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Andre45 [30]
3 years ago
11

Heavy Metal Corporation is expected to generate the following free cash flows over the next five years.

Business
1 answer:
mezya [45]3 years ago
4 0

Answer:

Enterprise value of Heavy Metal= $1,080.766

Share price =  $18.945 per unit

Explanation:

<em>The value of a firm is the present value of the free cash flow discounted at the weighted average  cost of capital</em>

Year                                          PV

1        52.1 × 1.14^(-1)      = 45.70175439

2        68.6 × 1.14^(-2)  = 52.40073869

3          78.6 × 1.14^(-3) = 53.05276117

4         74.4×  1.14^(-4) = 44.05077264

5          81.1 ×  1.14^(-5) = 42.12079868

Year  and beyond

  81.1 × 1.04/(0.14-0.04) = 843.44

Total value =   45.70+ 52.40+53.052 + 44.050 +42.120+  843.44 = 1080.766826

Enterprise value of Heavy Metal= $1,080.766

Share price = Total value - Debt value / number of shares

=  (1,080.766  - 304 )/ 41 million units= $18.945 per unit

Share price =  $18.945 per unit

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Q 6.30: Rollins Technology is determining ending inventory. In the inventory process, Rollins inadvertently miscategorized a $9,
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The manager of a computer software company wishes to study the number of hours senior executives by type of industry spend at th
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Answer:

Since the calculated value of F= 5.733 falls in the critical region we reject the null hypothesis  and conclude all three means are not equal.

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The given data is

Banking          Retail            Insurance

12                      8                        10

10                     8                          8

10                     6                          6

12                     8                           8

<u>10                    10                          10</u>

The results of excel are:            

<u><em>Anova: Single Factor  </em></u>    

     

SUMMARY      

Groups    Count Sum Average         Variance  

Column 1 5 54            10.8              1.2  

Column 2 5 40               8               2  

Column 3 5 42              8.4                 2.8  

         

1) Let the null and alternate hypotheses be

H0: u1=u2=u3  i.e all the three means are equal and

Ha: Not all three means are equal

2) The significance level is set at ∝ =0.05

3)The test statistic to use is

F= sb²/ sw²

which has F distribution with v1= k-1 →3-1=2 and v2= n-k →15-3=12 degrees of freedom

After calculations the following table is obtained.

<u><em>ANOVA  </em></u>    

Source              SS            df         MS            F        P-value        F crit

of Variation

B/w Groups        22.93         2     11.467      5.733    0.01788        3.885

<u>Within Groups       24           12           2                                                       </u>      

<u>Total                     46.93      14                                                                </u>

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