<span>Ending Amt = Bgng Amt * e ^-0.03t
In this equation, the "-0.03" is the decay factor or "k"
We can now solve for half-life by this equation:
</span>t = <span>(<span>ln [y(t) ÷ a]<span>)<span> ÷ -k (we can say beginning amount = 200 and ending amount = 100
</span></span></span></span>t = <span>(<span>ln [200 ÷ 100]<span>)<span> ÷ -k
</span></span></span></span>t = <span>(<span>ln [2]<span>)<span> ÷ -k
</span></span></span></span>t = 0.69314718056<span> ÷ --.03
t =</span><span><span><span> 23.1049060187
</span>
about 23 years
</span></span>
Dude I don’t feel like working
Observations can you make from a comparison of the number of water supply stoppages reported by owner-occupied units versus renter-occupied units 1.
Total no. of times owner-occupied units had a water supply stoppage lasting 6 or more hours
547000 + 5012000 + 6110000 + 2544000 + 557000 = 14770000
x = no. of times owner-occupied units had a water supply stoppage.
<h3>What is the probability at x =0?</h3>
P(x=0) = 547000/14770000
P(x=0) = 0.0370
Similarly, we have at x=1
P(x=1) = 5012000/14770000
P(x=1) = 0.3393
P(x=2) = 6110000/14770000
P(x=2) = 0.4137
P(x=3) = 2544000/14770000
P(x=3) = 0.1722
P(x=4) = 557000/14770000
P(x=4) = 0.0378
x f(x)
0 0.0370
1 0.3393
2 0.4137
3 0.1722
4 0.0378
Total 1
Observations can you make from a comparison of the number of water supply stoppages reported by owner-occupied units versus renter-occupied units 1.
To learn more about the probability distribution visit:
brainly.com/question/24756209
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Answer: 95.40
Step-by-step explanation:
7.95*12=95.40
Answer: (3,4)
Each input can only have one output for a function
The other pairs give a different output for the input so that would make it a relation not a function.
