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2 years ago

the sum of two number is 21. The larger number is 6 less than twice the smaller number. Find the two numbers.

Mathematics

1 answer:

Natalka [10]2 years ago

6 0

Answer:

7.5 is the large number and 6.75 is the smaller number

Step-by-step explanation:

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1 year ago

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Answer:

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2 years ago

Read 2 more answers

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:

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

8 0

3 years ago

What is the equivalent of <br>2/3x -9 - 2x + 2 = 1 after combining like terms?

elena-s [515]

Answer:

Step-by-step explanation:

5 0

3 years ago

I have no idea how to do this question

Korvikt [17]

I know this doesn't really help but... Good luck on the test!! And also I'm pretty sure they have youtube vids on how to do this stuff. Well, anyway bye!!! Good luck!!!

4 0

3 years ago