The answer would be A. When using Cramer's Rule to solve a system of equations, if the determinant of the coefficient matrix equals zero and neither numerator determinant is zero, then the system has infinite solutions. It would be hard finding this answer when we use the Cramer's Rule so instead we use the Gauss Elimination. Considering the equations:
x + y = 3 and <span>2x + 2y = 6
Determinant of the equations are </span>
<span>| 1 1 | </span>
<span>| 2 2 | = 0
</span>
the numerator determinants would be
<span>| 3 1 | . .| 1 3 | </span>
<span>| 6 2 | = | 2 6 | = 0.
Executing Gauss Elimination, any two numbers, whose sum is 3, would satisfy the given system. F</span>or instance (3, 0), <span>(2, 1) and (4, -1). Therefore, it would have infinitely many solutions. </span>
F(x)=(-2/((x+y-2)^(1/2))-(x+y+2)^(1/2)
the only irrational part of this expression is the (x+y-2)^(1/2) in the denominator, so, to rationalize this, you multiply the numerator and denominator by the denominator, as well as the other parts of the expression
also, you must multiply the -sqrt(x+y+2) by sqrt(x+y-2)/sqrt(x+y-2) to form a common denominator
(-2)/(x+y-2)^(1/2)-(x+y+2)^(1/2)(x+y-2)^(1/2)/(x+y-2)^(1/2)
(common denominator)
(-2-(x^2+xy+2x+xy+y^2+2y-2x-2y-4))/(x+y-2)^(1/2)
(FOIL)
(-2-x^2-y^2-2xy+4)/(x+y-2)^(1/2)
(Distribute negative)
(-x^2-y^2-2xy+2)/(x+y-2)^(1/2)
(Simplify numerator)
(-x^2-y^2-2xy+2)(x+y-2)^(1/2)/(x+y-2)^(1/2)(x+y-2)^(1/2)
(Rationalize denominator by multiplying both top and bottom by sqrt)
(-x^2-y^2-2xy+2)((x+y-2)^(1/2))/(x+y-2)
(The function is now rational)
=(-x^2-y^2-2xy+2)(sqrt(x+y-2))/(x+y-2)
The answer is clearly b makes the most sense
Answer: a) (0.755, 0.925)
Step-by-step explanation:
Let p be the population proportion of 8th graders are involved with some type of after school activity.
As per given , we have
n= 100
sample proportion: 
Significance level : 
Critical z-value : 
(using z-value table)
Then, the 98% confidence interval that estimates the proportion of them that are involved in an after school activity will be :-
i.e. 
i.e. 
i.e. 
Hence, the 98% confidence interval that estimates the proportion of them that are involved in an after school activity : a) (0.755, 0.925)
