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yulyashka [42]
3 years ago
15

By use of technology, we investigated Mary’s investment and created the model, M(x) = 3.03(1.28)2x, in thousands of dollars. Wha

t was Mary’s initial investment?
Mathematics
2 answers:
Allisa [31]3 years ago
5 0

Answer: Mary’s initial investment = $,3030

Step-by-step explanation:

Given: By use of technology, we investigated Mary’s investment and created the model, M(x) = 3.03(1.28)^{2x}

To find the initial investment, we substitute x =0, we get

M(0) = 3.03(1.28)^{2(0)}=3.03\text{ thousands of dollars}\\\\=3.03\times1000=\$3,030

Hence,  Mary’s initial investment = $,3030

blsea [12.9K]3 years ago
4 0
In this item, we are given with the exponential equation that predicts the amount that Mary is able to obtain through her initial investment. The characters are well identifiable. 3.03 is the initial investment, 28% is the rate and x is the number of years. Thus, the answer is 3.03. 
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Which expression is equivalent to (4^-3)^-6<br><br> a.) 4^3<br> b.) 4^-9<br> c.) 4^-18<br> d.) 4^18
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Suppose the standard deviation of a normal population is known to be 3, and H0 asserts that the mean is equal to 12. A random sa
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z=\frac{12.95-12}{\frac{3}{\sqrt{36}}}=1.9  

p_v =P(z>1.9)=0.0287  

If we compare the p value and the significance level given \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 true mean is higher than 12 at 5% of signficance.  

Step-by-step explanation:

Data given and notation  

\bar X=12.95 represent the sample mean

\sigma=3 represent the population standard deviation for the sample  

n=36 sample size  

\mu_o =12 represent the value that we want to test  

\alpha=0.05 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)  

State the null and alternative hypotheses.  

We need to conduct a hypothesis in order to check if the mean is equal to 12, the system of hypothesis would be:  

Null hypothesis:\mu \leq 12  

Alternative hypothesis:\mu > 12  

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".  

Calculate the statistic  

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

z=\frac{12.95-12}{\frac{3}{\sqrt{36}}}=1.9  

P-value  

Since is a right tailed test the p value would be:  

p_v =P(z>1.9)=0.0287  

Conclusion  

If we compare the p value and the significance level given \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 true mean is higher than 12 at 5% of signficance.  

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