Michel wants to buy his mother handmade chocolates for Mothers Day. He has $15.oo to spend. The price of each chocolate is given below.
Fudge $0.60 NutClusters $0.75 Truffles $0.80
a)create and solve an equation for each type of chocolate to determine how many Michel can buy with $15.00, if he buys one type.
B) Michel decides to buy all 3 types, instead of choosing just one.
a)If Michel buys the same number of all three types, how many can he buy and stay within his budget?
b)Exactly how much will Michel spend?
c)Write an algebraic expression that represents the total amount Michel will spend. Then substitute the variable ad evaluate your expression. How does this amount compare with your amount in part(b)?
.6f = 15
f = 25
.75n = 15
n = 20
.8t = 15
t = 18.75, or just 18 whole truffles
If he buys n of each,
(.6 + .75 + .8)n <= 15
2.15n <= 15
n <= 6.97
so, he can only buy 6 of each
buying 6, he will spend 6*2.15 = 12.90
Thank you : )
How do you do letter c
a) To determine how many chocolates of each type Michel can buy with $15.00, we need to divide the available budget by the price of each chocolate.
For the Fudge:
Let x be the number of Fudge chocolates Michel can buy.
The cost of one Fudge chocolate is $0.60.
Therefore, the equation can be set up as: 0.60x = 15.00
Solving the equation, we can find the value of x:
x = 15.00 / 0.60
x = 25
So, Michel can buy 25 Fudge chocolates with his $15.00 budget.
Similarly, we can set up equations for the other types of chocolates:
For NutClusters:
Let y be the number of NutClusters Michel can buy.
The cost of one NutClusters chocolate is $0.75.
Therefore, the equation can be set up as: 0.75y = 15.00
Solving the equation, we can find the value of y:
y = 15.00 / 0.75
y = 20
So, Michel can buy 20 NutClusters chocolates with his $15.00 budget.
For Truffles:
Let z be the number of Truffles Michel can buy.
The cost of one Truffles chocolate is $0.80.
Therefore, the equation can be set up as: 0.80z = 15.00
Solving the equation, we can find the value of z:
z = 15.00 / 0.80
z = 18.75
Since Michel cannot buy a fraction of a chocolate, we will round down z to the nearest whole number.
So, Michel can buy 18 Truffles chocolates with his $15.00 budget.
b) If Michel decides to buy the same number of all three types of chocolates, we need to determine the maximum number of chocolates of each type Michel can buy while staying within his budget.
Let n be the number of chocolates of each type Michel can buy.
Using the price of each chocolate, we can set up the following equations:
0.60n + 0.75n + 0.80n = 15.00
2.15n = 15.00
Dividing both sides of the equation by 2.15, we can find the value of n:
n = 15.00 / 2.15
n ≈ 7
Therefore, Michel can buy 7 chocolates of each type and stay within his budget.
c) To determine how much Michel will spend in total, we can multiply the price of each chocolate by the number of chocolates he buys.
Total amount spent = (0.60 + 0.75 + 0.80) * n
Total amount spent = 2.15 * n
Total amount spent = 2.15 * 7
Total amount spent ≈ $15.05
The total amount Michel will spend is approximately $15.05.
Comparing to part (b), we can see that the amounts are slightly different due to rounding when dividing the budget equally among the three types of chocolates.
why did you put (n) is it considered as an (x) value
If he buys n of each,
(.6 + .75 + .8)n <=
also, how did you get 6.97
did you subtract a number
Michel’s final choice costs $14.80. It is a combination of all three types of chocolates.
Determine four possible combinations of chocolates he could have selected.
Create and evaluate an algebraic expression to show that each of your combinations works.
1fudge 4 nuts 13 truffles
2 fudge, 16 nuts, 1 truffles
3 fudge, 12 nuts, 4 truffles
4 fudge, 8 nuts, 7 truffles
Create and evaluate an algebraic expression to show that each of your combinations works.