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3 years ago

In how many ways can five children posing for a photograph line up in a row?

Mathematics

1 answer:

Shtirlitz [24]3 years ago

5 0

Answer:

number of ways = 120

Step-by-step explanation:

The number of ways five children can pose for a photograph line up in a row is given by the number of permutations of 5 elements in 5 different positions (positions in the line), then

number of ways = number of permutations of 5 elements = 5! = 5 * 4 * 3 * 2 * 1 = 120

Since the first children that occupies the line can be on any of the positions (5 positions) , but then the second one can choose any of the 4 remaining positions (since the first children had already occupied one) , the third can choose 3 ... and so on.

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Determine which transformation took place from f(x) to g(x).

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3 years ago

The square root of 36 is what

Alekssandra [29.7K]

Answer:

Step-by-step explanation:

The square root of 36 is 6!

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3 years ago

Read 2 more answers

If the ice shop has only enough cocoa to make 4 gallons of chocolate ice cream, how many gallons of strawberry ice cream can the

tankabanditka [31]

The complete questions is:

The ice cream shop needs 5 pounds of strawberries for every gallon of strawberry ice cream. the shop chooses to produce 4 gallons of chocolate ice cream. how many pounds of strawberries should the shop purchase?

Pounds of strawberry ice cream needed to be bought by the shop exists 20 pounds.

<h3>How many gallons of strawberry ice cream can the shop efficiently produce?</h3>

The ice cream shop requires 5 pounds of strawberries for every gallon of strawberry ice cream and the shop chooses to make 4 gallons of ice cream,

Quantity of chocolate ice cream created in the shop = 4 gallons

Quantity of strawberries needed for each gallon of ice cream = 5 pounds

Pounds of strawberries needed to be bought by the shop = Quantity of chocolate ice cream created by the shop $*$ Quantity of strawberries needed for each gallon of ice cream

$$= 4 * 5$

= 20 pounds.

Pounds of strawberries needed to be bought by the shop exists 20 pounds.

To learn more about the word problem refer to:

brainly.com/question/12333642

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2 years ago

3ax-n/5=-4 solve for x

ollegr [7]

The solution for x in the equation 3ax - n/5 = -4 is x = -4/3a + n/15a.

<h3>What is the solution for x in the given equation?</h3>

3ax - n/5 = -4

To solve for x, isolate the term that contains x.

3ax - n/5 = -4

3ax = -4 + n/5

Divide each term by the coefficient of x and simplify

3ax/3a = -4/3a + (n/5)/3a

x = -4/3a + (n/5)/3a

multiply (n/5) by the reciprocal of 3a

x = -4/3a + (n/5) × 1/3a

x = -4/3a + n/15a

Therefore, the solution for x in the equation 3ax - n/5 = -4 is x = -4/3a + n/15a.

Learn more about equations here: brainly.com/question/14686792

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2 years ago

You have been invited to be part of the planning committee for the County Fair. In order to be an effective member you will need

kakasveta [241]

Firstly let's find the dimension of this large rectangle:(given)

Area of Rectangle = 660 x 66 =43,560 ft²

And we know that 1 acre = 43,560 ft², then each rectangle has an area of 1 acre & the 20 acres will correspond to 20 x 43560 = 871,200 ft²

We know that the 20 acres form a rectangle. We need to know what is their disposition:

1) We would like to know the layout of the rectangles since we have 4 possibilities FOR THE LAYOUTS

Note that W=66 & L=666 = 43,956 ft²/ unit )

lay out shape could be either:(in ft)
1 W by 20 L (Final shape Linear 66 x 13320 = 879,120) or
2 W by 10 L (Final shape Stacked 132 x 6660 = 879,120) or
4 W by 5 L (Final shape Stacked 264 x 3330 = 879,120) or

2) We would like to know the number of participants so that to allocate equal space as well as the pedestrian lane, if possible, if not we will calculated the reserved space allocated for pedestrian/visitors)

3) Depending on the shape given we will calculate the visitor space & we will deduct it from the total space to distribute the remaining among the exhibitors.

4) (SUGGESTION) Assuming it's linear, we will reserve
20ft x 13320 ft = = 266,400 ft² and the remaining 612,720 ft² for exhibitors
5) Depending on the kind of the exhibition, we will divide the 612,720 ft² accordingly
6) How can we select the space allocated for each exhibitor:
the 617,720 ft² could be written as a product of prime factors:
612720 = 2⁴ x 3² x 5 x 23 x 37
If you chose each space will be185 ft² , then we can accommodate up to 3,312 exhibitors.
Obviously you can choose any multiple of the prime factors to specify the area allocated & to calculate the number of exhibitors accordingly

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3 years ago