La Mamá de la mamá de la mamá de la mamá de la mamá
Answer:
$34
Step-by-step explanation:
I'd work backwards.
For the third book, she paid all her remaining money. The problem says she paid "1/2 her leftover money + $3". This means that: (let m = money used to buy book 3)
m = 1/2m + 3
1/2m = 3
m = 6
For the second book: (let n = money before book 2)
n - m (money left after book 2) = 1/2n + 2
1/2n +2 is money used up for book, which is the same as n-m.
n = 1/2n + 2 + m
1/2n = 2 + m = 2 + 6
1/2n =8
n = 16
For the first book, she spent 1/2 her money + 1. If o = money before book 1 (or the whole allowance):
o - n = 1/2o + 1
o = 1/2o + n + 1 = 1/2o + 16 + 1
1/2o = 17
o = 34
Check!
Spent 17 (half 34) + 1 on book 1
16 left
Spent 8 (half 16) +2 on book 2
6 left
Spent 3 (half 6) + 3 on book 3
0 left
<span> The lower and upper bounds of the confidence intervals must be equally distanced from the mean
so it will be
</span><span>70.9 - 73.1
</span>hope it helps
The 90% confidence interval for the population mean of the considered population from the given sample data is given by: Option C: [130.10, 143.90]
<h3>
How to find the confidence interval for population mean from large samples (sample size > 30)?</h3>
Suppose that we have:
- Sample size n > 30
- Sample mean =

- Sample standard deviation = s
- Population standard deviation =

- Level of significance =

Then the confidence interval is obtained as
- Case 1: Population standard deviation is known

- Case 2: Population standard deviation is unknown.

For this case, we're given that:
- Sample size n = 90 > 30
- Sample mean =
= 138 - Sample standard deviation = s = 34
- Level of significance =
= 100% - confidence = 100% - 90% = 10% = 0.1 (converted percent to decimal).
At this level of significance, the critical value of Z is:
= ±1.645
Thus, we get:
![CI = \overline{x} \pm Z_{\alpha /2}\dfrac{s}{\sqrt{n}}\\CI = 138 \pm 1.645\times \dfrac{34}{\sqrt{90}}\\\\CI \approx 138 \pm 5.896\\CI \approx [138 - 5.896, 138 + 5.896]\\CI \approx [132.104, 143.896] \approx [130.10, 143.90]](https://tex.z-dn.net/?f=CI%20%3D%20%5Coverline%7Bx%7D%20%5Cpm%20Z_%7B%5Calpha%20%2F2%7D%5Cdfrac%7Bs%7D%7B%5Csqrt%7Bn%7D%7D%5C%5CCI%20%3D%20138%20%5Cpm%201.645%5Ctimes%20%5Cdfrac%7B34%7D%7B%5Csqrt%7B90%7D%7D%5C%5C%5C%5CCI%20%5Capprox%20138%20%5Cpm%205.896%5C%5CCI%20%5Capprox%20%5B138%20-%205.896%2C%20138%20%2B%205.896%5D%5C%5CCI%20%5Capprox%20%5B132.104%2C%20143.896%5D%20%5Capprox%20%5B130.10%2C%20143.90%5D)
Thus, the 90% confidence interval for the population mean of the considered population from the given sample data is given by: Option C: [130.10, 143.90]
Learn more about confidence interval for population mean from large samples here:
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