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RUDIKE [14]
3 years ago
12

suppose 7,500 is invested at 1.5%, compounded continuously. what will the account balance be in 16 years?​

Mathematics
1 answer:
viktelen [127]3 years ago
7 0

Answer:

\boxed {\boxed {\sf \$9,534.37}}

Step-by-step explanation:

The formula for continuously compounded is:

A=Pe^{rt}

where P is the principal amount, r is the interest rate as a decimal, and t is the amount of time.

The principal amount is $7,500 and time is 16 years. The interest rate is 1.5%. Convert this to a decimal- divide by 100 or move the decimal places two spots to the left.

  • 1.5/100= 0.015         or           1.5 --> 0.15 --> 0.015

Now we know all the values and can substitute them into the formula.

P= 7500 \\r= 0.015 \\t= 16

A= 7500(e^{0.015*16})

Solve the multiplication in the exponent first.

A= 7500(e^{24})

Solve the exponent.

A= 7500 (1.27124915)

A=9534.368627

Round to the nearest cent (hundredth place). The 8 in the thousandth place tells us to the round the 6 to a 7.

A=9534.37

The account balance will be <u>$9,534.37</u> after 16 years.

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-2/3m = -12 please explain how to solve.
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Answer:

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Step-by-step explanation:

-2/3m = -12

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3 0
3 years ago
Read 2 more answers
A sample of 81 account balances of a credit company showed an average balance of $1,200 with a standard deviation of $126.
TEA [102]

Answer:

(a) Null Hypothesis, H_0 : \mu = $1,150  

    Alternate Hypothesis, H_A : \mu \neq $1,150

(b) The test statistic is 3.571.

(c) We conclude that the mean of all account balances is significantly different from $1,150.

Step-by-step explanation:

We are given that a sample of 81 account balances of a credit company showed an average balance of $1,200 with a standard deviation of $126.

We have test the hypothesis to determine whether the mean of all account balances is significantly different from $1,150.

<u><em>Let </em></u>\mu<u><em> = mean of all account balances</em></u>

(a)So, Null Hypothesis, H_0 : \mu = $1,150     {means that the mean of all account balances is equal to $1,150}

Alternate Hypothesis, H_A : \mu \neq $1,150    {means that the mean of all account balances is significantly different from $1,150}

The test statistics that will be used here is <u>One-sample t test statistics</u> as we don't know about population standard deviation;

                               T.S.  = \frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }  ~ t_n_-_1

where, \bar X = sample average balance = $1,200

             s = sample standard deviation = $126

             n = sample of account balances = 81

(b) So, <u>test statistics</u>  =  \frac{1,200-1,150}{\frac{126}{\sqrt{81} } }  ~ t_8_0

                               =  3.571

The value of the sample test statistics is 3.571.

(c) <u>Now, P-value of the test statistics is given by the following formula;</u>

          P-value = P( t_8_0 > 3.571) = <u>Less than 0.05%</u>

Because in the t table the highest critical value for t at 80 degree of freedom is given between 3.460 and 3.373 at 0.05% level.

Now, since P-value is less than the level of significance as 5% > 0.05%, so we sufficient evidence to reject our null hypothesis.

Therefore, we conclude that the mean of all account balances is significantly different from $1,150.

8 0
4 years ago
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I think the answer is C not to sure
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