Quarterly withdrawls of $650 for 6 years; interest rate is 4.5% compounded quarterly.Find the amount necessary to fund the given withdrawls.

Answers

Answer 1
Answer:

Answer:

The amount needed to finance this level of withdrawals is $ 9422.06

Step-by-step explanation:

For a fixed amount periodically withdrawn, during n periods, with an interest rate i, there is a present value of:

V_(p) = A [(1- (1 + i) ^(- n))/(i)]. From where,

V_(p) = 650 [(1- (1 + 0.045)^(-24))/(0.045)] = 9422.06

The amount needed to finance this level of withdrawals is $ 9422.0609


Related Questions

Scott is a software salesman. Let represent his total pay (in dollars). Let represent the number of copies of History is Fun he sells. Suppose that and are related by the equation 80x+2100=y Answer the questions below.Note that a change can be an increase or a decrease.For an increase, use a positive number. For a decrease, use a negative number.What is the change in Scott's total pay for each copy of History is Fun he sells?What is Scott's total pay if he doesn't sell any copies of History is Fun?
If the exterior angel of a triangle measures 130 degrees, then what is the sum of the non-adjacent interior angles of that triangle. please answer this question
Ninety-nine is what percent of 150?
Simplify.7.52 - 2.9 × 1 + 0.8
Decorating your house for Christmas takes 106ft of lights. You use two lengths of lights,the short length of lights is 5 ft and the long length of lights is 11 ft. The total number ofstrings of lights you use is 14. How many of each type of light do you have?

A motorboat moves across a lake. It begins at 50km from shore after 9 minutes it is 14 km from shore. What’s it’s distance from shore in a function

Answers

Answer:

y = -4x + 50

Step-by-step explanation:

To write the function we'll use y = mx + b, where y is the distance the boat is away from shore, m is how many km/minute the boat moves at, x is how many minutes it's been, and b is how many km away from shore the boat was at the start. We already know the value of b is 50, so we can put that into the equation:

y = mx + 50

To find how many km the boat moves per minute, let's use the given amount of minutes and km (9, 14) and put them into the equation:

14 = m(9) + 50

Now let's solve for m:

14 = 9m + 50

Subtract 50 from both sides to isolate the 9m:

14 - 50 = 9m + 50 - 50

- 36 = 9m

divide both sides by 9 to isolate the m:

-36/9 = 9m/9

-4 = m

The boat moves at -4 km/minute, which is the same thing as saying that  after 1 minute passes, the boat gets 4km closer to the shore. Now we can input this value into the equation and we have our answer:

y = -4x + 50

Answer:

List all of the solutions.

50km=

9minutes=

14km=

Step-by-step explanation:

quiz 4-1 : classifying and solving for sides/angles in triangles gina wilson (all things algebra) URGENT

Answers

Answer:

14.

x = 11  m<B = 108   m<C= 36   m<D = 36

15.

x = 4     WX = 25     XY = 17     WY = 25

Step-by-step explanation:

Solving (14):

BC = BD

m<B = 13x - 35

m<C = 5x - 19

m<D = 2x + 14

First, we solve for x.

Since sides BC = BD, then

m<C = m<D

This is so because the theorem of ASA (Angle-Side-Angle)

So, we have:

5x - 19 = 2x + 14

Collect Like Terms

5x - 2x = 19 + 14

3x = 33

Divide both sides by 3

x = 11

Solving for m<B, m<C and m<D.

We simply substitute 11 for x in their respective expressions.

m<B = 13x - 35

m<B = 13*11 - 35

m<B = 143 - 35

m<B = 108

m<C = 5x - 19

m<C= 5*11 - 19

m<C= 55 - 19

m<C= 36

m<D = 2x + 14

m<D = 2*11 + 14

m<D = 22 + 14

m<D = 36

Solving (15):

<X = <Y

WX = 9x - 11

XY = 4x + 1

WY = 7x - 3

First, we solve for x.

Since <X = <Y, then

WX = WY

This is so because the theorem of SAS (Side-Angle-Side)

So, we have:

9x - 11 = 7x - 3

Collect Like Terms

9x - 7x = 11-3

2x = 8

Divide both sides by 2

x = 4

Solving for WX, WY and XY

We simply substitute 4 for x in their respective expressions.

WX = 9x - 11

WX = 9 * 4 - 11

WX = 36 - 11

WX = 25

XY = 4x + 1

XY = 4 * 4 + 1

XY = 16 + 1

XY = 17

WY = 7x - 3

WY = 7 * 4 - 3

WY = 28 - 3

WY = 25

The map shows the intersection of three roads. Malcom Way intersects Sydney Street at an angle of 162°. Park Road intersects Sydney Street at an angle of 87°. Find the angle at which Malcom Way intersects Park Road.

Answers

Answer:

75

Step-by-step explanation:

A rectangle has a length of 12 m and a width of 400cm. What is the perimeter of the rectangle?

Answers

Answer:

800 + 2400 = 3200

Step-by-step explanation:

Because 12 m = 1200 cm

A = L(2) +  W(2)

A = 1200 (2) + 400 (2)

A = 2400 + 800

A = 3200

6 apple trees fir every 4 pears trees How many apple trees would there be if there were 42 pear trees

Answers

Answer: 63 apple trees
Explanation: divide 42 by 4 to get that amount of groups (10.5) then multiply that number by 6 so (10.5 • 6) and that should give the apple tree total.

