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denis23 [38]
2 years ago
15

Please help me.....!!!!!

Mathematics
2 answers:
sertanlavr [38]2 years ago
8 0

The value of the investment of $2500 for 3 years will be $3703.86.

<h3>What is compound interest?</h3>

Compound interest is applicable when there will be a change in principle amount after the given time period.

For instance, if you give someone $500 at a 10% yearly rate, $500 is considered your primary sum. After a year, the interest will be $50, making the principle amount $550. Moving forward, the interest will be $550 rather than $500.

Given,

Principle amount (P) = $2500

Rate of interest (R) = 14%.

Time period (T) = 3 years.

Compound interest formula

A = P[1 + R/100]^{T}

So,

A = 2500[1 + 14/100]³

A = $3703.86

Hence the investment of $2500 at a rate of 14% for 3 years will be $3703.86.

For more information about compound interest

brainly.com/question/26457073

#SPJ1

Marianna [84]2 years ago
8 0

Answer:

Step-by-step explanation:

2500(1.14)^3 ≈ $3704

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A student at a four-year college claims that average enrollment at four-year colleges is higher than at two-year colleges in the
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Answer:

a) Since we have sample data and we don't know the population deviation we can use the t distribution in order to test the hypothesis

t would represent the statistic

b) t=\frac{(\bar X_{2}-\bar X_{1})-\Delta}{\sqrt{\frac{s^2_{1}}{n_{1}}+\frac{s^2_{2}}{n_{2}}}} (1)  

The degrees of freedom are given by df=n_1 +n_2 -2=35+35-2=68  Replacing the info given we got:

t=\frac{(5216-5069)-0}{\sqrt{\frac{4773^2}{35}+\frac{8141^2}{35}}}}=0.0921  

c) p_v =P(t_{68}>0.0921)=0.463  

Since the p value is higher than the significance level we don't have enough evidence to conclude that the true mean of enrollment is significantly higher for four year college than for two year college

Step-by-step explanation:

Information given

\bar X_{1}=5069 represent the mean for two year colleges

\bar X_{2}=5216 represent the mean for four year college

s_{1}=4773 represent the sample standard deviation for 1  

s_{2}=8141 represent the sample standard deviation for 2  

n_{1}=35 sample size for the group 2  

n_{2}=35 sample size for the group 2  

\alpha=0.05 Significance level provided

Part a

Since we have sample data and we don't know the population deviation we can use the t distribution in order to test the hypothesis

t would represent the statistic

Part b

We want to test if the mean of enrollment for four year college is higher than for two year college, the system of hypothesis would be:  

Null hypothesis:\mu_{2}-\mu_{1} \leq 0  

Alternative hypothesis:\mu_{2} - \mu_{1} > 0  

The statistic is given by:

t=\frac{(\bar X_{2}-\bar X_{1})-\Delta}{\sqrt{\frac{s^2_{1}}{n_{1}}+\frac{s^2_{2}}{n_{2}}}} (1)  

The degrees of freedom are given by df=n_1 +n_2 -2=35+35-2=68  Replacing the info given we got:

t=\frac{(5216-5069)-0}{\sqrt{\frac{4773^2}{35}+\frac{8141^2}{35}}}}=0.0921  

Part c

The p value would be given by:

p_v =P(t_{68}>0.0921)=0.463  

Since the p value is higher than the significance level we don't have enough evidence to conclude that the true mean of enrollment is significantly higher for four year college than for two year college

6 0
4 years ago
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