A fruit stand sells bags of apples and tangerines. • Each bag has 3 fewer apples than tangerines. 2 • In each bag, the number of apples is the number of tangerines. Let A represent the number of apples in each bag and T represent the number of tangerines in each bag. Which system of equations can be used to solve for A and T?​

The system we need to use is:

A = T - 3

A = (2/3)*T

Solving this, we get:

In a bag, there are 9 tangerines and 6 apples.

Step-by-step explanation:

The question seems to be incomplete, by a quick online search i've found the complete question:

"A fruit stand sells bags of apples and tangerines. • Each bag has 3 fewer apples than tangerines. 2 • In each bag, the number of apples is 2/3 the number of tangerines..."

We know that:

A = number of apples in one bag

T = number of tangerines in one bag

We also know that:

"Each bag has 3 fewer apples than tangerines"

This can be written as:

A = T - 3

"In each bag, the number of apples is 2/3 the number of tangerines"

This can be written as:

A = (2/3)*T

Then we have a system of equations:

A = T - 3

A = (2/3)*T

This is the system of equations that can be used to solve for A and T, now let's solve this.

Because A means the same thing in both equations, then we can write:

A = T - 3 = A = (2/3)*T

Then:

T - 3 = (2/3)*T

We can solve this for T.

T - 3 = (2/3)*T

T - (2/3)*T = 3

(1 - 2/3)*T = 3

(1/3)*T = 3

T = 3/(1/3) = 3*3 = 9

This means that in each bag we have 9 tangerines in a bag.

Now we can use any of the two equations in the system to find the value of A.

I will use the first one:

A = T - 3

Now let's replace T by 9, then:

A = 9 - 3 = 6

A = 6

So there are 6 apples in a bag.

wholesale price is the initial price before markup. retail price comes after the markup.

a price of 8 bucks that's marked up by 100%   is

formula: initial price + (percent markup in fraction/decimal)(initial price)

8+(1)(8)

8+8

16

retail price is $16.

you just switch the numbers basically so it would be 3/17