2.) What amount presently must be invested earning 5.25% compounded continuouslyso that it will grow up to be worth $25,000 12 years from now?

Answers

Answer 1

Answer:

The amount A resulting from a principal amount P being invested at rate r compounded continuously for time t is given by

... A = P·e^(rt)

FIll in your given values and solve for P.

... 25000 = P·e^(0.0525·12) = P·e^0.63

... P = 25000/e^0.63 ≈ 13314.80 . . . . . divide by the coefficient of P

The amount that must be invested is $13,314.80.

Answer 2

Answer:

An initial investment (P) compounded continuously with a rate of interest (r) in time (t) will grow to amount (Q) is given by:

Q = P * e^(rt)

Q=25000, r=0.0525, t=12

25000 = P * e^(0.0525*12)

1.8776P = 25000

P = 13314.8

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A fair die is rolled 8 times. What is the probability that the die comes up 6 exactly twice? What is the probability that the die comes up an odd number exactly five times? Find the mean number of times a 6 comes up. Find the mean number of times an odd number comes up. Find the standard deviation of the number of times a 6 comes up. Find the standard deviation of the number of times an odd number comes up.

A four pack of batteries cost 5.16 at this price what is the cost of

Answers

Answer:

1.29 for 1 pack

Step-by-step explanation:

5.16/4=1.29

5. Taking a cruise is a costly discretionary expense. In a recent year, the top fivecruise lines in the world had this many passengers:
4,133,000 2,369,000 1,295,000 928,000 679,000
Round your answers to the nearest integer.
a. The computations will be easier to work if you view this problem in terms
of thousands of passengers. Represent each number in terms of thousands
of passengers.
b. What is the mean number of passengers for these five cruise lines? (Give
the full number.)
c. What is the range? (Give the full number.)
d. What is the standard deviation? (Give the full number.)

Answers

Final answer:

The numbers in thousands are: 4133, 2369, 1295, 928, 679. The mean is 1881 thousand passengers, the range is 3454 thousand passengers, and the standard deviation is approximately 1218 thousand passengers.

Explanation:

Part A: To represent each number in terms of thousands of passengers, we simplify them as follows: 4,133,000 is 4,133 thousands of passengers, 2,369,000 is 2,369 thousands, 1,295,000 is 1,295 thousands, 928,000 is 928 thousands, and 679,000 is 679 thousands.

Part B: To calculate the mean number of passengers, you would add up all the passenger numbers and then divide by the number of values (5 in this case). This adds up to 9,404,000 passengers or, in terms of thousands, 9,404. Divided by 5, this gives a mean of 1,880,800 passengers, or 1,881 thousands of passengers.

Part C: The range is calculated by subtracting the smallest number from the largest. In this case, that would be 4,133,000 - 679,000 = 3,454,000, which is also 3,454 thousands of passengers.

Part D: Calculating the standard deviation involves multiple steps. First, for each value, subtract the mean and square this result. Then, calculate the mean of these squared differences. Finally, take the square root of this mean. Doing so, the standard deviation is approximate 1,217,982 passengers, or 1,218 thousands of passengers.

Learn more about Statistics here:

brainly.com/question/31538429

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In the checksums, you validated that the average SAT score for all of the records is 1,059.07. When we work with the data more rigorously, several tests will require us to transform NULL values. If you were to transform the NULL SAT values into 0, what would happen to the average (would it stay the same, decrease, or increase?

Answers

Answer:

It would decrease.

Step-by-step explanation:

The null almost always acts like a placeholder where there is nothing to record for that entry, we can say that it is a dummy character.

The difference between null and zero is that zero has a value in a context whereas null does not have any value.

So given the example in the question, if we replace the null values with zeros, that would affect the average because of the fact that zero values are going to get included in the average, null values were not included. Therefore this will result in the average being decreased.

I hope this answer helps.

Test the following numbers for divisibility by 3, 5, 6 and 11. (Do not divide or factorise.) a) 68 979

Answers

Final answer:

The number 68,979 is divisible by 3 but not by 5, 6, or 11.

Explanation:

To test a number for divisibility by 3, we sum up its individual digits. If the sum is divisible by 3, then the number itself is divisible by 3. For the number 68,979, the sum of its digits is 6 + 8 + 9 + 7 + 9 = 39. Since 39 is divisible by 3, 68,979 is also divisible by 3.

To test a number for divisibility by 5, we simply check if the last digit is either 0 or 5. In the number 68,979, the last digit is 9, which is neither 0 nor 5. Therefore, 68,979 is not divisible by 5.

To test a number for divisibility by 6, we need to check if it is divisible by both 2 and 3. From the previous tests, we already know that 68,979 is divisible by 3, so we now need to check if it is divisible by 2. Since the last digit is 9, which is an odd number, 68,979 is not divisible by 2, and therefore not divisible by 6.

To test a number for divisibility by 11, we take the alternating sum of its digits. In this case, we have 6 - 8 + 9 - 7 + 9 = 9. Since 9 is not divisible by 11, 68,979 is not divisible by 11.