Answer:
$48
Step-by-step explanation:
To start off, we know that Avery put 20% of the $80 in her bank account. We can figure out how much 20% of 80 is by multiplying 80 by 20% and then subtracting that number from 80. Remember to switch 20% to its decimal form (20% -> .2).
Step 1) 80 x .2 = 16
Step 2) 80 - 16 = 64
Next, we have to figure out how much 1/4 of the remaining $64 would be. We just simply divide 64 by 4 to figure out how much Avery spent.
Step 3) 64 / 4 = 16
Lastly, since we know Avery spent 16 dollars, we can subtract the 16 from 64 to get our final answer.
Step 4) 64 - 16 = 48
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➷ 'm' is the slope or gradient of the line
'c' is the y intercept (where the line crosses the y axis)
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Answer:
the sum is a non terminating and a non repeating decimal
Answer: x=4
Step-by-step explanation:
The perimeter´s formula for this kind of polygons is
The triangle perimeter would be:
(x+2)+(x+3)+(x+3)
And the rectangle:
(x+2)+(x+2)+x+x
For them to have the same value, we should equal them:
(x+2)+(x+3)+(x+3)=(x+2)+(x+2)+x+x
Working with this:
3x+8=4x+4
x=4
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]
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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:
brainly.com/question/13770164