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Fofino [41]
2 years ago
11

Wezi has sixty-three $5-bills to the bank to exchange them for $20-bills. How many $20-bills does Wezi get? Is there money left

over? explain
Mathematics
1 answer:
In-s [12.5K]2 years ago
5 0

Answer:

A. Wezi will get fifteen (15) $20-bills.

B. There will be $15 left.

Step-by-step explanation:

A. We have been given that Wezi has sixty-three $5 bills.

Let us find the total amount of money that Wezi has by multiplying 63 by 5.

\text{Total amount of money Wezi has}=63\times\$5

\text{Total amount of money Wezi has}=\$315

Therefore, Wezi has $315.

Since Wezi wants to exchange her total money of $5 bills with $20 bills, so let us divide total amount of money by 20.

\text{The number of \$20 bills}=\frac{\$315}{\$20}

\text{The number of \$20 bills}=15.75

Since we cannot 0.75 of a $20 bill, so Wezi will get 15 $20-bills.

We can see that 315 is not completely divisible by 20, so let us subtract greatest multiple of 20 from 315.

We can see 15\times 20=300, so upon subtracting 300 from 315 we will get,

\$315-\$300=\$15

Therefore, there will be $15 left.

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The rate of transmission in a telegraph cable is observed to be proportional to x2ln(1/x) where x is the ratio of the radius of
sergij07 [2.7K]

Answer:

The value of x that gives the maximum transmission is 1/√e ≅0.607

Step-by-step explanation:

Lets call f the rate function f. Note that f(x) = k * x^2ln(1/x), where k is a positive constant (this is because f is proportional to the other expression). In order to compute the maximum of f in (0,1), we derivate f, using the product rule.

f'(x) = k*((x^2)'*ln(1/x) + x^2*(ln(1/x)')) = k*(2x\,ln(1/x)+x^2*(\frac{1}{1/x}*(-\frac{1}{x^2})))\\= k * (2x \, ln(1/x)-x)

We need to equalize f' to 0

  • k*(2x ln(1/x) - x) = 0 -------- We send k dividing to the other side
  • 2x ln(1/x) - x = 0 -------- Now we take the x and move it to the other side
  • 2x ln(1/x) = x -- Now, we send 2x dividing (note that x>0, so we can divide)
  • ln(1/x) = x/2x = 1/2 -------  we send the natural logarithm as exp
  • 1/x = e^(1/2)
  • x = 1/e^(1/2) = 1/√e ≅ 0.607

Thus, the value of x that gives the maximum transmission is 1/√e.

7 0
2 years ago
For each exercise draw circle O with radius 12. Then draw radii OA and OB to form an angle with the measure named. Find the leng
guajiro [1.7K]
<span>You are asked to draw circle O with radius 12. Then draw radii OA and OB to form an angle with the measure named. You are asked to find the length of AB. The answers for each of the following measures are:
1)Measure of AOB=90, AB = 16.97 units
2)measure of AOB=180, AB = 24 units
3) measure of AOB=60, AB = 10.39 units
3) measure of AOB=120, AB = 10.39 units</span>
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2 years ago
A driver brings a car to a full stop in 2.0s. It’s acceleration is -11 m/s^2. How far did the car travel while braking
tensa zangetsu [6.8K]

Answer:

  22 m

Step-by-step explanation:

The distance traveled is ...

  d = (1/2)at^2

where "a" is the magnitude of the acceleration.

  d = (1/2)(11 m/s^2)(2 s)^2

  d = 22 m

The car traveled 22 meters while braking.

6 0
2 years ago
Read 2 more answers
Davison Electronics manufactures two LCD television monitors, identified as model A andmodel B. Each model has its lowest possib
shtirl [24]

The question is incomplete. The complete question is :

Davison Electronics manufactures two LCD television monitors, identified as model A and model B. Each model has its lowest possible production cost when produced on Davison’s new production line. However, the new production line does not have the capacity to handle the total production of both models. As a result, at least some of the production must be routed to a higher-cost, old production line. The following table shows the minimum production requirements for next month, the production line capacities in units per month, and the production cost per unit for each production line:

                      Production per unit                         Minimum production

Model        New Line        Old Line                              requirements

A                   $30                $50                                          50,000

B                   $25                $40                                           70,000

Production  80000          60000

line capacity.

Let

AN- Units of model A produced on the new production line

AO Units of model A produced on the old production line

BN Units of model B produced on the new production line

BO Units of model B produced on the old production line

Davison's objective is to determine the minimum cost production plan. The computer solution is shown in Figure 3.21.

a. Formulate the linear programming model for this problem using the following four constraints:

Constraint 1: Minimum production for model A

Constraint 2: Minimum production for model B

Constraint 3: Capacity of the new production line

Constraint 4: Capacity of the old production line  

Solution :

<u>Linear programming model</u>:

Linear programming is defined as a mathematical model where the linear function is either minimize or maximize when they are subjected to some constraints.

The linear programming model is determined as follows :

Minimum : 30 AN + 50 AO + 25 BN + 40 BO

This is subject to the constraints as :

AN+AO \geq 60,000

BN+BO \geq 70,000

AN+BN \leq 80,000

AO+BO \leq 60,000

Learn more :

https://brainly.in/question/15044395

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2 years ago
A car uses 1 liter of gas every 12 kilometers.
Alexxandr [17]
The answer is 36,000 because 1 km=1,000 m. Since 1 liter is used every 12 km and you are finding out how many m you will go, you change the km to m and you will multiply it by 3.

12,000x3=36,000 m
4 0
2 years ago
Read 2 more answers
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