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Serjik [45]
3 years ago
14

Suppose you invest $50 a month in an annuity that earns 4% APR compounded monthly. How much money will you have in this account

after 3 years?
Mathematics
2 answers:
skelet666 [1.2K]3 years ago
4 0
To solve this we are going to use formula for the future value of an ordinary annuity: FV=P[ \frac{(1+ \frac{r}{n} )^{nt} -1}{ \frac{r}{n} } ]
where
FV is the future value
P is the periodic payment
r is the interest rate in decimal form
n is the number of times the interest is compounded per year
t is the number of years

We know from our problem that the periodic payment is $50 and the number of years is 3, so P=50 and t=3. To convert the interest rate to decimal form, we are going to divide the rate by 100%
r= \frac{4}{100}
r=0.04
Since the interest is compounded monthly, it is compounded 12 times per year; therefore, n=12.
Lets replace the values in our formula:
FV=P[ \frac{(1+ \frac{r}{n} )^{nt} -1}{ \frac{r}{n} } ]
FV=50[ \frac{(1+ \frac{0.04}{12} )^{(12)(3)} -1}{ \frac{0.04}{12} } ]
FV=1909.08

We can conclude that after 3 years you will have $1909.08 in your account.
slava [35]3 years ago
3 0

Answer:

APEX answer is $1909.08

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Step-by-step explanation:

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2 years ago
1) y = |x+4| - 1<br><br><br><br> This needs to be graphed, please help!
otez555 [7]

Answer:

Attachment

Step-by-step explanation:

x = 0

y = | 0 + 4 | - 1

y = 4 - 1

y = 3

y = 0

0 = | x + 4 | - 1

1 = | x + 4 |

x + 4 = 1 => x₁ = 1 - 4 => x₁ =  - 3

x + 4 = - 1 => x₂ = - 1 - 4 => x₂ = - 5

7 0
4 years ago
Suppose x has a distribution with μ = 10 and σ = 9. (a) If a random sample of size n = 43 is drawn, find μx, σ x and P(10 ≤ x ≤
JulsSmile [24]

Answer:

P(10 ≤ x ≤ 12) = 0.4274

Step-by-step explanation:

Population mean = u = 10

Population Standard Deviation = \sigma = 9

Sample size = n = 43

Sample mean(\mu_{x}) is equal to the population mean. So,

Sample mean = \mu_{x} = 10

Sample standard deviation(\sigma_{x}) is equal to population standard deviation divided by square root of sample size. So,

Sample standard deviation = \sigma_{x} = \frac{9}{\sqrt{43}}=1.372

We have to find the probability that for a random sample of n = 43, the value lies between 10 and 12 i.e. P(10 ≤ x ≤ 12)

P(10 ≤ x ≤ 12) = P(x ≤ 12) - P( x ≤ 10)

We can find P(x ≤ 12 ) and P(x ≤ 10) by converting these values to z scores.

The formula for z score is:

z=\frac{x-\mu_{x}}{\sigma_{x}}

For x =12, we get:

z=\frac{12-10}{1.3725}=1.457

For x =10, we get:

z=\frac{10-10}{1.3725}=0

So,

P(x ≤ 12) - P( x ≤ 10) = P(z ≤ 1.457) - P(z ≤ 0)

From the z table,

P(z ≤ 1.457) = 0.9274

P(z ≤ 0) = 0.5

So,

P(x ≤ 12) - P( x ≤ 10) = P(z ≤ 1.458) - P(z ≤ 0) = 0.9274 - 0.5 = 0.4274

So,

P(10 ≤ x ≤ 12) = P(x ≤ 12) - P( x ≤ 10) = 0.4274

Therefore,

The probability that for a random sample of size 43, the mean lies between 10 and 12 is 0.4274.

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