The correct answer is 55 1/50. Hope this helps!!
Ok so let us label our equations first
1- 3x+2y+z=7
2- 5x+5y+4z=3
3- 3x+2y+3z=1
subtracting equation 3 from equation 1
3x+2y+3z=1
(-). 3x+2y+z= 7
----------------------
2z=-6----->z=-3
since we already chose equation 1 and 3 we must involve equation 2
substituting the value if z in the second equation
5x+5y-12=3
5x+5y=15
choosing either the first or the third
3x+2y-3=7
3x+2y=10
solving the system
5x+5y=15
3x+2y=10
multipling the first equation by 2 and the second equation by 5
10x+10y=30
15x+10y=50
subtracting the two equations
-5x=-20--->x=4 substituting for the value of x
40+10y=30--->10y=-10y--->y=-1
so our soultion is x=4,y=-1,z==-3 or (4,-1,-3)
I think it would be (8)(-1) = -8. by adding the two equations you get 2x = 16 so x = 8 then y = 7 - x = -1. Therefore xy = -8.
Hope that helps
Answers and steps are in the picture below!
Answer:
Step-by-step explanation:
a) The formula for determining the standard error of the distribution of differences in sample proportions is expressed as
Standard error = √{(p1 - p2)/[(p1(1 - p1)/n1) + p2(1 - p2)/n2}
where
p1 = sample proportion of population 1
p2 = sample proportion of population 2
n1 = number of samples in population 1,
n2 = number of samples in population 2,
From the information given
p1 = 0.77
1 - p1 = 1 - 0.77 = 0.23
n1 = 58
p2 = 0.67
1 - p2 = 1 - 0.67 = 0.33
n2 = 70
Standard error = √{(0.77 - 0.67)/[(0.77)(0.23)/58) + (0.67)(0.33)/70}
= √0.1/(0.0031 + 0.0032)
= √1/0.0063
= 12.6
the standard error of the distribution of differences in sample proportions is 12.6
b) the sample sizes are large enough for the Central Limit Theorem to apply because it is greater than 30