three dozen muffins cost $9.00 determine the cost per muffin determine the cost per muffin.

Question

which process will you use to solve this problem?
a. 9÷12
b. 9÷3
c. 9÷(3x12)
d. 12÷3x9
e. 9÷3x12

Answer 1

To find the cost per muffin when three dozen muffins cost $9.00, you first need to determine the total number of muffins in three dozen.

Since one dozen equals 12 muffins, three dozen equals: \[ 3 \times 12 = 36 \text{ muffins} \]

Next, to find the cost per muffin, you need to divide the total cost by the total number of muffins: \[ \text{Cost per muffin} = \frac{\text{Total Cost}}{\text{Total Muffins}} = \frac{9.00}{36} \]

Now, let's simplify this: \[ 9.00 \div 36 = 0.25 \]

So, the cost per muffin is $0.25.

Now, looking at the options you provided:

• a. \( 9 \div 12 \) – doesn't give total muffins
• b. \( 9 \div 3 \) – doesn't give total muffins
• c. \( 9 \div (3 \times 12) \) – this is \( 9 \div 36 \), which is the correct process to find the cost per muffin
• d. \( 12 \div 3 \times 9 \) – doesn't correctly calculate the cost per muffin
• e. \( 9 \div 3 \times 12 \) – doesn’t calculate the cost per muffin

The correct answer is c. \( 9 \div (3 \times 12) \).

Answer 2

which three equations represent non-proportional relationships?
a. y=-15x
b. y=x2
c. y=2/3x
d. y=10x+10
e. y=5/x-9
(the 2 in (b.) is an exponent)

Answer 3

To determine which equations represent non-proportional relationships, we need to identify those that do not fit the form of \( y = kx \), where \( k \) is a constant. Proportional relationships pass through the origin (0,0), while non-proportional relationships do not.

Let's analyze each equation:

a. \( y = -15x \)
This is a linear equation in the form \( y = kx \) (with \( k = -15 \)). Proportional relationship.

b. \( y = x^2 \)
This is a quadratic equation. It does not have a constant ratio between \( y \) and \( x \), especially since \( y(0) = 0^2 = 0 \) but the relationship is not linear. Non-proportional relationship.

c. \( y = \frac{2}{3}x \)
This is also in the form \( y = kx \) (with \( k = \frac{2}{3} \)). Proportional relationship.

d. \( y = 10x + 10 \)
This is a linear equation but includes a constant term \( +10 \), which means it does not pass through the origin. Non-proportional relationship.

e. \( y = \frac{5}{x} - 9 \)
This is a rational equation, and it has a variable in the denominator. It does not represent a constant ratio between \( y \) and \( x \). Non-proportional relationship.

In summary, the three equations that represent non-proportional relationships are:

• b. \( y = x^2 \)
• d. \( y = 10x + 10 \)
• e. \( y = \frac{5}{x} - 9 \)

Answer 4

i forgot to write down
f y=4.08x

Answer 5

Let's reassess all the equations including the new equation f:

1. a. \( y = -15x \)
   This is in the form \( y = kx \) (with \( k = -15 \)). Proportional relationship.

2. b. \( y = x^2 \)
   This is a quadratic equation and does not represent a constant ratio between \( y \) and \( x \). Non-proportional relationship.

3. c. \( y = \frac{2}{3}x \)
   This is in the form \( y = kx \) (with \( k = \frac{2}{3} \)). Proportional relationship.

4. d. \( y = 10x + 10 \)
   This includes a constant term \( +10 \), meaning it does not pass through the origin. Non-proportional relationship.

5. e. \( y = \frac{5}{x} - 9 \)
   This equation has a variable in the denominator and does not maintain a constant ratio between \( y \) and \( x \). Non-proportional relationship.

6. f. \( y = 4.08x \)
   This is in the form \( y = kx \) (with \( k = 4.08 \)). Proportional relationship.

In summary, the three equations that represent non-proportional relationships are:

• b. \( y = x^2 \)
• d. \( y = 10x + 10 \)
• e. \( y = \frac{5}{x} - 9 \)

The proportional relationships are:

• a. \( y = -15x \)
• c. \( y = \frac{2}{3}x \)
• f. \( y = 4.08x \)

If you need anything else, feel free to ask!

Answer 6

consider the graph
( the graph shows (1,5), (3,15), and (6,30). )
what is the constant of proportionality (unit rate) ?
a. 1
b. 0
c.1/5
d. 5

Answer 7

To determine the constant of proportionality (or the unit rate) from the given points in the graph—(1, 5), (3, 15), and (6, 30)—we look for a consistent ratio of \( y \) to \( x \).

The constant of proportionality \( k \) can be calculated by dividing \( y \) by \( x \).

Let's calculate this for the given points:

1. For the point \( (1, 5) \): \[ k = \frac{y}{x} = \frac{5}{1} = 5 \]

2. For the point \( (3, 15) \): \[ k = \frac{y}{x} = \frac{15}{3} = 5 \]

3. For the point \( (6, 30) \): \[ k = \frac{y}{x} = \frac{30}{6} = 5 \]

Since the ratio \( k \) is consistently 5 for all points, the constant of proportionality is 5.

So the answer is: d. 5.

Answer 8

The amount of money that Sarah earns varies directly with the number of hours worked. If Sarah earns $320 for working 40 hours, determine the constant of proportionality. $__ per hour (remember to include the hundredths place, two digits after the decimal)

Answer 9

To determine the constant of proportionality (the unit rate) for Sarah's earnings, we can use the formula for the unit rate:

\[ k = \frac{\text{Total Earnings}}{\text{Total Hours Worked}} \]

From the information given:

• Total Earnings = $320
• Total Hours Worked = 40 hours

Now, we can calculate the constant of proportionality \( k \):

\[ k = \frac{320}{40} = 8 \]

This means that Sarah earns $8 per hour.

In terms of the question, we need to express this amount to two decimal places:

\[ k = 8.00 \]

So, the constant of proportionality is $8.00 per hour.

Answer 10

the table represents the time it took bill to complete sets of math problems which equation represents this relationship?
time(in minuets) number of problems in set
___________________________________
| 30 | 36 |
| 40 | 48 |
| 50 | 60 |
| 60 | 72 |
___________________________________
a. y=x+1.2
b. y=1.2x
c. y=x+12
d. y=12x

Answer 11

To find the equation that represents the relationship between the time (in minutes) and the number of problems in the set, let’s analyze the given data.

We have:

• When time is 30 minutes, problems = 36
• When time is 40 minutes, problems = 48
• When time is 50 minutes, problems = 60
• When time is 60 minutes, problems = 72

Let's observe the relationship between the time (x) and the number of problems (y).

1. Find the rate of problems per minute:

You can calculate the ratio of the number of problems to the time for each entry:

• For 30 minutes: \( \frac{36}{30} = 1.2 \)
• For 40 minutes: \( \frac{48}{40} = 1.2 \)
• For 50 minutes: \( \frac{60}{50} = 1.2 \)
• For 60 minutes: \( \frac{72}{60} = 1.2 \)

The ratio is consistently \( 1.2 \) problems per minute.

2. Write the equation:

The relationship can be represented as: \[ y = 1.2x \] Where:

• \( y \) is the number of problems,
• \( x \) is the time in minutes, and
• \( 1.2 \) is the number of problems completed per minute.

So, the equation that represents this relationship is: b. \( y = 1.2x \).