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nalin [4]
2 years ago
10

You plan to work for 40 years and then retire using a 25-year annuity. You want to arrange a retirement income of $4000 per mont

h. You have access to an account that pays an APR of 6.0% compounded monthly. This requires a nest egg of $620,827.46.
What monthly deposits are required to achieve the desired monthly yield at retirement? (Round your answer to the nearest cent.)

Mathematics
1 answer:
Molodets [167]2 years ago
7 0

Answer:

  $311.74

Step-by-step explanation:

A financial calculator computes the payment amount to be $311.74.

___

Your graphing calculator may have the capability to do this. Certainly, such calculators are available in spreadsheet programs and on the web.

___

The appropriate formula is the one for the sum of terms of a geometric series.

  Sn = a1·((1+r)^n -1)/(r) . . . . . where r is the monthly interest rate (0.005) and n is the number of payments (480). Filling in the given numbers, you have ...

  $620827.46 = a1·(1.005^480 -1)/.005 = 1991.4907·a1

Then ...

  $620827.46/1991.4907 = a1 ≈ $311.74

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erin has $50. she wants to purchase a cell phone ($20) and spend the rest in music cds each cd costs $8. write an inepuality for
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1 year ago
J.J.Bean sells a wide variety of outdoor equipment and clothing. The company sells both through mail order and via the internet.
melamori03 [73]

Answer:

99% confidence interval for the true mean difference between the mean amount of mail-order purchases and the mean amount of internet purchases is [$(-31.82) , $12.02].

Step-by-step explanation:

We are given that a random sample of 16 sales receipts for mail-order sales results in a mean sale amount of $74.50 with a standard deviation of $17.25.

A random sample of 9 sales receipts for internet sales results in a mean sale amount of $84.40 with a standard deviation of $21.25.

The pivotal quantity that will be used for constructing 99% confidence interval for true mean difference is given by;

                      P.Q.  =  \frac{(\bar X_1-\bar X_2)-(\mu_1-\mu_2)}{s_p \times \sqrt{\frac{1}{n_1}+\frac{1}{n_2} } }  ~ t__n_1_+_n_2_-_2

where, \bar X_1 = sample mean for mail-order sales = $74.50

\bar X_2 = sample mean for internet sales = $84.40

s_1 = sample standard deviation for mail-order purchases = $17.25

s_2 = sample standard deviation for internet purchases = $21.25

n_1 = sample of sales receipts for mail-order purchases = 16

n_2 = sample of sales receipts for internet purchases = 9

Also,  s_p =\sqrt{\frac{(n_1-1)\times s_1^{2}+(n_2-1)\times s_2^{2} }{n_1+n_2-2} }  =  \sqrt{\frac{(16-1)\times 17.25^{2}+(9-1)\times 21.25^{2} }{16+9-2} } = 18.74

The true mean difference between the mean amount of mail-order purchases and the mean amount of internet purchases is represented by (\mu_1-\mu_2).

Now, 99% confidence interval for (\mu_1-\mu_2) is given by;

             = (\bar X_1-\bar X_2) \pm t_(_\frac{\alpha}{2}_)  \times s_p \times \sqrt{\frac{1}{n_1} +\frac{1}{n_2}}

Here, the critical value of t at 0.5% level of significance and 23 degrees of freedom is given as 2.807.

          = (74.50-84.40) \pm (2.807  \times 18.74 \times \sqrt{\frac{1}{16} +\frac{1}{9}})

          = [$-31.82 , $12.02]

Hence, 99% confidence interval for the true mean difference between the mean amount of mail-order purchases and the mean amount of internet purchases is [$(-31.82) , $12.02].

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Step-by-step explanation:

<em>aint no body got time for dat!!</em>

Hope this helps!

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(21/5)/(7/3)=(21/5)(3/7)=63/35=9/5=1 and 4/5
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