Answer:
mean score of class B = 1778/25 = 71.12
Step-by-step explanation:
This was your question : Class A has 12 pupils and class B has 25 pupils. Both classes sit the same maths test. The mean score for class A is 80. The mean score for both classes is 74. What is the mean score (rounded to 2 DP) in the maths test for class B?
mean of class A = Σfx/Σf
mean of class A = 80
Σfx = 80 × 12 = 960
Mean score for both classes = 74
where
b = Σfx of class B
960 + b/37 = 74
cross multiply
960 + b = 2738
b = 2738 - 960
b = 1778
mean score of class B = Σfx/Σf
Σfx = 1778
Σf = 25
Therefore,
1778/25 = 71.12
An interesting question!
Instead of solving each set of non-linear system of equations 6 times, I will try a simpler way, which is more adapted to multiple choice questions. However, please be warned that this method will not improve immensely math skills, but will help with reasoning, and possibly a broader understand in system of equations.
For simplicity, I will denote the sets of system of non-linear equations by S={A,B,C,D,E,F} from left to right.
1. given solutions {(-2,3),(7,-6)}
We first check each member of S, i.e. A,B,C,D,E,F for the linear conditions.
A: x+y=-2+3=1 ≠ 3 so no
B: x-y=-2-3=-5 ≠ 1 so no
C: 2x+y=-2(2)+3=-1 ≠ 1 so no
D: x+2y=-2+3(2)=4 ≠ 2 so no
E: -x+y=2+3=5 ≠ 1 so no
F: x+y=-2+3=1 ≠ 3 so YES now check the other solution 7-6=-1 OK
Now check the non-linear condition for F:
S1: (-2,3)
y-15=3-15=-12
-x^2+4x=-(-2)^2+4*(-2)=-12 good
S2:(7,-6)
y-15=-6-15=-21
-x^2+4x=-(7^2)+4(7)=-49+28=-21 also good,
So Solution set (1) matches tile F
(2) Given solutions {(-5,8),(3,0)}
We proceed in a similar way, to find that
-5+8=3 & 3+0=3 (matches A) and nothing else.
Check non-linear conditions (optionally, I transposed the terms to make comparison easier)
A. x^2+x-y=(-5)^2+(-5)-8=25-5-8=12 (looks good)
x^2+x-y=(3)^2+(3)-0=9+3=12 (looks even better)
So we determined that {(-5,8),(3,0)} is the solution for tile A.
(3) For solution set {(-2,5),(3,-5)}
Check linear conditions:
x+y=3 and -3, so does not satisfy A,& F.
x-y=-7 and 8, so does not satisfy B
2x+y=5 and 1 so satisfies C
x+2y=8 & -7 so does not satisfy D
-x+y=7 & -8 so does not satisfy E
Thus C is our only set. NOTE: if two distinct points satisfy one linear condition (say C), then both points cannot satisfy another non-equivalent linear condition, i.e. we didn't really have to check points D & E (which are not equivalent linear conditions as C) once we have found C.
Now check non-linear conditions:
x^2-3x-y=(-2)^2-3(-2)-5=5 ok
x^2-3x-y=(3)^2-3(3)-(-5)=5 ok
So solution set (3) matches tile C.
@officiallyqueenz, for your benefit, I will leave one for you as exercise. Please proceed to find the system for solution set (4). Post a your work as comments if you get stuck, or your answer for verification if you wish.
For those who consider this answer as incomplete, please feel free to report the above answer. At his/her discretion, moderator may delete answer as incomplete, may delete the whole question as too many questions, or then again, may leave the answer intact for the better learning of the user who posed the question.
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Answer:
FALSE
Step-by-step explanation:
246=2*3*41
If n=2 then 123=3*41 is not a multiple of n
FALSE
For this shape there are four points. The plotted ordered pairs are:
M = (3, 9)
N = (6, 0)
O = (3, 6)
P = (-3, 0)
All you have to do is multiply the BOTH x- and y-values by 1/3 on EACH ordered pair. And if it’s easier, you can divide by 3 as well. It’s the same thing.
M = (3, 9) = (3*1/3, 9*1/3) = (1, 3)
N = (6, 0) = (6*1/3, 0*1/3) = (2, 0)
O = (3, 6) = (3*1/3, 6*1/3) = (1, 2)
P = (-3, 0) = (-3*1/3, 0*1/3) = (-1, 0)
(1, 3) , (2, 0) , (1, 2) , and (-1, 0) are your points.
<span>adjacent angles
m<BAC and m<CAD
answer: </span><BAC and <CAD (second choice)
