(x - 4) * (x + 3) = 0
***************************************************************
Answer: tea = 15 rupees per kg
sugar= 3 rupees per kg
Step-by-step explanation:
Hi, to answer this question we have to write a system of equations with the information given:
<em>"Two kg of tea and 3 kg of sugar cost rupees 39 in january 1997":
</em>
2 t + 3 s =39 (a)
Where:
- t= price of 1 kg of tea
- s = price of 1 kg of sugar
<em>"in march 1997 the price of the tea increased by 25% (1.25)and the price of the sugar increased by 20%(1.20) and the same quantity of tea and sugar cost rupees 48.30.
"</em>
2(t1.25)+3(s1.2) = 48.30 (b)
- <em>Solving for t in (b)
</em>
2t =39-3s
t = (39 -3s)/2
t = 19.5-1.5s
- <em>Replacing the value of t in (b)
</em>
2 x ((19.5-1.5s)1.25)+ 3 ( 1.2s) =48.30
2x ( 24.375 -1.875s) +3.6s =48.30
48.75 -3.75s+3.6s= 48.30
48.75-48.30 = 3.75s-3.6s
0.45= 0.15s
0.45/0.15 =s
3 =s
- <em>Replacing the value of s in (a)
</em>
2 t + 3 (3) =39
2 t + 9 =39
2 t =39 -9
2 t =30
t = 30/2
t= 15
Prices in january:
tea = 15 rupees per kg
sugar= 3 rupees per kg
Feel free to ask for more if needed or if you did not understand something.
Answer:
You are correct,y = 0.
Step-by-step explanation:
y = 4 x -3 + 12
y = -12 + 12
y = 0
Hope that helps!
Answer:
9.60 ; - 60.96
Step-by-step explanation:
Given the function :
F(x)=6(x+1) /25, x=0, 1, 2, 3, 4.
x = 0
F(0)=6(0+1)/25 = 6/25 = 0.24
x = 1
F(1)=6(1+1)/25 = 12/25 = 0.48
x = 2
F(2)=6(2+1)/25 = 18/25 = 0.72
x = 3
F(2)=6(3+1)/25 = 24/25 = 0.96
x = 4
F(2)=6(4+1)/25 = 30/25 = 1.2
X ______0 _____ 1 ______ 2 ______ 3 ____ 4
P(x) ___ 0.24 __ 0.48 ___ 0.72 ____ 0.96 __ 1.2
Mean, μ = Σx*p(x) :
(0*0.24) + (1*0.48) + (2*0.72) + (3*0.96) + (4*1.2)
= 9.60
Variance : Σx²*p(x) - μ²
(0^2*0.24) + (1^2*0.48) + (2^2*0.72) + (3^2*0.96) + (4^2*1.2) - 9.6^2
= 31.2 - 92.16
= - 60.96
