Answer:
(15, 12)
Step-by-step explanation:
Let's generate two systems of equations that fit this scenario.
Number of trips to the airport = x
Number of trips from the airport = y
Total number of trips to and from the airport = 27
Thus:
=> equation 1.
Total price for trips to the Airport = 14*x = 14x
Total price of trips from the airport = 7*y = 7y
Total collected for the day = $294
Thus:
=> equation 2.
Multiply equation 1 by 7, and multiply equation 2 by 1 to make both equations equivalent.
7 ×
1 ×
Thus:
=> equation 3
=> equation 4
Subtract equation 4 from equation 3
-7x = -105
Divide both sides by -7
x = 15
Substitute x = 15 in equation 1
Subtract both sides by 15
The ordered pair would be (15, 12)
Answer:
1 4/25 - 3/5, or 29/25 - 3/5, or
it doesn't really matter they all mean the same
Step-by-step explanation:
1 2/5m - 3/5(2/5m + 1)
multiply first
1 2/5m - 6/25m - 3/5
multiply 1 2/5 only so the denominator is 25 as we can't simplify 6/25, it's so we can add them together, we can't add 3/5 tho bc it's a different term so there's no need to multiply that.
1 10/25m - 6/25m - 3/5
add/subtract
1 4/25m - 3/5
Answer:
a) 0.25
b) 0.25
c) 0.0625
Step-by-step explanation:
The complete question is:
Do you remember when breakfast cereal companies placed prizes in boxes of cereal? Possibly you recall that when a certain prize or toy was particularly special to children, it increased their interest in trying to get that toy. How many boxes of cereal would a customer have to buy to get that toy? Companies used this strategy to sell their cereal.
One of these companies put one of the following toys in its cereal boxes: a block (B), a toy watch (W), a toy ring (R), and a toy airplane (A). A machine that placed the toy in the box was programmed to select a toy by drawing a random number of 1 to 4. If a 1 was selected, the block (or B) was placed in the box; if a 2 was selected, a watch (or W) was placed in the box; if a 3 was selected, a ring (or R) was placed in the box; and if a 4 was selected, an airplane (or A) was placed in the box. When this promotion was launched, young children were especially interested in getting the toy airplane.
What is the probability of getting an airplane in the first cereal box?
Since the machine randomly selects toys, each toy has the same probability of being obtained in a cereal box.
Then, the total outcomes are 4 and the probability of getting an airplane in the first cereal box is 0.25 (25%).
What is the probability of getting an airplane in the second cereal box?
Two independent events do not change the probability of occurrence of one event or another.
The probability of getting an airplane in the second cereal box is 0.25 (25%).
What is the probability of getting airplanes in both cereal boxes?
P(1°∩2°)= P(1°) × P(2°) =
P(1°∩2°)= 0.0625 = 6.25%