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2 years ago

A circle inside of a square. The square has side lengths of 5 meters.

Mathematics

2 answers:

Veseljchak [2.6K]2 years ago

7 0

Answer:

the area is 25/4 π

Step-by-step explanation:

diameter is 5 m

area will be 25/4π

jolli1 [7]2 years ago

5 0

Answer:

17.9 m²

Step-by-step explanation:

The area of the deck is the difference between the area of the square and the area of the circular hot tub.

<h3>square area</h3>

The area of the square is given by the formula ...

A = s² . . . . where s is the side length

For a side length of 5 m, the area is ...

A = (5 m)² = 25 m²

<h3>circle area</h3>

The area of a circle is found from the formula ...

A = (π/4)d² . . . . where d is the diameter

For a diameter of 3 m, the area is ...

A = (π/4)(3 m)² = 9/4π m² ≈ 7.069 m²

The area of the deck is the difference between the square area and the circle area:

deck area = 25 m² -7.069 m² ≈ 17.9 m²

The area of the deck surrounding the hot tub is about 17.9 m².

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Apply each of the four counting/sampling methods (with replacement and with ordering, without replacement and with ordering, wit

dusya [7]

Answer:

Step-by-step explanation:

I will illustrate this solution with a unique birthday party situation between Jane (the celebrant) and her 10 friends.

In the said party , we will assume that Jane only has 5 chocolates to share among her friends

i. Assuming that the 5 chocolates are of the same type, if she doesn't want to give any friend more than one piece of chocolate, the situation here is said to be WITHOUT REPLACEMENT & UN-ORDERED

n = 10 and k = 5

Total number of possible combinations becomes,

$(\left {n} \atop {k}} \right. )=(\left {10} \atop {5}} \right. )=252$

ii. Assuming that the 5 pieces of chocolate are of the same type and she is willing to give a friend more than one piece of chocolate, the situation is said to be WITH REPLACEMENT & UN-ORDERED

n = 10 and k = 5

Total number of possible combinations becomes,

$(\left {n+k-1} \atop {k}} \right. )=(\left {14} \atop {5}} \right. )=2002$

iii. Assuming that the 5 pieces of chocolate are of different types and she isn't willing to give any friend more than one piece of chocolate, the situation is said to be WITHOUT REPLACEMENT & ORDERED

n = 10 & k = 5

Total number of possible combinations becomes,

$P\left {n} \atop {k}} \right=P\left {10} \atop {5}} \right. =30240$

iv. Assuming the the 5 pieces of chocolate are of different types, if she is willing to give a friend more than one piece of chocolate, the situation is said to be WITH REPLACEMENT & ORDERED

n = 10 AND k = 5

Total number of combinations becomes

$n^k=10^5=100,000$