Solve for xxx. 12x+7 4012x+7 4012, x, plus, 7, is less than, minus, 11, start color #ed5fa6, start text, space, O, R, space, end
gregori [183]
Answer:
<h2>
x < -3/2 and x>48/5</h2>
Step-by-step explanation:
Given the inequality function 12x+7< -11 and 5x-8>40, we are to solve for the value of x. To solve for x, the following steps must be followed.
For the inequality 12x+7< -11
Step 1: Subtract 7 from both sides of the inequality
12x+7-7< -11-7
12x < -18
Step 2: Divide both sides of the inequality by 12
12x/12 < -18/12
x < -3/2 .............. 1
For the inequality 5x-8> 40;
Step 1: Add 8 to both sides of the inequality
5x-8+8 > 40+8
5x > 48
Step 2: Divide both sides of the inequality by 5
5x/5 > 48/5
x>48/5....... 2
Combining equation 1 and 2, we will have;
x < -3/2 and x>48/5
If x>48/5 then 48/5<x
Combining 48/5<x with x < -3/2 will give 48/5<x<-3/2
Answer:
In point slope form, this is y = (-5/2)x + 2. In standard form, it's (5/2)x + y = 2.
Step-by-step explanation:
y-6=-5/2 (x+4) has already been written in point-slope form. The slope of the line is -5/2 and the point on the line is (-4,6).
In slope-intercept form: add 6 to both sides of the original equation, obtaining y = 6 - (5/2)(x + 4), or y = 2 - (5/2)x, or y = (-5/2)x + 2
<span>Winning Probablity = 0.2, hence Losing Probability = 0.8
Probablity of winning atmost one time, that means win one and lose four times or lose all the times. So p(W1 or W0) = p (W1) + p(W0)
Winning once W1 is equal to L4, winning zero times is losing 5 times.
p(W1) = p(W1&L4) and this happens 5 times; p(W0) = p(L5);
p (W1) + p(W0) = p(L4) + p(L5)
p(L4) + p(L5) = (5 x 0.2 x 0.8^4) + (0.8^5) => 0.8^4 + 0.8^5
p(W1 or W0) = 0.4096 + 0.32768 = 0.7373</span>
Answer:
b+9=20-12
Step-by-step explanation:
Hopefully this helped, if not HMU and I will try my best to get you a better answer!
I worked out the problem hope it helps
