6x^2 -9x -2 just add common variables
Answer:
i think probably 80°
Step-by-step explanation:
PQ = RS
2x+3 = 4x-3
x = 3
PTQ = RTS
9y-3x-1 = 5x+7y-5
2y-8x = -4
y-4x = -2
y-4(3) = -2
y-12 = -2
y = -2+12
y = 10
so PTQ = 9y-3x-1
9(10)-3(3)-1
90-9-1
80°
Let
be the 20 marks of the boys, and
be the 10 marks of the girls.
We know that the global mean was 70, meaning that
![\dfrac{b_1+b_2+\ldots+b_{20}+g_1+g_2+\ldots+g_{10}}{30}=70](https://tex.z-dn.net/?f=%5Cdfrac%7Bb_1%2Bb_2%2B%5Cldots%2Bb_%7B20%7D%2Bg_1%2Bg_2%2B%5Cldots%2Bg_%7B10%7D%7D%7B30%7D%3D70)
Multiplying both sides by 30 we deduce that the sum of the scores of the whole classroom is
![b_1+b_2+\ldots+b_{20}+g_1+g_2+\ldots+g_{10}=2100](https://tex.z-dn.net/?f=b_1%2Bb_2%2B%5Cldots%2Bb_%7B20%7D%2Bg_1%2Bg_2%2B%5Cldots%2Bg_%7B10%7D%3D2100)
By the same logic, we work with the marks of the boys alone: we know the average:
![\dfrac{b_1+b_2+\ldots+b_{20}}{20}=62](https://tex.z-dn.net/?f=%5Cdfrac%7Bb_1%2Bb_2%2B%5Cldots%2Bb_%7B20%7D%7D%7B20%7D%3D62)
And we deduce the sum of the marks for the boys:
![b_1+b_2+\ldots+b_{20}=1240](https://tex.z-dn.net/?f=b_1%2Bb_2%2B%5Cldots%2Bb_%7B20%7D%3D1240)
Which implies that the sum of the marks of the girls is ![2100-1240=860](https://tex.z-dn.net/?f=2100-1240%3D860)
And finally, the mean for the girls alone is
![\dfrac{860}{10}=86](https://tex.z-dn.net/?f=%5Cdfrac%7B860%7D%7B10%7D%3D86)
The estimate of the sum of 202 and 57 is 260. You add the numbers together and than round up.