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Debora [2.8K]
3 years ago
6

You want to be able to withdraw $4000 a month for 30 years how much would you need to have in your account with an APR of 3.4% t

o accomplish this goal
Mathematics
1 answer:
tigry1 [53]3 years ago
8 0

Answer:

  $904,510.28

Step-by-step explanation:

If we assume the withdrawals are at the beginning of the month, we can use the annuity-due formula.

  P = A(1 +r/n)(1 -(1 +r/n)^(-nt))/(r/n)

where r is the APR, n is the number of times interest is compounded per year (12), A is the amount withdrawn, and t is the number of years.

Filling in your values, we have ...

  P = $4000(1 +.034/12)(1 -(1 +.034/12)^(-12·30))/(.034/12)

  P = $904,510.28

You need to have $904,510.28 in your account when you begin withdrawals.

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c=6

Step-by-step explanation:

c+25-19=100

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Reyesburg Corporation is contemplating using a new, more expensive glue in the construction of its laminated veneer lumber. Of i
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Answer:

The mean carrying load of the beams tested is 908.7 pounds per square foot

Step-by-step explanation:

1. Data class 1: 860-879

Midpoint 1: M1=(860+879)/2=1,739/2→M1=869.5

Number of values in data class 1: n1=2

2. Data class 2: 880-899

Midpoint 2: M2=(880+899)/2=1,779/2→M2=889.5

Number of values in data class 2: n2=5

3. Data class 3: 900-919

Midpoint 3: M3=(900+919)/2=1,819/2→M3=909.5

Number of values in data class 3: n3=10

4. Data class 4: 920-939

Midpoint 4: M4=(920+939)/2=1,859/2→M4=929.5

Number of values in data class 4: n4=6

5. Data class 5: 940-959

Midpoint 5: M5=(940+959)/2=1,899/2→M5=949.5

Number of values in data class 5: n5=1


Mean=Sum(Mi*ni)/Sum(ni)

Mean=(M1*n1+M2*n2+M3*n3+M4*n4+M5*n5)/(n1+n2+n3+n4+n5)

Mean=(869.5*2+889.5*5+909.5*10+929.5*6+949.5*1)/(2+5+10+6+1)

Mean=(1,739+4,447.5+9,095+5,577+949.5)/(24)

Mean=(21,808)/(24)

Mean=908.6666666

Rounding to one decimal place:

Mean=908.7

3 0
3 years ago
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