Answer:
can you be more clear?
Step-by-step explanation:
Answer:
The 90% confidence interval for the average monthly residential water usage for all households in this city is between 4401.3 gallons and 4598.7 gallons.
Step-by-step explanation:
We have that to find our 
level, that is the subtraction of 1 by the confidence interval divided by 2. So:
Now, we have to find z in the Ztable as such z has a pvalue of 
.
So it is z with a pvalue of 
, so 
Now, find the margin of error M as such
In which 
is the standard deviation of the population and n is the size of the sample.
The lower end of the interval is the sample mean subtracted by M. So it is 4500 - 98.7 = 4401.3 gallons
The upper end of the interval is the sample mean added to M. So it is 4500 + 98.7 = 4598.7 gallons
The 90% confidence interval for the average monthly residential water usage for all households in this city is between 4401.3 gallons and 4598.7 gallons.
The equation that represents the amount saved by Alyssa after tt weeks is AA = 100 + $7tt
The equation that represents the amount saved by Colton after tt weeks is CC = $55 + $10tt
The amount they would have when they have same amount of money is $205.
<h3>What equation represents the total amount saved?</h3>
The total amount saved: inital amount + (amount saved per week x total number of weeks
AA = 100 + $7tt
CC = $55 + $10tt
<h3>What would be the amount when they have the same amount of money?</h3>
$55 + $10tt = 100 + $7tt
Combine similar terms and solve for tt
100 - 55 = 10tt - 7tt
45 = 3tt
tt = 15 weeks
100 + $7(15) = $205
To learn more about linear functions, please check: brainly.com/question/26434260
Answer:
- number of multiplies is n!
- n=10, 3.6 ms
- n=15, 21.8 min
- n=20, 77.09 yr
- n=25, 4.9×10^8 yr
Step-by-step explanation:
Expansion of a 2×2 determinant requires 2 multiplications. Expansion of an n×n determinant multiplies each of the n elements of a row or column by its (n-1)×(n-1) cofactor determinant. Then the number of multiplies is ...
mpy[n] = n·mp[n-1]
mpy[2] = 2
So, ...
mpy[n] = n! . . . n ≥ 2
__
If each multiplication takes 1 nanosecond, then a 10×10 matrix requires ...
10! × 10^-9 s ≈ 0.0036288 s ≈ 0.004 s . . . for 10×10
Then the larger matrices take ...
n=15, 15! × 10^-9 ≈ 1307.67 s ≈ 21.8 min
n=20, 20! × 10^-9 ≈ 2.4329×10^9 s ≈ 77.09 years
n=25, 25! × 10^-9 ≈ 1.55112×10^16 s ≈ 4.915×10^8 years
_____
For the shorter time periods (less than 100 years), we use 365.25 days per year.
For the longer time periods (more than 400 years), we use 365.2425 days per year.
