<u>Answer:
</u>
Expression x + 2my + z represents cost of order where x, y, z are cost of small , medium and large drinks (in dollars) respectively.
<u>Solution:
</u>
Given that
Juan’s family ordered a small drink and m medium drinks.
Alex family ordered m medium drinks and a large drink.
Need to write an algebraic expression which shows total cost of both order in dollars.
Let’s assume cost of one small drink = x
And assume cost of one medium drink = y
And assume cost of one large drink = z
So now cost of order of Juan’s family is equal to cost of 1 small drink + cost of m medium drinks = 1 
x + m y
= x + my
And cost of order of Alex family is equal to cost of m medium drinks + cost of one large drink
= m x y + 1 x z
=my + z
So total cost of both order in dollars = x + my + my + z = x + 2my + z
Hence expression x + 2my + z represents cost of order where x , y , z are cost of small , medium and large drinks (in dollars) respectively.
Answer:
r = 6
Step-by-step explanation:
Hello!
-8 - 2(7r + 1) = -94
Add 8 to both sides
-2(7r + 1) = -86
Divide both sides by -2
7r + 1 = 43
Subtract 1 from both sides
7r = 42
Divide both sides by 7
r = 6
Hope this Helps
Answer:
According to my calculations, 1.6 goes into 10.8 a total of 6 times with a remainder of 1.1999999999999993. If you continue the long division beyond the decimal point, the result would be 6.75.
Step-by-step explanation:
First, isolate the variable
X+0=75-5
X+0=70
X=70
Hope I helped!
Good Luck!
Answer:
probability that contractor 1,2 and 3 win is 33%,50% and 17% respectively
Step-by-step explanation:
assuming that there are no other contractors then:
probability that 1 , 2 or 3 win = 1
denoting as X= probability that contractor 3 wins , then assuming that only one wins , we have
probability that 1 , 2 or 3 win = 1
probability that contractor 1 wins + probability that contractor 2 wins + probability that contractor 3 wins = 1
2*P(X) + 3*P(X) + P(X) = 1
6*P(X) = 1
P(X) =1/6
then
-probability that contractor 1 wins = 2/6 (33%)
-probability that contractor 3 wins = 3/6 (50%)
-probability that contractor 3 wins = 1/6 (17%)
