Answer:
Step-by-step explanation:
3
5h - w = 170
for h = 77
w = 5h - 170
w = 5*77 - 170
w = 385 - 170
<span>w = 215
Shoutout to @himanshuchaturvedi
Your answer is : w = 215
Have an amazing day and stay hopeful!
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We are not given tables, so will just use the amortization formula.

where
P=amount to be deposited today, to be found
A=amount withdrawn each year=18000
i=Annual interest=9%
n=number of years = 20
Substituting values,

=164313.82 to the nearest cent
Answer:
(a) 1825 = 2.25x + (2.25-1)(x -108)
(b) 560 mi/h
Step-by-step explanation:
(a) distance = speed·time
The first plane's speed is x. The distance it travels in 2.25 hours is 2.25x.
The second plane's speed is x-108. It travels only 1.25 hours (since it started an hour later). The distance it travels is then (2.25 -1)(x -108).
The problem statement tells us the total of the distances traveled by the two planes is 1825 miles, so we can write the equation ...
... 1825 = 2.25x + (2.25 -1)(x -108)
(b) Simplifying the equation gives ...
... 1825 = 3.50x -135
To solve this 2-step equation, we add 135, then divide by 3.50.
.. 1960 = 3.50x
... 1960/3.50 = x = 560
The first airplane's speed is 560 mph.
<u>Check</u>
In 2.25 hours, the first plane travels (560 mi/h)·(2.25 h) = 1260 mi.
In 1.25 hours, the second plane travels (452 mi/h)·(1.25 h) = 565 mi.
Then 2.25 hours after the first plane leaves, the planes are 1260 +565 = 1825 miles apart, as given in the problem statement.
The histogram is shown below.
You'll have 10 bars. Under each bar is a label from 0 to 9. The height of each bar represents the frequency of each units digit.
The histogram shows two bars that are relatively large compared to the rest. These bars have frequency of 16 and 14 (for units digits of 0 and 5 respectively). The distribution is nearly bimodal. If the two frequencies mentioned were the same, say both 16, then it would be exactly bimodal. As you can probably guess, bimodal means "two modes".
The rest of the bar heights are nearly the same, so the remaining portion of the distribution is nearly uniform. A uniform distribution has every bar the same height. Collectively, all the bars of a uniform distribution forms a larger rectangle.