Y = -5x + 28
will be parallel to your equation because it has the same coefficient but will also pass through (3, 13)
Answer:
I’m pretty sure that’s - 11/3 .
Step-by-step explanation:
B. First off , standard form of a 2nd degree equation is Ax^2 + Bx + C. So look at the coefficient of Ax^2 which is -2.
If positive, the parabola opens up and has a minimum.
If negative, the parabola opens down and has a maximum.
A. To find the vertex (in this case maximum),
Graph the equation -OR—
make a table. — OR—
Find the zeroes and find the middle x-value
-2x^2 - 4x + 6
-2(x^2 +2x - 3 = 0
-2 (x - 1) ( x + 3)=0
x - 1 = 0. x + 3 = 0
x = 1. x = -3. So halfway would be at (-1, __).
Sub in -1 into original equation -2x^2 -4x + 6 … -2(-1)^2 -4(-1) + 6 = -2 +4 +6 = 8
So the vertex is (-1,8)
Answer:
x=17
Step-by-step explanation:
3x-15=20+16
3x-15=36
3x=36+15
3x=51
x=17
Answer:
a) F(2)=0.30
b)P(X>3)=0.7
c)P(2≤X≤5)=0.67
d)P(2<X<5)=0.71
Step-by-step explanation:
a) F(2)=P(X≤2)=0.30
F(2) is given in the probability distribution
b) P(X>3)=1-P(X≤2)=1-F(2)=1-0.3=0.7
P(X>3)=0.7
c)P(2≤X≤5)=?
As we know that P(a≤X≤b)=F(b)-F(a)=P(X≤b)-P(X≤a)
So,
P(2≤X≤5)=P(X≤5)-P(X≤2)=F(5)-F(2)=0.97-0.3=0.67
P(2≤X≤5)=0.67
d)P(2<X<5)=P(1≤X≤4)=P(X≤4)-P(X≤1)=F(4)-F(1)=0.91-0.2=0.71.
P(2<X<5)=0.71
