Answer:
1.
8x + 6y = 28
9x + 3y = 39
Set to equal y.
8x + 6y = 28
6y = -8x + 28
y = -8/6x + 28/6
y = -4/3x + 14/3
9x + 3y = 39
3y = -9x + 39
y = -9/3x + 39/3
y = -3x + 13
Set equal to each other.
-4/3x + 14/3 = -3x + 13
Combine like terms.
5/3x = 25/3
Multiply by 3/5
x = 5
Plug x in.
y = -3(5) + 13
y = -15 + 13
y = -2
(5, -2)
2.
-2x - 9y = -3
-7x - 9y= -33
Set equal to y.
-2x - 9y = -3
-9y = 2x - 3
y = -2/9x + 3/9
y = -2/9x + 1/3
-7x - 9y= -33
-9y = 7x - 33
y = -7/9x + 33/9
y = -7/9x + 11/3
Set equal to each other
-2/9x + 1/3 = -2/9x + 1/3
0 = 0
infinitely many solutions
Step-by-step explanation:
<h2>
<em><u>You can solve this using the binomial probability formula.</u></em></h2><h2>
<em><u>You can solve this using the binomial probability formula.The fact that "obtaining at least two 6s" requires you to include cases where you would get three and four 6s as well.</u></em></h2><h2>
<em><u>You can solve this using the binomial probability formula.The fact that "obtaining at least two 6s" requires you to include cases where you would get three and four 6s as well.Then, we can set the equation as follows:</u></em></h2><h2>
<em><u>You can solve this using the binomial probability formula.The fact that "obtaining at least two 6s" requires you to include cases where you would get three and four 6s as well.Then, we can set the equation as follows: </u></em></h2><h2>
<em><u>You can solve this using the binomial probability formula.The fact that "obtaining at least two 6s" requires you to include cases where you would get three and four 6s as well.Then, we can set the equation as follows: P(X≥x) = ∑(k=x to n) C(n k) p^k q^(n-k) </u></em></h2><h2>
<em><u>You can solve this using the binomial probability formula.The fact that "obtaining at least two 6s" requires you to include cases where you would get three and four 6s as well.Then, we can set the equation as follows: P(X≥x) = ∑(k=x to n) C(n k) p^k q^(n-k) n=4, x=2, k=2</u></em></h2><h2>
<em><u>You can solve this using the binomial probability formula.The fact that "obtaining at least two 6s" requires you to include cases where you would get three and four 6s as well.Then, we can set the equation as follows: P(X≥x) = ∑(k=x to n) C(n k) p^k q^(n-k) n=4, x=2, k=2when x=2 (4 2)(1/6)^2(5/6)^4-2 = 0.1157</u></em></h2><h2>
<em><u>You can solve this using the binomial probability formula.The fact that "obtaining at least two 6s" requires you to include cases where you would get three and four 6s as well.Then, we can set the equation as follows: P(X≥x) = ∑(k=x to n) C(n k) p^k q^(n-k) n=4, x=2, k=2when x=2 (4 2)(1/6)^2(5/6)^4-2 = 0.1157when x=3 (4 3)(1/6)^3(5/6)^4-3 = 0.0154</u></em></h2><h2>
<em><u>You can solve this using the binomial probability formula.The fact that "obtaining at least two 6s" requires you to include cases where you would get three and four 6s as well.Then, we can set the equation as follows: P(X≥x) = ∑(k=x to n) C(n k) p^k q^(n-k) n=4, x=2, k=2when x=2 (4 2)(1/6)^2(5/6)^4-2 = 0.1157when x=3 (4 3)(1/6)^3(5/6)^4-3 = 0.0154when x=4 (4 4)(1/6)^4(5/6)^4-4 = 0.0008</u></em></h2><h2>
<em><u>You can solve this using the binomial probability formula.The fact that "obtaining at least two 6s" requires you to include cases where you would get three and four 6s as well.Then, we can set the equation as follows: P(X≥x) = ∑(k=x to n) C(n k) p^k q^(n-k) n=4, x=2, k=2when x=2 (4 2)(1/6)^2(5/6)^4-2 = 0.1157when x=3 (4 3)(1/6)^3(5/6)^4-3 = 0.0154when x=4 (4 4)(1/6)^4(5/6)^4-4 = 0.0008Add them up, and you should get 0.1319 or 13.2% (rounded to the nearest tenth)</u></em></h2>
Y= -2x²-3x+2, (form y = ax² +b .x + c
The parabola opens downward & hence it's Vertex is a MAXIMUM.
We know that the axis of symmetry is -b/2a ==>3/(-4) ==>
x= - 3/4 (axis of symmetry. At x=-3/4 , y = 25/8 = 3.125
Hence the axis of symmetry is x= - 3/4 & the vertex is at (-3/4, 25/8)
Anything between -3≤x<3
Those integers can be -2,-1,0,1 or 2
Answer:
y=K/4z
Step-by-step explanation:
K=4yz
y=K/4z
