Scott has $500 in the bank. He withdraws the same amount of money each week for 12 weeks, at which point he has $20 left in the bank. How much money did he withdraw each week?

Answers

Answer 1

Answer: Scott knew that the only chance we would have to answer this problem would be if he kept his money in a bank that pays no interest, so that's what he did. The bank simply stored his money in a mayonnaise jar under a tightly-guarded mattress, and he was allowed to come in and take some out of it each week.

He started with $500. He withdrew (500 - 20) = $480 in 12 weeks.

So he must have come in and picked up 480/12 = $40 each week.

Related Questions

A wildflower bouquet costs $4.25 at the farmer's market. A vase can be purchased for an additional $3.50. Kelly has $20 to buy a bouquet for her mother with a vase. She'd also like to buy just a bouquet for herself and another for her best friend.If x represents the number of bouquets Kelly can buy, which inequality will tell Kelly if she has enough money to make her purchase?A) $4.25x + $3.50 ≤ $20 B) $4.25x - $3.50 ≤ $20 C) $4.25x + $3.50 ≥ $20 EliminateD) $3.50x + $4.25 ≤ $20;
DAILY MATH QUESTINCAN YOU SOLVE THE PROBLEM?
Which system of inequalities is shown below in the graph
Factor the expression 3x3 + 2x2y + 3xy2 + 2y3
A bakery sold 57 vanilla cupcakes in a day, which was 95% of the total number of cupcakes sold that day. How many total cupcakes did the bakery sell that day?

Isolating variables abx=cy the variable is x. [please show me how)

Answers

abx = cy

We want to get rid of 'ab', and it is being multiplied to 'x', so we divide it to both sides.

x = (cy)/(ab)

$abx=cy\nx=(cy)/(ab)$

The mean radius of earth is 6,371.0 kilometers and the mean radius of earths moon is 1,737.5 kilometers. What is the approximate difference in the mean conferences, in kilometers, of earth and earths moon? Round your answer to the nearest tenth of a kilometer.

Answers

6,371.0- 1,737.5 = 4,633.5 kilometres.
the approximate difference is 4,633.5 kilometres

When we divide 2√3 by 2√3 it gives 3 how ?(2√3/2√3)

Answers

Only if the individual doing the calculation makes a substantial mistake.

Any fraction with the same quantity in the numerator and denominator
always has the value of ' 1 ', (provided only that the quantity is not zero).

It doesn't.
$(2\sqrt3)/(2\sqrt3)=1$

1. Each of 9 friends chooses her favorite positive integera. The median of the chosen number is 91, what is the smallest the average of the 9 chosen numbers could be?

b. The median of the chosen number is 91, is there an limit to how large the aerge of the chosen numbers can be? If so, what is the largest the average can be?

c. The average of the chosen number is 91, what is the smallest the median of the 9 chosen numbers could be?

d. The average of the chosen numbers is 91. What is the largest the median of the chosen numbers could be?

Answers

Answer:

a) 1

b) There is no limit to which the largest number can be because we are only given information about the median.

c) 1

d) 90

Final answer:

The smallest average is 49 and the largest average is 91. The smallest median is 91 and the largest median is also 91.

Explanation:

a. Since the median is 91, at least 5 friends must choose numbers greater than or equal to 91, and at most 4 friends can choose numbers less than 91. To minimize the average, let's assume the four friends choose the smallest possible numbers less than 91 (1, 2, 3, and 4). The remaining five friends can then choose 91, 91, 91, 91, and 91. The average of the nine chosen numbers is (1 + 2 + 3 + 4 + 91 + 91 + 91 + 91 + 91)/9 = 49.

b. There is no limit to how large the average of the chosen numbers can be. The nine friends can all choose the same number, such as 91, which would make the average 91.

c. Since the average is 91, let's assume the eight friends choose the smallest possible numbers less than 91 (1, 2, 3, ..., 8). The remaining friend can then choose a number greater than or equal to 91. To minimize the median, the friend can choose the smallest possible number greater than or equal to 91, which is 91. So, the smallest median would be 91.

d. Since the average is 91, let's assume the eight friends choose the largest possible numbers less than 91 (84, 85, ..., 91). The remaining friend can then choose a number greater than or equal to 91. To maximize the median, the friend can choose the largest possible number greater than or equal to 91, which is 91. So, the largest median would also be 91.

Learn more about median and average of chosen numbers here:

brainly.com/question/33899535

#SPJ2

Michelle wants to make a long-term investment. She wants to triple an initial investment of$2,000. If she invests at a simple interest rate of 12% per year, approximately how long will it
take her principal to triple?