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borishaifa [10]

3 years ago

14

Write a real world situation for the following equations:

1 answer:

Jet001 [13]3 years ago

3 0

2. Snap gym costs $120 for registration fee and $25 each month. Awesome gym costs $45 each month with no registration fee. After how many months will the cost of the two gyms be equal ? 3. Awesome video game has a total of 100 points. If you are out, you miss 6 points. Cool video game has a total of 160 points. If you are out, you miss 10 points. After how many times of getting out will you have the same number of points on both the games ?

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Which statement describes how El Niño contributes to climate trends on Earth?

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I think it is C but im not real sure

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3 years ago

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The following equation will have two solutions, y = x2 - 2x - 15?<br> True<br> False

motikmotik

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3 years ago

Hamburger Hut sells regular hamburgers as well as a larger burger. Either type can include cheese, relish, lettuce, tomato, must

Studentka2010 [4]

Answer:

a) 40 different hamburgers can be ordered with exactly three extras

b) 20 different regular hamburgers can be ordered with exactly three extras

c) 7 different regular hamburgers can be ordered with at least five extras

Step-by-step explanation:

The order in which the extras are ordered is not important. So we use the combinations formula to solve this question.

Combinations formula:

$C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

In this problem:

2 options of hamburger(regular or larger)

6 options of extras(cheese, relish, lettuce, tomato, mustard, or catsup.).

(a) How many different hamburgers can be ordered with exactly three extras?

1 hamburger type, from a set of 2.

3 extras, from a set of 6. So

$C_{2,1}*C_{6,3} = \frac{2!}{1!(2-1)!}*\frac{6!}{3!(6-3)!} = 2*20 = 40$

40 different hamburgers can be ordered with exactly three extras

(b) How many different regular hamburgers can be ordered with exactly three extras?

3 extras, from a set of 6. So

$C_{6,3} = \frac{6!}{3!(6-3)!} = 20$

20 different regular hamburgers can be ordered with exactly three extras

(c) How many different regular hamburgers can be ordered with at least five extras?

Five extras:

5 extras, from a set of 6. So

$C_{6,5} = \frac{6!}{5!(6-5)!} = 6$

Six extras:

6 extras, from a set of 6. So

$C_{6,6} = \frac{6!}{6!(6-6)!} = 1$

6 + 1 = 7

7 different regular hamburgers can be ordered with at least five extras

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3 years ago

Watch one piece

viktelen [127]

Is one piece good? and how long is it?

3 0

3 years ago

Hi there! Can someone help with this?

Sergeu [11.5K]

4 dozen because if you add all dozens and subtract
Or 48 cookies

4 0

3 years ago