Cherries cost $4/lb. grapes cost $2.50/lb. you can spend no more than $15 on fruit, and you need at least 5 lb in all. create a graph showing the amount of each fruit you can buy.

The quantity of cherries bought is 1.67 lb

The quantity of grapes bought is 3.33 lb

Step-by-step explanation:

Given as :

The cost of cherries = $4 per lb

The cost of grapes = $2.50 per lb

Total money spend on fruits = $15

The quantity of both fruits to bought = 5  lb

Let The quantity of cherries bought = c  lb

Let The quantity of grapes bought = g    lb

Now, According to question

quantity of both fruits to bought = quantity of cherries bought + quantity of grapes bought

i.e c + g = 5 lb              ........1

And

Total money spend on fruits = cost of cherries × quantity of cherries bought + cost of grapes × quantity of grapes bought

Or , c   lb ×  $4 per lb +  g  lb  ×  $2.50 per lb = $15

Or, 4 c + 2.50 g = 15           .......2

Now, Solving equation 1 and 2

So, (4 c + 2.50 g) - 2.50 × (c + g) = 15 - 5 × 2.50

Or, (4 c - 2.50 c) + (2.50 g - 2.50 g) = 15 - 12.50

Or, 1.5 c + 0 = 2.5

∴ c =

I.e c = 1.67  lb

So, The quantity of cherries bought = c  = 1.67 lb

Putting the value of c in eq 1

Since , c + g = 5  lb

Or, g = 5  lb - c

Or, g = 5  lb - 1.67  lb

i.e  g = 3.33  lb

So, The quantity of grapes bought = g  = 3.33 lb

Hence, The quantity of cherries bought is 1.67 lb

and The quantity of grapes bought is 3.33 lb   Answer

the answer to the question is a

step-by-step explanation:

it has the same slope and the same y-intercept.

y + 11 = -7/9(x - 18)

y + 11 =-7/9x + 14

b

y = -7/9x + 3

y-intercept (0, 3)

the data of the detroit pistons in numerical order from least to greatest is 0.30, 0.37, 0.55, 0.62, 0.69, 0.72, 0.73, 0.75, 0.77, 0.78, 0.80, 0.81, 0.81, 0.83, 0.84, 0.89

the data of the miami heat in numerical order from least to greatest is 0.50, 0.50, 0.56, 0.57, 0.67, 0.72, 0.72, 0.72, 0.75, 0.75, 0.76, 0.81, 0.82, 0.90, 1.00, 1.00

the detroit pistons median : 1.52

the miami heat median: 1.47

the iqr of this data is 0.05