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Veronika [31]
3 years ago
7

Match each cost with the investment type to which it relates.

Mathematics
2 answers:
Luden [163]3 years ago
7 0

Matching each cost with the investment type to which it relates we have :

The bond market is where investors go to trade debt securities and bonds. The stock market is a place where investors go to trade common stocks and derivatives.

So under Stocks and bonds column we have - commissions and investment advisory cost

A mutual fund is an investment fund that pools money from many investors to purchase securities.

So under Mutual funds column we have - administrative costs and hourly fees.


hammer [34]3 years ago
4 0
The Costs and the type of investment is relates is as follows:
Stocks and bonds                  Mutual Funds
Commissions                          administrative costs
investment advisory costs      Hourly fees
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The quotient of a number and 2 is larger than 3
Lostsunrise [7]
There are many for this ,but one sample is: 2 divided by 2is 4 X 3 which is 12.
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Find the measure of 44.<br> 41<br> A2<br> 43<br> 144<br> 970<br> 830<br> *4 = [?]
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Angle 4 is 83 degrees
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Select the expression that can be used to find the volume of this rectangular prism. A. ( 6 × 3 ) + 15 = 33 i n . 3 (6×3)+15=33
Usimov [2.4K]

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D

Step-by-step explanation:

(3x6)x15=270 in

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6 0
3 years ago
Read 2 more answers
The price of a notebook was
Pie

Old price: p1 = $ 3.90;

New price: p2 = $ 3.40.


The percentual decrease is given by

d(%) = [ (p2 - p1) / p1 ] * 100 %

d(%) = [ (3.40 - 3.90) / 3.90 ] * 100 %

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I hope this helps. =)

4 0
3 years ago
NarStor, a computer disk drive manufacturer, claims that the median time until failure for their hard drives is more than 14,400
Ne4ueva [31]

Answer:

The test statistics is  t  =  -1.727

Step-by-step explanation:

From the question we are told that

The data given is  

   330 620 1870 2410 4620 6396 7822 81028309 12882 14419 16092 18384 20916 23812 25814

 The population mean is  \mu  =  14400

    The  sample  size is  n =  16

  The  null hypothesis is  \mu \le  14400

    The  alternative hypothesis is  H_a  :  \mu > 14400

The sample mean is mathematically evaluated as

  \= x  =  \frac{\sum x_i}{n}

So

   \= x  =  \frac{330+ 620+ 1870 +2410+ 4620+ 6396+ 7822+ 8102+8309+ 12882+ 14419+ 16092+ 18384 +20916+ 23812+ 25814 }{16}

=>  \= x = 10799.9

The  standard deviation is mathematically represented as

      \sigma =\sqrt{\frac{ \sum (x_i - \=x)^2}{n}}

So

\sigma =\sqrt{\frac{(330- 10799.9)^2 + (620- 10799.9)^2+ (1870- 10799.9)^2 +(2410- 10799.9)^2 + (4620- 10799.9)^2 +(6396- 10799.9)^2 +(7822- 10799.9)^2 }{16}}  \ ..

   ..\sqrt{ \frac{(8102 - 10799.9)^2 +(8309 - 10799.9)^2 + (12882 - 10799.9)^2 + (14419 - 10799.9)^2 + (16092 - 10799.9)^2 + (18384 - 10799.9)^2 +(20916 - 10799.9)^2  }{16}} \ ...

  \ ... \sqrt{\frac{(23812 - 10799.9)^2 +(25814 - 10799.9)^2 }{16}}

=>  \sigma  =  8340

  Generally the test statistic is mathematically represented as

  t =  \frac{10799.9- 14400}{ \frac{8340}{\sqrt{16} } }

t  =  -1.727

From the z-table  the p-value is  

     p-value  = P(Z > t) =  P(Z >  -1.727) =  0.95792

 From the values obtained we see that

        p-value  >  \alpha  so we fail to reject the null hypothesis

Which implies that the claim of the NarStor is wrong

5 0
3 years ago
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