Ask question

Ask question

All categories

antiseptic1488 [7]

3 years ago

11

A deposit of $500 is made in an account that earns interest at an annual rate of 4%. How long will it take for the balance to do

uble when the interest is compounded (a) annually, (b) monthly, (c) daily, and (d) continuously?

1 answer:

kompoz [17]3 years ago

6 0

Answer:

(a) annually = 17.67 years

(b) monthly = 17.36 years

(c) daily = 17.34 years

(d) continuously = 17.328 years

Step-by-step explanation:

given data

principal = $500

annual rate = 4%

solution

we know here amount formula that is

amount = principal × $(1+\frac{r}{n})^{n*t}$ ..................1

put here value for compound annually

1000 = 500 × $(1+\frac{0.04}{1})^{t}$

take ln both side

ln 2 = ln ${1.04}^{t}$

t = 17.67 years

and

put value now in equation 1 for monthly

amount = principal × $(1+\frac{r}{n})^{n*t}$

1000 = 500 × $(1+\frac{0.04}{12})^{12*t}$

take ln both side

ln 2 = 12t × ln(1.003333)

t = 17.36 years

and

put value now in equation 1 for daily

amount = principal × $(1+\frac{r}{n})^{n*t}$

1000 = 500 × $(1+\frac{0.04}{365})^{365*t}$

take ln both side

ln 2 = 365 t × ln (1.0001095)

t = 17.34 years

and

for compound continuously

amount = principal × $e^{r*t}$ .................2

put here value

1000 = 500 × $e^{0.04*t}$

t = 17.328 years

You might be interested in

A carnival game has 3 identical boxes. The boxes have the following amounts of money: one has $30.00, one has $6.00, and one has

garri49 [273]

$200 divided by 100 customers = 2$.

So, if he wants to make a profit of $200, he would have to make each customer pay $2. :)

3 0

3 years ago

A population of rabbits is doubling every 20 days. On day 15, there are 42 rabbits. How many rabbits did the population start wi

Nimfa-mama [501]

Answer: So, if it just magically doubles every 20 days, and its only been 15 days, then it would've started with 42, but I'm not sure since I can't see the original problem

Step-by-step explanation:

5 0

3 years ago

HELP!! PLEASE EXPLAIN

yanalaym [24]

Answer:

10 pieces

Step-by-step explanation:

basically you divide 7.5 by .75 which equals 10.

You do that because you are seeing how many 3/4 inch pieces are in 7 and a half inches of ribbon and 7 1/2 = 7.5 and 3/4= .75

3 0

2 years ago

-2(x+2)=-12 .............

Dima020 [189]

Answer:

x=4

Step-by-step explanation:

-2(x+2)=-12

x+2=-12/-2

x+2=6

x=6-2

x=4

6 0

3 years ago

A random sample of n = 64 observations is drawn from a population with a mean equal to 20 and standard deviation equal to 16. (G

dezoksy [38]

Answer:

a) The mean of a sampling distribution of $\\ \overline{x}$ is $\\ \mu_{\overline{x}} = \mu = 20$ . The standard deviation is $\\ \frac{\sigma}{\sqrt{n}} = \frac{16}{\sqrt{64}}=2$ .

b) The standard normal z-score corresponding to a value of $\\ \overline{x} = 16$ is $\\ Z = -2$ .

c) The standard normal z-score corresponding to a value of $\\ \overline{x} = 23$ is $\\ Z = 1.5$ .

d) The probability $\\ P(\overline{x}$ .

e) The probability $\\ P(\overline{x}>23) = 1 - P(Z$ .

f) $\\ P(16 < \overline{x} < 23) = P(-2 < Z < 1.5) = P(Z$ .

Step-by-step explanation:

We are dealing here with the concept of <em>a sampling distribution</em>, that is, the distribution of the sample means $\\ \overline{x}$ .

We know that for this kind of distribution we need, at least, that the sample size must be $\\ n \geq 30$ observations, to establish that:

$\\ \overline{x} \sim N(\mu, \frac{\sigma}{\sqrt{n}})$

In words, the distribution of the sample means follows, approximately, a <em>normal distribution</em> with mean, $\mu$ , and standard deviation (called <em>standard error</em>), $\\ \frac{\sigma}{\sqrt{n}}$ .