In a sample of 1200 U.S.​ adults, 191 dine out at a resaurant more than once per week. Two U.S. adults are selected at random from the population of all U.S. adults without replacement. Assuming the sample is representative of all U.S.​ adults, complete parts​ (a) through​ (d). ​Required:a. Find the probability that both adults dine out more than once per week. b. Find the probability that neither adult dines out more than once per week. c. Find the probability that at least one of the two adults dines out more than once per week. d. Which of the events can be considered unusual? Explain.

Answers

Answer:

a) The probability that both adults dine out more than once per week = 0.0253

b) The probability that neither adult dines out more than once per week = 0.7069

c) The probability that at least one of the two adults dines out more than once per week = 0.2931

d) Of the three events described, the event that can be considered unusual because of its low probability of occurring, 0.0253 (2.53%), is the event that the two randomly selected adults both dine out more than once per week.

Step-by-step explanation:

In a sample of 1200 U.S. adults, 191 dine out at a restaurant more than once per week.

Assuming this sample.is a random sample and is representative of the proportion of all U.S. adults, the probability of a randomly picked U.S. adult dining out at a restaurant more than once per week = (191/1200) = 0.1591666667 = 0.1592

Now, assuming this probability per person is independent of each other.

Two adults are picked at random from the entire population of U.S. adults, with no replacement, thereby making sure these two are picked at absolute random.

a) The probability that both adults dine out more than once per week.

Probability that adult A dines out more than once per week = P(A) = 0.1592

Probability that adult B dines out more than once per week = P(B) = 0.1592

Probability that adult A and adult B dine out more than once per week = P(A n B)

= P(A) × P(B) (since the probability for each person is independent of the other person)

= 0.1592 × 0.1592

= 0.02534464 = 0.0253 to 4 d.p.

b) The probability that neither adult dines out more than once per week.

Probability that adult A dines out more than once per week = P(A) = 0.1592

Probability that adult A does NOT dine out more than once per week = P(A') = 1 - P(A) = 1 - 0.1592 = 0.8408

Probability that adult B dines out more than once per week = P(B) = 0.1592

Probability that adult B does NOT dine out more than once per week = P(B') = 1 - P(B) = 1 - 0.1592 = 0.8408

Probability that neither adult dines out more than once per week = P(A' n B')

= P(A') × P(B')

= 0.8408 × 0.8408

= 0.70694464 = 0.7069 to 4 d.p.

c) The probability that at least one of the two adults dines out more than once per week.

Probability that adult A dines out more than once per week = P(A) = 0.1592

Probability that adult A does NOT dine out more than once per week = P(A') = 1 - P(A) = 1 - 0.1592 = 0.8408

Probability that adult B dines out more than once per week = P(B) = 0.1592

Probability that adult B does NOT dine out more than once per week = P(B') = 1 - P(B) = 1 - 0.1592 = 0.8408

The probability that at least one of the two adults dines out more than once per week

= P(A n B') + P(A' n B) + P(A n B)

= [P(A) × P(B')] + [P(A') × P(B)] + [P(A) × P(B)]

= (0.1592 × 0.8408) + (0.8408 × 0.1592) + (0.1592 × 0.1592)

= 0.13385536 + 0.13385536 + 0.02534464

= 0.29305536 = 0.2931 to 4 d.p.

d) Which of the events can be considered unusual? Explain.

The event that can be considered as unusual is the event that has very low probabilities of occurring, probabilities of values less than 5% (0.05).

And of the three events described, the event that can be considered unusual because of its low probability of occurring, 0.0253 (2.53%), is the event that the two randomly selected adults both dine out more than once per week.

Hope this Helps!!!

Other Questions
Consider a random sample of ten children selected from a population of infants receiving antacids that contain aluminum, in order to treat peptic or digestive disorders. The distribution of plasma aluminum levels is known to be approximately normal; however its mean u and standard deviation o are not known. The mean aluminum level for the sample of n = 10 infants is found to be X = 37.20 ug/l and the sample standard deviation is s = 7.13 ug/1. Furthermore, the mean plasma aluminum level for the population of infants not receiving antacids is known to be only 4.13 ug/1.(a) Formulate the null hypothesis and complementary alternative hypothesis, for a two-sided test of whether the mean plasma aluminum level of the population of infants receiving antacids is equal to the mean plasma aluminum level of the population of infants not receiving antacids.(b) Construct a 95% confidence interval for the true mean plasma aluminum level of the population of infants receiving antacids.(c) Calculate the p-value of this sample (as best as possible), at the a=.05 significance level.(d) Based on your answers in parts (b) and (c), is the null hypothesis rejected in favor of the alternative hypothesis, at the a = .05 significance level? Interpret your conclusion: What exactly has been demonstrated, based on the empirical evidence?(e) With the knowledge that significantly elevated plasma aluminum levels are toxic to human beings, reformulate the null hypothesis and complementary alternative hypothesis, for the appropriate one-sided test of the mean plasma aluminum levels. With the same sample data as above, how does the new p-value compare with that found in part (c), and what is the resulting conclusion and interpretation?