The system shown at the right has no solution, as the grahs never intersect.
On the other hand, the line and the parab. at the left do intersect, and the points of intersection are (-3,0) and (6,6).
Given: 11-pound mixture of peanuts, almonds, and raisins
Cost:
peanuts - 1.5 per pound
almonds - 3 per pound
raisins - 1.5 per pound
mixture:
twice as many peanuts as almond; total cost of mixture is 21.
a + p + r = 11 lbs
a + 2a + r = 11 lbs
3a + r = 11
r = 11 - 3a
1.5(2a) + 3a + 1.5r = 21
3a + 3a + 1.5r = 21
6a + 1.5r = 21
6a + 1.5(11-3a) = 21
6a + 16.5 - 4.5a = 21
6a - 4.5a = 21 - 16.5
1.5a = 4.5
1.5a/1.5 = 4.5/1.5
a = 3
almonds = 3 lbs
peanuts = 2a = 2(3) = 6lbs
raisins = 11 - 3a = 11 - 3(3) = 11 - 9 = 2 lbs
<span>My answer is: C. 6 lbs peanuts, 3 lbs almonds, 2 lbs raisins </span>
Answer:use money she planned to spend on her medical bills to pay her credit card
I think 8. Plug in -4 for x. 1/2-4^2
Using pemdas -4^2 is 16 and 1/2•16 is 8. If there were parenthesis around 1/2x then 1/2•-4 would be -2 and -2^2 is 4 but since there aren’t parentheses then 8
Correct question:
An urn contains 3 red and 7 black balls. Players A and B withdraw balls from the urn consecutively until a red ball is selected. Find the probability that A selects the red ball. (A draws the first ball, then B, and so on. There is no replacement of the balls drawn).
Answer:
The probability that A selects the red ball is 58.33 %
Step-by-step explanation:
A selects the red ball if the first red ball is drawn 1st, 3rd, 5th or 7th
1st selection: 9C2
3rd selection: 7C2
5th selection: 5C2
7th selection: 3C2
9C2 = (9!) / (7!2!) = 36
7C2 = (7!) / (5!2!) = 21
5C2 = (5!) / (3!2!) = 10
3C2 = (3!) / (2!) = 3
sum of all the possible events = 36 + 21 + 10 + 3 = 70
Total possible outcome of selecting the red ball = 10C3
10C3 = (10!) / (7!3!)
= 120
The probability that A selects the red ball is sum of all the possible events divided by the total possible outcome.
P( A selects the red ball) = 70 / 120
= 0.5833
= 58.33 %
