Answer: The rate of change in the value of the guitar is modeled by 0.95, which represents a decrease of 5% per year.
Step-by-step explanation:
Hi, to answer this question we have to apply an exponential decay function:
A = P (1 - r) t
Where:
p = original price
r = decrease rate (decimal form)
t= years
A = price after t years
Replacing with the values given:
A(t)=145(0.95)t.
Where
0.95 = 1-r
0.95-1 = r
-0.05 =-r
0.05 =r
Converting to percentage form:
0.05 x 100 = 5%
Feel free to ask for more if needed or if you did not understand something.
One side let it be 5 1/2, so 5 units and a half of one and then the other side make it 7 units wide.
Then connect the two end points to make a triangle, the area would be 19 1/4
Now you would need two.. just do the exact same thing but flip it so it doesn’t look alike
Remember you said the scale shows you 10% heavier.
So that's an increase of 10%
So value = 10% + 100% = 110%
So scale would show 110% of 82
110% of 82
= (110/100) * 82
= 1.1 * 82
= 90.2
OR
We can simply find 10% of 82 and add it to the original 82.
10% of 82
(10/100) * 82
0.1 * 82 = 8.2
8.2 + 82 = 90.2 Same answer as before.
So scale shows 90.2 kg.
To get the greatest possible quotient, you need the biggest possible
dividend and the smallest possible divisor.
96,432 divided by 225 = 428.5666... <== greatest possible
23,469 divided by 522 = 44.9597... <== smallest possible
Answer:
Using a formula, the standard error is: 0.052
Using bootstrap, the standard error is: 0.050
Comparison:
The calculated standard error using the formula is greater than the standard error using bootstrap
Step-by-step explanation:
Given
Sample A Sample B
![x_B = 50](https://tex.z-dn.net/?f=x_B%20%3D%2050)
![n_B =250](https://tex.z-dn.net/?f=n_B%20%3D250)
Solving (a): Standard error using formula
First, calculate the proportion of A
![p_A = \frac{x_A}{n_A}](https://tex.z-dn.net/?f=p_A%20%3D%20%5Cfrac%7Bx_A%7D%7Bn_A%7D)
![p_A = \frac{30}{100}](https://tex.z-dn.net/?f=p_A%20%3D%20%5Cfrac%7B30%7D%7B100%7D)
![p_A = 0.30](https://tex.z-dn.net/?f=p_A%20%3D%200.30)
The proportion of B
![p_B = \frac{x_B}{n_B}](https://tex.z-dn.net/?f=p_B%20%3D%20%5Cfrac%7Bx_B%7D%7Bn_B%7D)
![p_B = \frac{50}{250}](https://tex.z-dn.net/?f=p_B%20%3D%20%5Cfrac%7B50%7D%7B250%7D)
![p_B = 0.20](https://tex.z-dn.net/?f=p_B%20%3D%200.20)
The standard error is:
![SE_{p_A-p_B} = \sqrt{\frac{p_A * (1 - p_A)}{n_A} + \frac{p_A * (1 - p_B)}{n_B}}](https://tex.z-dn.net/?f=SE_%7Bp_A-p_B%7D%20%3D%20%5Csqrt%7B%5Cfrac%7Bp_A%20%2A%20%281%20-%20p_A%29%7D%7Bn_A%7D%20%2B%20%5Cfrac%7Bp_A%20%2A%20%281%20-%20p_B%29%7D%7Bn_B%7D%7D)
