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Varvara68 [4.7K]
3 years ago
6

Each investment matures in 3 years. The interest compounds annually.

Mathematics
1 answer:
Sveta_85 [38]3 years ago
4 0

bearing in mind that 4¾ is simply 4.75.

\bf ~~~~~~ \textit{Compound Interest Earned Amount} \\\\ A=P\left(1+\frac{r}{n}\right)^{nt} \quad \begin{cases} A=\textit{accumulated amount}\\ P=\textit{original amount deposited}\dotfill &\$600\\ r=rate\to 5\%\to \frac{5}{100}\dotfill &0.05\\ n= \begin{array}{llll} \textit{times it compounds per year}\\ \textit{annually, thus once} \end{array}\dotfill &1\\ t=years\dotfill &3 \end{cases} \\\\\\ A=600\left(1+\frac{0.05}{1}\right)^{1\cdot 3}\implies A=600(1.05)^3\implies A=694.575 \\\\[-0.35em] ~\dotfill

\bf ~~~~~~ \textit{Compound Interest Earned Amount} \\\\ A=P\left(1+\frac{r}{n}\right)^{nt} \quad \begin{cases} A=\textit{accumulated amount}\\ P=\textit{original amount deposited}\dotfill &\$750\\ r=rate\to 4.75\%\to \frac{4.75}{100}\dotfill &0.0475\\ n= \begin{array}{llll} \textit{times it compounds per year}\\ \textit{annually, thus once} \end{array}\dotfill &1\\ t=years\dotfill &3 \end{cases} \\\\\\ A=750\left(1+\frac{0.0475}{1}\right)^{1\cdot 3}\implies A=750(1.0475)^3\implies A\approx 862.032

well, the interest for each is simply A - P

695.575 - 600 = 95.575.

862.032 - 750 = 112.032.

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Answer:

Null hypothesis:\mu \leq 8000  

Alternative hypothesis:\mu > 8000  

z=\frac{8300-8000}{\frac{1200}{\sqrt{64}}}=2  

p_v =P(Z>2)=0.0228  

Step-by-step explanation:

1) Data given and notation  

\bar X=8300 represent the sample mean  

\sigma=1200 represent the population standard deviation  

n=64 sample size  

\mu_o =800 represent the value that we want to test  

\alpha represent the significance level for the hypothesis test.  

z would represent the statistic (variable of interest)  

p_v represent the p value for the test (variable of interest)  

2) State the null and alternative hypotheses.  

We need to conduct a hypothesis in order to check if the mean is higher than 8000, the system of hypothesis are :  

Null hypothesis:\mu \leq 8000  

Alternative hypothesis:\mu > 8000  

Since we know the population deviation, is better apply a z test to compare the actual mean to the reference value, and the statistic is given by:  

z=\frac{\bar X-\mu_o}{\frac{\sigma}{\sqrt{n}}} (1)  

z-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".  

3) Calculate the statistic  

We can replace in formula (1) the info given like this:  

z=\frac{8300-8000}{\frac{1200}{\sqrt{64}}}=2  

4) P-value  

Since is a one-side upper test the p value would given by:  

p_v =P(Z>2)=0.0228  

5) Conclusion  

If we compare the p value and the significance level assumed, for example \alpha=0.05 we see that p_v so we can conclude that we have enough evidence to reject the null hypothesis, so we can conclude that the mean is significantly higher than $8000.  

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