)a confectioner sells two types of nut mixtures. the standard-mixture package contains

100 g of cashews and 200 g of peanuts and sells for $1.95. the deluxe-mixture package

contains 150 g of cashews and 50 g of peanuts and sells for $2.25. the confectioner has

15 kg of cashews and 20 kg of peanuts available. based on past sales, he needs to have

at least as many standard as deluxe packages available. how many bags of each mixture

should he package to maximize revenue? co

total of 15kg will be used each for cashew and peanut to produced 60bags each of standard package and deluxe package, to achieve maximum revenue

Step-by-step explanation:

types A = the standard-mixture package contains  

we need to find the percentage of cashew & nut inside each package

100 g of cashews =100/15000 =0.66%

200 g of peanuts =200/20000 = 1%

total gran is 300g

sells for $1.95.

types B = The deluxe-mixture package    contains

150 g of cashews=150/15000 = 1%

50 g of peanuts = 50/20000 = 0.25%

total gran is 200g

sells for $2.25.

therefore the total percentage of cashew needed for the two package = 0.66+1= 1.66% = 250g of cashew in a package of cashew and nt

the total percentage of nut needed for the two package = 0.25+1= 1.25% = 250g of cashew in a package of cashew and nut

The confectioner has

 15 kg of cashews = 15000/250g = 60bags

20 kg of peanuts

total Kg of nut is 35kg, for a standard and deluxe package its 500g.

therefore almost 70bags in total should be achieved but the cashew nut wont be enough to make 70bags, so, will go for 60bags to achieve revenue

for standard package = 300g * 60bags =18000g

for deluxe package = 200g * 60bags =12000g

for each package

standard =

cashew 100g * 60=6kg

nut = 200*60 =12kg

deluxe =

cashew 150g * 60=9kg

nut = 50*60 =3kg

total of 15kg will be used each for cashew and peanut to produced 60bags each of standard package and deluxe package, to achieve maximum revenue

(0,-3) see attached graph

step-by-step explanation:

a. to solve the system, graph the two equations.

x = y+3 becomes y=x-3 has m=1/1 and b=-3. start at (0,-3) on the y-axis and plot a point. then move up 1 unit and 1 unit to the right. plot a point and connect.

y=-4x+3 has m =-4/1 and b=-3. start at (0,-3) on the y-axis and plot a point. then move down four units and to the right 1 unit. plot a point and connect.

the solution is the point (x,y) where the two lines intersect. by the graph below it is (0,-3).

b. to solve by substitution, substitute y = -4x-3 into x= y+3

x= (-4x-3) +3

x=-4x

5x=0

x=0

now substitute x = 0 back in to find y.

y= -4(0)-3

y=0-3

y=-3

solution is (0,-3)

c. use elimination by adding the equations together.

x-y=3

4x+y=-3

5x=0

x=0

then substitute into one equation to find y.

y=-4(0)-3

y=-3

54 km / hours

step-by-step explanation:

we need to convert seconds to hours

60 seconds = 1 minutes

60 minutes = 1hour

15 m         60 second   60 minutes

  * *         = 54000 meters/ hour

second     1 minute             1 hour

now we need to convert meters to km

1000 m = 1 km

54000 m       1 km

*   = 54 km / hours

hour               1000m