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Karolina [17]
3 years ago
11

3.

Mathematics
1 answer:
irina1246 [14]3 years ago
5 0
3)
Principal Amount = P = $7200
Interest rate = r = 4% = 0.04
Time = t = 8 years

Formula for compounding is:

A=P (1+r)^{t}

Using the values, we get:

A=7200 (1+0.04)^{8}=9853.70

Therefore, the compounded amount will be $9853.70 

4)
Principal Amount = P = $7200
Compounded Amount = A = $9853.70

Interest Earned = Compounded Amount - Principal Amount
So,
Interest Earned = 9853.70 - 7200 = $2653.70

Therefore, $2653.70 will be earned as interest over a period of 8 years.
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Answer:

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Step-by-step explanation:

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$3000 is invested at 3% annual interest for 3 years.<br> a. Simple Interest b. Compound Interest
stich3 [128]

Answer:

a. $270

b. $3,278.18

Step-by-step explanation:

Given that

The principal amount is $3,000

Annual rate of interest is 3%

And, the time period is  3 years

We need to find out the simple interest & compound interest

The following formulas should be used

a. For simple interest

= Principal × rate of interest × time period

= $3,000 × 3% × 3 years

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b. For compound interest

= Principal × (1 + rate of interest)^time period

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2 years ago
The following prices, in dollars, of 7.5-cubic-foot refrigerators were recorded from a random sample. 314 305 344 283 285 310​ 3
Mariana [72]

Answer:

t=\frac{310.9-300}{\frac{31.09}{\sqrt{10}}}=1.109      

The degrees of freedom are given by:

df=n-1=10-1=9  

And the p value would be given by:

p_v =P(t_{9}>1.109)=0.148  

Since the p value is higher than the significance level of 0.05 we don't have enough evidence to conclude that the true mean is significantly higher than $300 because we FAIL to reject the null hypothesis.

Step-by-step explanation:

Information given

314 305 344 283 285 310​ 383​ 285​ 300​ 300

We can calculate the sample mean and deviation with the following formula:

\bar X =\frac{\sum_{i=1}^n X_i}{n}

s =\sqrt{\frac{\sum_{i=1}^n (x_i -\bar X)^2}{n-1}}

\bar X=310.9 represent the sample mean      

s=31.09 represent the standard deviation for the sample      

n=10 sample size      

\mu_o =300 represent the value to test

\alpha=0.05 represent the significance level

t would represent the statistic

p_v represent the p value

Hypothesis to test

We want to verify if the true mean is greater than 300, the system of hypothesis would be:      

Null hypothesis:\mu \leq 300      

Alternative hypothesis:\mu > 300      

The statistic is given by:

t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}} (1)      

Replacing the info given we got:

t=\frac{310.9-300}{\frac{31.09}{\sqrt{10}}}=1.109      

The degrees of freedom are given by:

df=n-1=10-1=9  

And the p value would be given by:

p_v =P(t_{9}>1.109)=0.148  

Since the p value is higher than the significance level of 0.05 we don't have enough evidence to conclude that the true mean is significantly higher than $300 because we FAIL to reject the null hypothesis.

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Answer:

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Variance for a sample set of data = ∑ ( x - X ) ^2 / n - 1

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