Answer:
When the eqn cuts the x-axis , y=0
3x=12
x=12/3
x=4
So,at x=4 is the point at which the linear equation cuts the x-axis
If you are evaluating the answer is -41
5x - 3y = 11 ⇒ 5x - 3y = 11 ⇒ 5x - 3y = 11
x - 2y = 2 ⇒ -5(x - 3y) = 2 ⇒ <u>-5x + 15y = 2</u>
<u>12y</u> = <u>13</u>
12 12
y = 1 1/12
5x - 3(1 1/12) = 11
5x - 3 1/4 = 11
<u> +3 1/4 +3 1/4</u>
<u> 5x</u> = <u>14 1/4</u>
5 5
x = 2 17/20
(x, y) = (2 17/20, 1 1/12)
Answer:
Kindly check explanation
Step-by-step explanation:
Given the following :
P(brown) = 12% = 0.12
P(Yellow) = 15% = 0.15
P(Red) = 12% = 0.12
P(blue) = 23% = 0.23
P(orange) = 23% = 0.23
P(green) = 15% = 0.15
A.) Compute the probability that a randomly selected peanut M&M is not yellow.
P(not yellow) = P(Yellow)' = 1 - P(Yellow) = 1 - 0.15 = 0.85
B.) Compute the probability that a randomly selected peanut M&M is brown or red.
P(Brown) or P(Red) :
0.12 + 0.12 = 0.24
C.) Compute the probability that three randomly selected peanut M&M’s are all brown.
P(brown) * P(brown) * P(brown)
0.12 * 0.12 * 0.12 =0.001728
D.) If you randomly select three peanut M&M’s, compute that probability that none of them are blue.
P(3 blue)' = 1 - P(3 blue)
P(3 blue) = 0.23 * 0.23 * 0.23 = 0.012167
1 - P(3 blue) = 1 - 0.012167 = 0.987833
If you randomly select three peanut M&M’s, compute that probability that at least one of them is blue.
P(1 blue) + p(2 blue) + p(3 blue)
(0.23) + (0.23*0.23) + (0.23*0.23*0.23)
0.23 + 0.0529 + 0.012167
= 0.295067