![SE_{p_A-p_B} = \sqrt{\frac{0.30 * (1 - 0.30)}{100} + \frac{0.20* (1 - 0.20)}{250}}](https://tex.z-dn.net/?f=SE_%7Bp_A-p_B%7D%20%3D%20%5Csqrt%7B%5Cfrac%7B0.30%20%2A%20%281%20-%200.30%29%7D%7B100%7D%20%2B%20%5Cfrac%7B0.20%2A%20%281%20-%200.20%29%7D%7B250%7D%7D)
![SE_{p_A-p_B} = \sqrt{\frac{0.30 * 0.70}{100} + \frac{0.20* 0.80}{250}}](https://tex.z-dn.net/?f=SE_%7Bp_A-p_B%7D%20%3D%20%5Csqrt%7B%5Cfrac%7B0.30%20%2A%200.70%7D%7B100%7D%20%2B%20%5Cfrac%7B0.20%2A%200.80%7D%7B250%7D%7D)
![SE_{p_A-p_B} = \sqrt{\frac{0.21}{100} + \frac{0.16}{250}}](https://tex.z-dn.net/?f=SE_%7Bp_A-p_B%7D%20%3D%20%5Csqrt%7B%5Cfrac%7B0.21%7D%7B100%7D%20%2B%20%5Cfrac%7B0.16%7D%7B250%7D%7D)
![SE_{p_A-p_B} = \sqrt{0.0021+ 0.00064}](https://tex.z-dn.net/?f=SE_%7Bp_A-p_B%7D%20%3D%20%5Csqrt%7B0.0021%2B%200.00064%7D)
![SE_{p_A-p_B} = \sqrt{0.00274}](https://tex.z-dn.net/?f=SE_%7Bp_A-p_B%7D%20%3D%20%5Csqrt%7B0.00274%7D)
![SE_{p_A-p_B} = 0.052](https://tex.z-dn.net/?f=SE_%7Bp_A-p_B%7D%20%3D%200.052)
Solving (a): Standard error using bootstrapping.
Following the below steps.
- Open Statkey
- Under Randomization Hypothesis Tests, select Test for Difference in Proportions
- Click on Edit data, enter the appropriate data
- Click on ok to generate samples
- Click on Generate 1000 samples ---- <em>see attachment for the generated data</em>
From the randomization sample, we have:
Sample A Sample B
![x_B = 57](https://tex.z-dn.net/?f=x_B%20%3D%2057)
![n_B =250](https://tex.z-dn.net/?f=n_B%20%3D250)
![p_A = 0.228](https://tex.z-dn.net/?f=p_A%20%3D%200.228)
So, we have:
![SE_{p_A-p_B} = \sqrt{\frac{p_A * (1 - p_A)}{n_A} + \frac{p_A * (1 - p_B)}{n_B}}](https://tex.z-dn.net/?f=SE_%7Bp_A-p_B%7D%20%3D%20%5Csqrt%7B%5Cfrac%7Bp_A%20%2A%20%281%20-%20p_A%29%7D%7Bn_A%7D%20%2B%20%5Cfrac%7Bp_A%20%2A%20%281%20-%20p_B%29%7D%7Bn_B%7D%7D)
![SE_{p_A-p_B} = \sqrt{\frac{0.23 * (1 - 0.23)}{100} + \frac{0.228* (1 - 0.228)}{250}}](https://tex.z-dn.net/?f=SE_%7Bp_A-p_B%7D%20%3D%20%5Csqrt%7B%5Cfrac%7B0.23%20%2A%20%281%20-%200.23%29%7D%7B100%7D%20%2B%20%5Cfrac%7B0.228%2A%20%281%20-%200.228%29%7D%7B250%7D%7D)
![SE_{p_A-p_B} = \sqrt{\frac{0.1771}{100} + \frac{0.176016}{250}}](https://tex.z-dn.net/?f=SE_%7Bp_A-p_B%7D%20%3D%20%5Csqrt%7B%5Cfrac%7B0.1771%7D%7B100%7D%20%2B%20%5Cfrac%7B0.176016%7D%7B250%7D%7D)
![SE_{p_A-p_B} = \sqrt{0.001771 + 0.000704064}](https://tex.z-dn.net/?f=SE_%7Bp_A-p_B%7D%20%3D%20%5Csqrt%7B0.001771%20%2B%200.000704064%7D)
![SE_{p_A-p_B} = \sqrt{0.002475064}](https://tex.z-dn.net/?f=SE_%7Bp_A-p_B%7D%20%3D%20%5Csqrt%7B0.002475064%7D)
![SE_{p_A-p_B} = 0.050](https://tex.z-dn.net/?f=SE_%7Bp_A-p_B%7D%20%3D%200.050)