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A person invests $1,150 in an account that earns 5% annual interest compounded continuously. Find when the value of the investme

AfilCa [17]

Answer:

11.1 years

Step-by-step explanation:

The formula for interest compounding continuously is:

$A(t)=Pe^{rt}$

Where A(t) is the amount after the compounding, P is the initial deposit, r is the interest rate in decimal form, and t is the time in years. Filling in what we have looks like this:

$2000=1150e^{{.05t}$

We will simplify this first a bit by dividing 2000 by 1150 to get

$1.739130435=e^{.05t}$

To get that t out the exponential position it is currently in we have to take the natural log of both sides. Since a natural log has a base of e, taking the natual log of e cancels both of them out. They "undo" each other, for lack of a better way to explain it. That leaves us with

ln(1.739130435)=.05t

Taking the natural log of that decimal on our calculator gives us

.5533852383=.05t

Now divide both sides by .05 to get t = 11.06770477 which rounds to 11.1 years.

7 0

2 years ago

Please help!!! I will give you BRIANLEIST AND A HEARTED COMMENT AND 5 STARS if you can explain this:(

Agata [3.3K]

Answer:

I can do this give me a min

4 0

2 years ago

Polymer composite materials have gained popularity because they have high strength to weight ratios and are relatively easy and

Brums [2.3K]

Answer:

$t=\frac{51.6-48}{\frac{1.1}{\sqrt{10}}}=10.349$

$p_v =P(t_9>10.349)=1.34x10^{-6}$

If we compare the p value and the significance level given for example $\alpha=0.05$ we see that $p_v$ so we can conclude that we reject the null hypothesis, and the the actual mean is significantly higher than 48 MPa at 5% of significance.

Step-by-step explanation:

1) Data given and notation

$\bar X=51.6$ represent the sample mean

$s=1.1$ represent the standard deviation for the sample

$n=10$ sample size

$\mu_o =48$ represent the value that we want to test

$\alpha=0.05$ represent the significance level for the hypothesis test.

t 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 determine if the mean score is higher than 48, the system of hypothesis would be:

Null hypothesis: $\mu \geq 48$

Alternative hypothesis: $\mu > 48$

We don't know the population deviation, so for this case is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:

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

t-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:

$t=\frac{51.6-48}{\frac{1.1}{\sqrt{10}}}=10.349$

Calculate the P-value

First we find the degrees of freedom:

$df=n-1=10-1=9$

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

$p_v =P(t_9>10.349)=1.34x10^{-6}$

Conclusion

If we compare the p value and the significance level given for example $\alpha=0.05$ we see that $p_v$ so we can conclude that we reject the null hypothesis, and the the actual mean is significantly higher than 48 MPa at 5% of significance.

5 0

3 years ago

What fraction is equivalent to 1/8??

ASHA 777 [7]

Answer:

2/16

5/40

Step-by-step explanation:

4 0

2 years ago

4x-1=3(x-1) Please help!

Sveta_85 [38]

Answer:

x=-2

Step-by-step explanation:

4x-1=3x-3

x=-2

5 0

2 years ago

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