The value of t that makes the factor e^(.032t) have the value of 2 can be found using logarithms.
2 = e^(0.032t)
ln(2) = ln(e^(0.032t)) = 0.032t
t = ln(2)/0.032 ≈ 21.66
It would take 21.66 years for the cost to double.
We need the half-life of C-14 which is 5,730 years.
Now, we will need a half-life equation:
elapsed time = half-life * log (bgng amt / ending amt) / log 2
We'll say beginning amount = 100 and ending amount = 41
elapsed time = 5,730 * log (100/41) / log 2
elapsed time = 5,730 * log (
<span>
<span>
<span>
2.4390243902
</span>
</span>
</span>
) / 0.30102999566
elapsed time = 5,730 * 0.38721614327 / 0.30102999566
elapsed time =
<span>
<span>
</span></span><span><span><span>5,730 * 1.2863041851
</span>
</span>
</span>
<span>elapsed time = 7,370.523 years
Source:
http://www.1728.org/halflife.htm </span>
By testing the hypothesis we can conclude that the bag does not contain 11 kg of food.
Given mean of 10.6 kg, population standard deviation of 0.6 and sample size of 25.
We are required to find whether the company is right in saying that their bags contain 11 kg of food.
First we have to make hypothesis.
:μ≠11
:μ=11
We have to use t statistic because the sample size is less than 30.
t=(X-μ)/s/
We will use s/
=0.6 because we have already given population standard deviation of weights.
t=(11-10.6)/0.6
=0.4/0.6
=0.667
Degree of freedom=n-1
=25-1
=24
T critical at 0.05 with degree of freedom 24=2.0639
T critical at 0.05 with degree of freedom is greater than calculated t so we will accept the null hypothesis.It concludes that the bags donot contain 11 kg of food.
Hence by testing the hypothesis we can conclude that the bag does not contain 11 kg of food.
Learn more about t test at brainly.com/question/6589776
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