2,0.025} \times \left( s \sqrt{\frac{1}{n_1}+\frac{1}{n_2}}$

$s^2 = \frac{(n_1-1)S_1^2+(n_2-1)S^2_2}{n_1+n_2-2}$

  $=\frac{38(3.49)^2+36(2.47)^2}{39+37-2}$

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s = 3.03

$t_{74,0.05} = 1.99$

$M.E. = 1.99 \times 3.03 \times \sqrt{\frac{1}{39}+\frac{1}{37}}$

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Therefore, 95% CI = [(25.90-27.70) - 1.4 , (25.90-27.70) + 1.4]

                              = [-3.2, -0.4]

Therefore, the lower bound is -3.2

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