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ziro4ka [17]
3 years ago
5

100 is deposited into an investment account on January 1, 1998. You are given the following information on investment activity t

hat takes place during the year:
April 19,1998 October 30, 1998
Value immediately prior to deposit 95 105
Deposit 2X X
The amount in the account on January 1, 1999 is 115. During 1998, The annual effective dollar weighted yield is 0%, and the annual effective time weighted yield is y. Calculate y.
Mathematics
1 answer:
klio [65]3 years ago
8 0

Answer:

y = - 0.681 % ≈ -0.7 %

Step-by-step explanation:

Given:

                                                               April 19,1998            October 30, 1998

Value immediately prior to deposit                95                               105

                     Deposit                                       2X                                 X

amount in the account on January 1, 1999 = 115

effective dollar weighted yield = 0%

annual effective time weighted yield = y

To find:

Calculate y

Solution:

Given that the dollar weighted return is 0%

100 is deposited into investment account on January 1, 1998. So, add 100 to the deposits 2X X

100 + 2x + x = 115

3x = 115 - 100

3x = 15

x = 15/3

x = 5

Compute y

1 + y = (95/100)(105/105)(115/110)

1 + y = 0.95 * 1 * 1.045

1 + y = 0.99318

y = 0.99318 - 1

y = - 0.0068 * 100

y = - 0.681 % ≈ -0.7 %

y = -0.7 %

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The data of players of volleyball tournament, the mean is 56.5, median is 53, range is 56 and interquartile range is 20

<h3>What is mean, median and mode of a data set?</h3>

The mean of the data is the average value of the given data. The mean of the data is the ratio of the sum of all the values of data to the total number of values of data.

The median of the data is the middle value of the data set when it arrange in ascending or descending order.

Mode of a data set is the value, which occurs most times for that data set. The value which has the highest frequency in the given set of data is known as the mode of that data set.

The stem-and-leaf plot shows the number of digs for the top 15 players at a volleyball tournament. The table is attached below.

The data according to the table is,

41,41,43,43,45,50,52,53,54,62,63,63,67,75,97

  • The mean is-

\mu=\dfrac{41+41+43+43+45+50+52+53+54+62+63+63+67+75+97}{15}\\\mu=\dfrac{849}{15}\\\mu=56.5

  • The median is.

The arranged data according to the table is,

41,41,43,43,45,50,52,53,54,62,63,63,67,75,97

Here, the middle value is 53. Thus, the median is 53.

  • The modes from least to greatest are , , and.

Mode is the data which occur most offen. Here the order of modes from least to greatest is,

41   41   43   43   45   50   52   53   54   62   63   63   67   75   97  

  • The range is.

Lowest value in data is 41 and highest is 97. Thus, the range is,

r=97-41\\r=56

  • The interquartile range is.

The third quartile of data is 63 while first quartile is 43. Thus, the interquartile range is.

IQR=63-23\\IQR=20

Thus, the data of players of volleyball tournament, the mean is 56.5, median is 53, range is 56 and interquartile range is 20.

Learn more about the mean, median and mode here;

brainly.com/question/14532771

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If f(x) = –x + 2, find f(4).
NeTakaya

Answer:

-2

Step-by-step explanation:

If you substitute x for 4, you have -4+2 to get -2.

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Henry drew the model below of his yard. He said the perimeter of the yard is 48 feet. He knows that one side is 9 feet wide. How
saw5 [17]

Answer:

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Step-by-step explanation:

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Perimeter, p = 48

Length of all sides

p = 2(l + w)

48 = 2(l + 9)

48 = 2l + 18

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Length of the sides = 15 feets

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