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Elden [556K]
3 years ago
5

The lodhl diner offers a meal combination consisting of an appetizer, a soup, a main course, and a dessert. There are five appet

izers, five soups, four main courses, and five desserts. Your diet restricts you to choosing between a dessert and an appetizer. (You cannot have both.) Given this restriction, how many three-course meals are possible?
Mathematics
1 answer:
kenny6666 [7]3 years ago
6 0

Answer:   100

Step-by-step explanation:

Given : The lodhl diner offers a meal combination consisting of an appetizer, a soup, a main course, and a dessert.

There are 5 appetizers, 5 soups, 4 main courses, and 5 desserts.

Also, a dessert and a appetizer are not allowed to take together.

By Fundamental counting principal ,

Number of three-course meals with dessert and without appetizer :

5\times4\times5=100      (1)

Number of three-course meals with appetizer and without dessert :

5\times5\times4=100      (2)

Now, the number of meals with either dessert or appetizer :-

100+100=200           [Add (1) and (2)]

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Shopping at Savers Mart, Lisa buys her children 4 shirts and 3 pairs of pants for $85.50. She returns the next day and buys 3 sh
maria [59]

Answer:

price for one shirt

=

$

7.50

price for one pair of pants

=

$

18.50

Explanation:

Start by letting variables

x

and

y

represent the pieces of clothing from the problem.

Let

x

be the price of one shirt.

Let

y

be the price of one pair of pants.

Equation

1

:

4

x

+

3

y

=

85.50

Equation

2

:

3

x

+

5

y

=

115.00

You can solve for each variable by using elimination or substitution. However, in this case, we will use use elimination. First, we will solve for

y

, the price of each pair of pants.

To isolate for

y

, we must eliminate

x

. We can do this by making the two equations have the same

x

values. First, we find the LCM of

4

and

3

, which is

12

. Next, multiply equation

1

by

3

and equation

2

by

4

so that

4

x

and

3

x

becomes

12

x

in both equations.

Equation

1

:

4

x

+

3

y

=

85.50

3

(

4

x

+

3

y

)

=

3

(

85.50

)

12

x

+

9

y

=

256.50

Equation

2

:

3

x

+

5

y

=

115.00

4

(

3

x

+

5

y

)

=

4

(

115.00

)

12

x

+

20

y

=

460.00

Now that we have two equations with

12

x

, we can subtract equation

2

from equation

1

to solve for

y

.

12

x

+

9

y

=

256.50

12

x

+

20

y

=

460.00

−

11

y

=

−

203.50

y

=

18.50

⇒

price for one pair of pants

Now that we know that a pair of pants is

$

18.50

, we can substitute this value into either equation

1

or

2

to find price for one shirt. In this case, we will choose equation

1

.

4

x

+

3

y

=

85.50

4

x

+

3

(

18.50

)

=

85.50

4

x

+

55.5

=

85.50

4

x

=

28

x

=

7.50

⇒

price for one shirt

∴

, the price for one shirt is

$

7.50

and the price for one pair of pants is

$

18.50

.

Step-by-step explanation:

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