Question

1. If $200 is invetsed at 6% annual interest compounded continuously, when will the investment be worth $300? Use A=pe^n
2. A surveyor finds that the angle between two merging roads is 120. Find the value of secant 120
3.The mean score of students is 65.3 and the standard deviation is 20.15.Find the two values between which most lie at least 75% of the data.

Answer 1

hhhhhheeeeeeeeeeelppppppppppppppppppp

Answer 2

1) \[A = 300, p =200, n=?\] You are trying to solve for n, so:\[300=200e^n\]\[\ln(300)=\ln(200e^n) \]\[\ln(300)=\ln(200)+\ln(e^n)\]\[\ln(300)=\ln(200)+n\]
\[n=\ln(300)-\ln(200)=\ln(\frac{300}{200}) \approx 0.4055 \]

Answer 3

r u sure you didn't forget the interest rate? Lol I just noticed this now it is not in your formula. I assume you meant \[A=pe^{rt}\]

Answer 4

\[A=pe^{rn}\]

Answer 5

otherwise you are assuming a 100% interest rate :)

Answer 6

2) \[\sec(120)=\frac{1}{\cos(120)}\]\[= \frac{1}{\cos(\frac{2 \pi}{3})}=\frac{1}{\frac{-1}{2}}=-2\]

Answer 7

for #1, to fix the problem using the interest rate , you do the same thing but just add the "r" next to the n, and the subsequent steps are:
\[\ln(300)=\ln(200)+\ln(e^{rn})=\ln(200)+rn\]\[rn = \ln(300)-\ln(200)=\ln(\frac{300}{200})\]\[n=\ln(\frac{300}{200})*\frac{1}{r} \approx 6.76\] by plugging in r = 0.06

Answer 8

#3) is an application of Cheyshev's rule:
75% of the measurements will fall within 2 standard deviations of the mean, so you look at the interval \[(\mu-2s,\mu+2s)\] where \[\mu\] is the mean, and s is the standard deviation, so:
\[(65.3-2(20.15),65.3+2(20.15))\]

Answer 9

\[31^{0}10\]