Answer:
A) sample mean = $1.36 million
B) standard deviation = $0.9189 million
C) confidence interval = ($1.93 million , $0.79 million)
*since the sample size is very small, the confidence interval is not valid.
Step-by-step explanation:
samples:
- $2.7 million
- $2.4 million
- $2.2 million
- $2 million
- $1.5 million
- $1.5 million
- $0.5 million
- $0.5 million
- $0.2 million
- $0.1 million
sample mean = $1.36 million
the standard deviation:
- $2.7 million - $1.36 million = 1.34² = 1.7956
- $2.4 million - $1.36 million = 1.04² = 1.0816
- $2.2 million - $1.36 million = 0.84² = 0.7056
- $2 million - $1.36 million = 0.64² = 0.4096
- $1.5 million - $1.36 million = 0.14² = 0.0196
- $1.5 million - $1.36 million = 0.14² = 0.0196
- $0.5 million - $1.36 million = -0.86² = 0.7396
- $0.5 million - $1.36 million = -0.86² = 0.7396
- $0.2 million - $1.36 million = -1.16² = 1.3456
- $0.1 million - $1.36 million = -1.26² = 1.5876
- total $8.444 million / 10 = $0.8444 million
standard deviation = √0.8444 = 0.9189
95% confidence interval = mean +/- 1.96 standard deviations/√n:
$1.36 million + [(1.96 x $0.9189 million)/√10] = $1.36 million + $0.57 million = $1.93 million
$1.36 million - $0.57 million = $0.79 million
476+23.8+11.9= 499 510 the answer is 511.70
Answer:
$18.5685 or $18.56
Step-by-step explanation:
The answer is $18.56 because 123.79 multiplied by 15% is $18.56
Answer:
Step-by-step explanation:
Given
![\sqrt[3]{217}](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B217%7D)
Required
Solve
Linear approximated as:
Take:
So:
![f(x) = \sqrt[3]{x}](https://tex.z-dn.net/?f=f%28x%29%20%3D%20%5Csqrt%5B3%5D%7Bx%7D)
Substitute 216 for x
![f(x) = \sqrt[3]{216}](https://tex.z-dn.net/?f=f%28x%29%20%3D%20%5Csqrt%5B3%5D%7B216%7D)
So, we have:
To calculate f'(x);
We have:
Rewrite as:
Differentiate
Split
Substitute 216 for x
So:
