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2 years ago

Which was the most important effect of the bessemer process

2 answers:

8 0

<span>Which was the most important effect of the bessemer process

d. it allowed factory production of steel

2. which was the immediate goal of the standard oil company when it lowers its prices?

b. to out complete rival businesses

</span>

Studentka2010 [4]2 years ago

6 0

Answer:

1. c

2. b

3. d

4. a

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The market and Stock J have the following probability distributions:

denis-greek [22]

Answer:

1) $E(M) = 14*0.3 + 10*0.4 + 19*0.3 = 13.9 \%$

2) $E(J)= 22*0.3 + 4*0.4 + 12*0.3 = 11.8 \%$

3) $E(M^2) = 14^2*0.3 + 10^2*0.4 + 19^2*0.3 = 207.1$

And the variance would be given by:

$Var (M)= E(M^2) -[E(M)]^2 = 207.1 -(13.9^2)= 13.89$

And the deviation would be:

$Sd(M) = \sqrt{13.89}= 3.73$

4) $E(J^2) = 22^2*0.3 + 4^2*0.4 + 12^2*0.3 =194.8$

And the variance would be given by:

$Var (J)= E(J^2) -[E(J)]^2 = 194.8 -(11.8^2)= 55.56$

And the deviation would be:

$Sd(M) = \sqrt{55.56}= 7.45$

Step-by-step explanation:

For this case we have the following distributions given:

Probability M J

0.3 14% 22%

0.4 10% 4%

0.3 19% 12%

Part 1

The expected value is given by this formula:

$E(X)=\sum_{i=1}^n X_i P(X_i)$

And replacing we got:

$E(M) = 14*0.3 + 10*0.4 + 19*0.3 = 13.9 \%$

Part 2

$E(J)= 22*0.3 + 4*0.4 + 12*0.3 = 11.8 \%$

Part 3

We can calculate the second moment first with the following formula:

$E(M^2) = 14^2*0.3 + 10^2*0.4 + 19^2*0.3 = 207.1$

And the variance would be given by:

$Var (M)= E(M^2) -[E(M)]^2 = 207.1 -(13.9^2)= 13.89$

And the deviation would be:

$Sd(M) = \sqrt{13.89}= 3.73$

Part 4

We can calculate the second moment first with the following formula:

$E(J^2) = 22^2*0.3 + 4^2*0.4 + 12^2*0.3 =194.8$

And the variance would be given by:

$Var (J)= E(J^2) -[E(J)]^2 = 194.8 -(11.8^2)= 55.56$

And the deviation would be:

$Sd(M) = \sqrt{55.56}= 7.45$

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Answer:

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Step-by-step explanation:

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Step-by-step explanation:

18/6÷3=6/2

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Answer:

Step-by-step explanation:

Answer is in below attachment

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