Answer:
2/7 or about 28% depending on what type of answer your teacher is looking for
The answer is 465 and you are welcome
No solution of the system of equations y = -2x + 5 and -5y = 10x + 20 ⇒ 2nd answer
Step-by-step explanation:
Let us revise the types of solutions of a system of linear equations
- One solution
- No solution when the coefficients of x and y in the two equations are equal and the numerical terms are different
- Infinitely many solutions when the coefficients of x , y and the numerical terms are equal in the two equations
∵ y = -2x + 5
- Add 2x to both sides
∴ 2x + y = 5 ⇒ (1)
∵ -5y = 10x + 20
- Subtract 10x from both sides
∴ -10x - 5y = 20
- Divide both sides by -5
∴ 2x + y = -4 ⇒ (2)
∵ The coefficient of x in equation (1) is 2
∵ The coefficient of x in equation (2) is 2
∴ The coefficients of x in the two equations are equal
∵ The coefficient of y in equation (1) is 1
∵ The coefficient of y in equation (2) is 1
∴ The coefficients of y in the two equations are equal
∵ The numerical term in equation (1) is 5
∵ The numerical term in equation (2) is -4
∴ The numerical terms are different
From the 2nd rule above
∴ No solution of the system of equations
No solution of the system of equations y = -2x + 5 and -5y = 10x + 20
Learn more:
You can learn more about the system of equations in brainly.com/question/6075514
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16. 5x^3 y^-5 • 4xy^3
20x^4y^-2
20x^4 • 1/y^2
=20x^4/y^2
17. -2b^3c • 4b^2c^2
= -8b^5c^3
18. a^3n^7 / an^4 (a^3 minus a = a^2 same as n^7 minus n^4 = n^3)
=a^2n^3
19. -yz^5 / y^2z^3
= -z^2/y
20. -7x^5y^5z^4 / 21x^7y^5z^2 (divide -7 to 21 and minus xyz)
= -z / 3x^2
21. 9a^7b^5x^5 / 18a^5b^9c^3
=a^2c^2 / 2b^4
22. (n^5)^4
n ^5 x 4
=n^20
23. (z^3)^6
z ^3 x 6
=z^18
Step-by-step explanation:
Given that,
a)
X ~ Bernoulli 
and Y ~ Bernoulli 
X + Y = Z
The possible value for Z are Z = 0 when X = 0 and Y = 0
and Z = 1 when X = 0 and Y = 1 or when X = 1 and Y = 0
If X and Y can not be both equal to 1 , then the probability mass function of the random variable Z takes on the value of 0 for any value of Z other than 0 and 1,
Therefore Z is a Bernoulli random variable
b)
If X and Y can not be both equal to 1
then,
or 
and 
c)
If both X = 1 and Y = 1 then Z = 2
The possible values of the random variable Z are 0, 1 and 2.
since a Bernoulli variable should be take on only values 0 and 1 the random variable Z does not have Bernoulli distribution
