Answer:
The upper boundary of the 95% confidence interval for the average unload time is 264.97 minutes
Step-by-step explanation:
We have the standard deviation for the sample, but not for the population, so we use the students t-distribution to solve this question.
The first step to solve this problem is finding how many degrees of freedom, we have. This is the sample size subtracted by 1. So
df = 35 - 1 = 35
95% confidence interval
Now, we have to find a value of T, which is found looking at the t table, with 34 degrees of freedom(y-axis) and a confidence level of ![1 - \frac{1 - 0.9}{2} = 0.975([tex]t_{975}](https://tex.z-dn.net/?f=1%20-%20%5Cfrac%7B1%20-%200.9%7D%7B2%7D%20%3D%200.975%28%5Btex%5Dt_%7B975%7D)
). So we have T = 2.0322
The margin of error is:
M = T*s = 2.0322*30 = 60.97
The upper end of the interval is the sample mean added to M. So it is 204 + 60.97 = 264.97
The upper boundary of the 95% confidence interval for the average unload time is 264.97 minutes
Answer:
∠B = 17°
Step-by-step explanation:
Vertical angles are congruent, thus
5x + 7 = x + 15 ( subtract x from both sides )
4x + 7 = 15 ( subtract 7 from both sides )
4x = 8 ( divide both sides by 4 )
x = 2
Hence
∠B = 5x + 7 = (5 × 2) + 7 = 10 + 2 = 17°
Answer:(9,-5)
Step-by-step explanation: trust me
Answer:
6 Years
Step-by-step explanation:
Orlando invests $1000 at 6% annual interest compounded daily.
Orlando's investment = 
Bernadette invests $1000 at 7% simple interest.
Bernadette's investment = A = 1000(1+0.07×t)
By trail and error method we will use t = 5
Bernadette's investment will be after 5 years
1000(1 + 0.07 × 5)
= 1000(1 + 0.35)
= 1000 × 1.35
= $1350
Orlando's investment after 5 years
= 
= 
= 1000(1.349826)
= 1349.825527 ≈ $1349.83
After 5 years Orlando's investment will not be more than Bernadette's.
Therefore, when we use t = 6
After 6 years Orlando's investment will be = $1433.29
and Bernadette's investment will be = $1420
So, after 6 whole years Orlando's investment will be worth more than Bernadette's investment.
Answer:
They lose about 2.79% in purchasing power.
Step-by-step explanation:
Whenever you're dealing with purchasing power and inflation, you need to carefully define what the reference is for any changes you might be talking about. Here, we take <em>purchasing power at the beginning of the year</em> as the reference. Since we don't know when the 6% year occurred relative to the year in which the saving balance was $200,000, we choose to deal primarily with percentages, rather than dollar amounts.
Each day, the account value is multiplied by (1 + 0.03/365), so at the end of the year the value is multiplied by about
... (1 +0.03/365)^365 ≈ 1.03045326
Something that had a cost of 1 at the beginning of the year will have a cost of 1.06 at the end of the year. A savings account value of 1 at the beginning of the year would purchase one whole item. At the end of the year, the value of the savings account will purchase ...
... 1.03045326 / 1.06 ≈ 0.9721 . . . items
That is, the loss of purchasing power is about ...
... 1 - 0.9721 = 2.79%
_____
If the account value is $200,000 at the beginning of the year in question, then the purchasing power <em>normalized to what it was at the beginning of the year</em> is now $194,425.14, about $5,574.85 less.
