Answer:
Step-by-step explanation:
Jill makes purses and backpacks.
Let x represent the number of foot of fabric required to make a purse.
Let y represent the number of foot of fabric required to make a backpack.
To make each purse,she uses 1 foot less than half the amount of fabric she uses to make a backpack. This means that
x = (y/2) - 1
Therefore, the amount of fabric that Jill needs to make backpack will be
y/2 = x-1
y = 2(x-1) = 2x - 2
The number of feet of fabric that she will use to make a purse is
x = (y/2) - 1
Answer: a) -$0.19, b) -$111.72 .
Step-by-step explanation:
Since we have given that
Number of free throws = 434
Number of throws made by them = 390
Amount for making the next 2 free throws = $40
Amount otherwise he has to pay = $169
a) Find the expected value of the proposition.
Expected value of success in next 2 free throws = 
Expected value would be

b) If you played this game 588 times how much would you expect to win or lose?
Number of times they played the game = 588
So, Expected value would be

Hence, a) -$0.19, b) -$111.72
Let's call the two numbers
and
.
Given these variables, we can say:
, based on the first sentence in the problem.
Also, remember that the reciprocal of a number is simply 1 divided by the number. Thus, we can say that:

To solve, we can simply substitute
in for
in the second equation and solve.


- Get terms on the left side to a common denominator for easier addition


- Cross multiplication (
)


- Subtract
from both sides of the equation

- Factor left side of the equation

Now, notice that we have found two solutions, but the problem is only asking for one. This <em>likely </em>means that one of our solutions is extraneous. Let's take a look. Remember that the smaller positive number is equal to 14 less than the larger number. However,
,
Since
is not positive in this case,
is not a solution.
Thus,
is our only solution. In this case,
,
which means that the smaller number is 14 and the larger number is 28.
Check the picture below.
make sure your calculator is in Degree mode.