Answer:
Check the solution below
Step-by-step explanation:
2) Given the equation
x +y =5... 1 and
x-y =3 ... 2
Add both equations
x+x = 5+3
2x = 8
x = 8/2
x = 4
Substitute x = 4 into 1:
From 1: x+y = 5
4+y= 5
y = 5-4
y = 1
3) Given
x+3y =15 ... 1
2x+7y=19 .... 2
From 2: x = 15-3y
Substitute into 2
2(15-3y)+7y = 19
30-6y+7y = 19
30+y = 19
y = 19-30
y = -11
Substitute y=-11 into x = 15-3y
x =15-3(-11)
x = 15+33
x = 48
The solution set is (48, -11)
4) given
x/2 +y/3 =0 and x+2y=1
From 1
(3x+2y)/6 = 0
3x+2y = 0.. 3
x+2y= 1... 4
From 4: x = 1-2y
Substutute
3(1-2y) +2y = 0
3-6y+2y = 0
3 -4y = 0
4y = 3
y = 3/4
Since x = 1-2y
x = 1-2(3/4)
x = 1-3/2
x= -1/2
The solution set is (-1/2, 3/4)
5) Given
5.x=1/2 and y =x +1 then solution is
We already know the vkue of x
Get y
y= x+1
y = 1/2 + 1
y = 3/2
Hence the solution set is (1/2, 3/2)
6) Given
3x +y =5 and x -3y =5
From 3; x = 5+3y
Substitute into 1;
3(5+3y)+y = 5
15+9y+y = 5
10y = 5-15
10y =-10
y = -1
Get x;
x = 5+3y
x = 5+3(-1)
x = 5-3
x = 2
Hence two solution set is (2,-1)
combine like terms: -5p > -10
divide by -5 on both sides: p > 2
First, we're going to see the candies per minute that machine C packs.
150 / 2 = 75
Now that we know that machine C packs 75 candies per minute, we're going to multiply the candies machine C makes by 11 minutes.
75 x 11 = 825
We're now gonna do the same with machine D.
130 x 11 = 1430
Then we're going to find the difference between machine C and machine D, we do this because the question basically asks how much more candies can machine D pack than machine C.
1430 - 825 = 605 candies.
This is how we find our final answer.
Our answer would be D) 605.
Hope this helps!~
The z-score for 36,000 is (36,000 - 40,000) / (15,000 / √25) = -1.333
The z-score for 42,000 is (42,000 - 40,000) / (15,000 / √25) = 0.6667
P(36,000 < x < 42,000) = P(-1.3333 < z < 0.6667)
P(z < 0.6667) - P(z < -1.3333)
0.7475 - 0.0913
0.6562