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Hunter-Best [27]
3 years ago
15

Which of the following hypotheses/theories suggests that investors regard a change in dividend payments as a signal that the fir

m's management expects future earnings to also change? a. Clientele effect theory b. Constant payout ratio hypothesis c. Dividend modification hypothesis d. Information content hypothesis e. Projected earnings hypothesi
Mathematics
1 answer:
lutik1710 [3]3 years ago
7 0

Answer:

C

Step-by-step explanation:

Dividend policy (hypothesis) of firms as stated by Musa (2009) is a cultural phenomenon that changes continuously according to environment and time, hence it is necessary to continuously modify dividend behavioural models to capture those factors that are peculiar to a particular period and environment, as well as changes in tax.

Hope this helps

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Find the area of the figure.<br> 6 in.<br> 6 in.<br> 8 in.<br> 4 in.<br> 10
expeople1 [14]

Answer:

68 in.

Step-by-step explanation:

Multiply 4 by 8 and get 32

Multiply 6 by 6 and get 36

Add 38 and 36

=68

8 0
3 years ago
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If a rectangular prism has a length of 3 units and a width of 3 units with a height of 3 2/3 units what is the volume?
ipn [44]

Step-by-step explanation:

3 2/3 units = 3.66

3×3×3.66 = 32.94

hope it helps!

3 0
3 years ago
What is $25; 10% increase
viktelen [127]
10% of $25 is $2.5 (:
4 0
3 years ago
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The amount of lateral expansion (mils) was determined for a sample of n = 7 pulsed-power gas metal arc welds used in LNG ship co
zvonat [6]

Answer:

95% Confidence interval for σ2 and for σ is (3.33 , 38.85) and (1.82 , 6.23) respectively.

Step-by-step explanation:

We are given that the amount of lateral expansion (mils) was determined for a sample of n = 7 pulsed-power gas metal arc welds used in LNG ship containment tanks. The resulting sample standard deviation was s = 2.83 mils.

Assuming data follows normal distribution.

So, firstly the pivotal quantity for 95% confidence interval for the population variance is given by;

        P.Q. = \frac{(n-1)s^{2} }{\sigma^{2} } ~ \chi^{2} __n_-_1

where, s = sample standard deviation = 2.83 mils

          \sigma^{2} = population variance

           \sigma = population standard deviation

           n = sample size = 7

<em>So, 95% confidence interval for population variance, </em>\sigma^{2} <em>is;</em>

P(1.237 < \chi^{2} __n_-_1 < 14.45) = 0.95 {As the table of at 6 degree of freedom

                                                     gives critical values of 1.237 & 14.45}

P(1.237 < \frac{(n-1)s^{2} }{\sigma^{2} } < 14.45) = 0.95

P( \frac{ 1.237}{(n-1)s^{2} } < \frac{1 }{\sigma^{2} } < \frac{ 14.45}{(n-1)s^{2} } ) = 0.95

P( \frac{ (n-1)s^{2}}{14.45 } < \sigma^{2} < \frac{ (n-1)s^{2}}{1.237 } ) = 0.95

95% confidence interval for \sigma^{2} = ( \frac{ (n-1)s^{2}}{14.45 }  , \frac{ (n-1)s^{2}}{1.237 }  )

                                                  = ( \frac{ (7-1) \times 2.83^{2}}{14.45 } , \frac{ (7-1) \times 2.83^{2}}{1.237 } )

                                                  = (3.33 , 38.85)

95% C.I. for population standard deviation, \sigma  = ( \sqrt{3.33} , \sqrt{38.85} )

                                                                            = (1.82 , 6.23)

Therefore, 95% confidence interval for the population variance (σ2) and population standard deviation (σ) are (3.33 , 38.85) and (1.82 , 6.23) respectively.

7 0
3 years ago
Divide<br> (i) 448 ÷ (-7)<br><br><br><br><br> I need it now <br> please someone
Anna11 [10]
=448/(_7
=-64
Therefore the correct answer is-64
7 0
2 years ago
Read 2 more answers
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