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hammer [34]
3 years ago
5

The measure of central angle XYZ is StartFraction 3 pi Over 4 EndFraction radians. What is the area of the shaded sector? 32Pi u

nits squared 85 96Pi units squared 256Pi units squared
Mathematics
2 answers:
Tems11 [23]3 years ago
5 0

Answer:

<em>96π units²</em>

Step-by-step explanation:

Find the diagram attached

Area of a sector is expressed as;

Area of a sector = θ/2π * πr²

Given

θ = 3π/4

r = 16

Substitute into the formula

area of the sector = (3π/4)/2π * π(16)²

area of the sector = 3π/8π * 256π

area of the sector = 3/8 * 256π

area of the sector = 3 * 32π

<em>area of the sector =96π units²</em>

xxTIMURxx [149]3 years ago
5 0

Answer:

96π units²

Step-by-step explanation:

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The area of right angled triangle with its legs are 6cm and 8cm is equal to​
aalyn [17]

Answer:

A = 24 cm^2

Step-by-step explanation:

The area of a triangle is

A = 1/2 bh

A = 1/2 (6)*8

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4 0
3 years ago
What is the probability that a group of five friends all have different birthdays? Show all of your work for full credit.
patriot [66]

Answer: 0.98630136986

Step-by-step explanation:

There are 365 possible birthdays. The key to assigning the probability is to think in terms of complements: “Two (or more) people share a birthday” is the complement of “All people in the group have different birthdays.” Each probability is 1 minus the other. What is the probability that any two people have different birthdays? The first person could have any birthday (p = 365÷365 = 1), and the second person could then have any of the other 364 birthdays (p = 364÷365). Multiply those two and you have about 0.9973 as the probability that any two people have different birthdays, or 1−0.9973 = 0.0027 as the probability that they have the same birthday. If you have a group of five, it would mean your equation would have to be (p=360÷365)

8 0
2 years ago
Q5: The probability that event A occurs is 5/7 and the probability that event B occurs is 2/3 . If A and B are independent event
bekas [8.4K]

Answer:

P(A \cap B)

And we can use the following formula:

P(A \cap B)= P(A)* P(B)

And replacing the info we got:

P(A \cap B) = \frac{5}{7} \frac{2}{3}= \frac{10}{21}=0.476

Step-by-step explanation:

We define two events for this case A and B. And we know the probability for each individual event given by the problem:

p(A) = \frac{5}{7}

p(B) = \frac{2}{3}

And we want to find the probability that A and B both occurs if A and B are independent events, who menas the following conditions:

P(A|B) = P(A)

P(B|A) = P(B)

And for this special case we want to find this probability:

P(A \cap B)

And we can use the following formula:

P(A \cap B)= P(A)* P(B)

And replacing the info we got:

P(A \cap B) = \frac{5}{7} \frac{2}{3}= \frac{10}{21}=0.476

7 0
3 years ago
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