The number of observations is n = 64.

We need also to remember that the random variable Z follows a <em>standard normal distribution</em> with $\\ \mu = 0$ and $\\ \sigma = 1$ .

$\\ Z \sim N(0, 1)$

The variable Z is

$\\ Z = \frac{\overline{x} - \mu}{\frac{\sigma}{\sqrt{n}}}$ [1]

With all this information, we can solve the questions.

Part a

The mean of a sampling distribution of $\\ \overline{x}$ is the population mean $\\ \mu = 20$ or $\\ \mu_{\overline{x}} = \mu = 20$ .

The standard deviation is the population standard deviation $\\ \sigma = 16$ divided by the root square of n, that is, the number of observations of the sample. Thus, $\\ \frac{\sigma}{\sqrt{n}} = \frac{16}{\sqrt{64}}=2$ .

Part b

We are dealing here with a <em>random sample</em>. The z-score for the sampling distribution of $\\ \overline{x}$ is given by [1]. Then

$\\ Z = \frac{\overline{x} - \mu}{\frac{\sigma}{\sqrt{n}}}$

$\\ Z = \frac{16 - 20}{\frac{16}{\sqrt{64}}}$

$\\ Z = \frac{-4}{\frac{16}{8}}$

$\\ Z = \frac{-4}{2}$

$\\ Z = -2$

Then, the <em>standard normal z-score</em> corresponding to a value of $\\ \overline{x} = 16$ is $\\ Z = -2$ .

Part c

We can follow the same procedure as before. Then

$\\ Z = \frac{\overline{x} - \mu}{\frac{\sigma}{\sqrt{n}}}$

$\\ Z = \frac{23 - 20}{\frac{16}{\sqrt{64}}}$

$\\ Z = \frac{3}{\frac{16}{8}}$

$\\ Z = \frac{3}{2}$

$\\ Z = 1.5$

As a result, the <em>standard normal z-score</em> corresponding to a value of $\\ \overline{x} = 23$ is $\\ Z = 1.5$ .

Part d

Since we know from [1] that the random variable follows a <em>standard normal distribution</em>, we can consult the <em>cumulative standard normal table</em> for the corresponding $\\ \overline{x}$ already calculated. This table is available in Statistics textbooks and on the Internet. We can also use statistical packages and even spreadsheets or calculators to find this probability.

The corresponding value is Z = -2, that is, it is <em>two standard units</em> <em>below</em> the mean (because of the <em>negative</em> value). Then, consulting the mentioned table, the corresponding cumulative probability for Z = -2 is $\\ P(Z$ .

Therefore, the probability $\\ P(\overline{x}$ .

Part e

We can follow a similar way than the previous step.

$\\ P(\overline{x} > 23) = P(Z > 1.5)$

For $\\ P(Z > 1.5)$ using the <em>cumulative standard normal table</em>, we can find this probability knowing that

$\\ P(Z1.5) = 1$

$\\ P(Z>1.5) = 1 - P(Z$

Thus

$\\ P(Z>1.5) = 1 - 0.9332$

$\\ P(Z>1.5) = 0.0668$

Therefore, the probability $\\ P(\overline{x}>23) = 1 - P(Z$ .

Part f

This probability is $\\ P(\overline{x} > 16)$ and $\\ P(\overline{x} < 23)$ .

For finding this, we need to subtract the cumulative probabilities for $\\ P(\overline{x} < 16)$ and $\\ P(\overline{x} < 23)$

Using the previous <em>standardized values</em> for them, we have from <em>Part d</em>:

$\\ P(\overline{x}$

We know from <em>Part e</em> that

$\\ P(\overline{x} > 23) = P(Z>1.5) = 1 - P(Z$

$\\ P(\overline{x} < 23) = P(Z1.5)$

$\\ P(\overline{x} < 23) = P(Z$

$\\ P(\overline{x} < 23) = P(Z$

Therefore, $\\ P(16 < \overline{x} < 23) = P(-2 < Z < 1.5) = P(Z$ .

5 0

3 years ago