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GuDViN [60]
3 years ago
6

if $4200 is invested in a savings account for which interest is compounded semiannually, and if the $4200 turns into $4900 in 4

years, what is the interest rate of the savings account?
Mathematics
2 answers:
nydimaria [60]3 years ago
6 0
4900=4200(1+r/2)^(2*4)
Solve for r
R=(4900/4200)^(1/8))-1)*2
R=((4,900÷4,200)^(1÷8)−1)×2
R=0.039*100=3.9%
wel3 years ago
5 0

Answer:

3.89

Step-by-step explanation:

did test

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Answer:

The 99% confidence interval = (126.93,157.67)

Step-by-step explanation:

The formula for Confidence Interval =

Mean ± z × Standard deviation /√n

Mean = 142. 3 mmHg

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n = 12

Z score for 99% confidence interval = 2.56

Confidence Interval =

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142.3 ± 15.371373567

Confidence Interval

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= 126.92862643

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142.3 + 15.371373567

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2 years ago
The projected rate of increase in enrollment at a new branch of the UT-system is estimated by E ′ (t) = 12000(t + 9)−3/2 where E
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Answer:

The projected enrollment is \lim_{t \to \infty} E(t)=10,000

Step-by-step explanation:

Consider the provided projected rate.

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Integrate the above function.

E(t) =\int 12000(t + 9)^{\frac{-3}{2}}dt

E(t) =-\frac{24000}{\left(t+9\right)^{\frac{1}{2}}}+c

The initial enrollment is 2000, that means at t=0 the value of E(t)=2000.

2000=-\frac{24000}{\left(0+9\right)^{\frac{1}{2}}}+c

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Therefore, E(t) =-\frac{24000}{\left(t+9\right)^{\frac{1}{2}}}+10,000

Now we need to find \lim_{t \to \infty} E(t)

\lim_{t \to \infty} E(t)=-\frac{24000}{\left(t+9\right)^{\frac{1}{2}}}+10,000

\lim_{t \to \infty} E(t)=10,000

Hence, the projected enrollment is \lim_{t \to \infty} E(t)=10,000

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2 years ago
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Step-by-step explanation:

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This means that the equation is:

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