We have that
<span>y=2x+4--------> equation 1
3x−6y=3-------> equation 2
step 1
</span>I substitute the value of y in equation 1 for the value of y in equation 2<span>
so
</span>3x−6*[2x+4]=3-------> 3x-12x-24=3
-9x=3+24
-9x=27------> 9x=-27
x=-27/9
x=-3
step 2
<span>I substitute the value of x in equation 1 to get the value of y</span>
y=2x+4--------> y=2*(-3)+4--------> y=-6+4
y=-2
the answer is
the solution is the point (-3,-2)
x=-3
y=-2
3:5 =3/5
3/5=4.16/x
using cross multiplication,
5(4.15)=3(x)
20.8=3x
x=108/15
thus, 4.16÷104/15 =3/5
Hello:<span>
the equation is : y = ax+b
the slope is a : a×(-3/2) = -1......(
perpendicular to a line with a slope of -3/2)
a = 2/3 y=(2/3)x+b
the line that passes through (-2, -2) :- 2 =
(2/3)(-2)+b
b = -2/3
<span> the equation is : y = (2/3)x-2/3</span></span>
<span><span>y =(2/3)(x-1)</span></span>
Answer:
20%
Step-by-step explanation:
Just did the question :)
Answer:
−35.713332 ; 0.313332
Step-by-step explanation:
Given that:
Sample size, n1 = 11
Sample mean, x1 = 79
Standard deviation, s1 = 18.25
Sample size, n2 = 18
Sample mean, x2 = 96.70
Standard deviation, s2 = 20.25
df = n1 + n2 - 2 ; 11 + 18 - 2 = 27
Tcritical = T0.01, 27 = 2.473
S = sqrt[(s1²/n1) + (s2²/n2)]
S = sqrt[(18.25^2 / 11) + (20.25^2 / 18)]
S = 7.284
(μ1 - μ2) = (x1 - x2) ± Tcritical * S
(μ1 - μ2) = (79 - 96.70) ± 2.473*7.284
(μ1 - μ2) = - 17.7 ± 18.013332
-17.7 - 18.013332 ; - 17.7 + 18.013332
−35.713332 ; 0.313332
