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ivann1987 [24]
3 years ago
5

Help me yall- its due tn but I'm going somewhere for the weekend.

Mathematics
1 answer:
Vanyuwa [196]3 years ago
6 0

Answer:

201.06

Step-by-step explanation:

Diameter: 12

12+4+4=20

R:10

Pi(10)^2

= 314.16

Subtract that from small circle

Pi(6)^2

=113.097

314.16-113.097

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Simplify -5(4x-3) please help?
Bond [772]

Answer:

-20x+15

Step-by-step explanation:

You use foil method so you distribute it and it becomes -5*4x amd -5*-3 so it would be -20x+15

8 0
3 years ago
Read 2 more answers
A football team carried out a report to see the impact of stretching on preventing injury. Of the 50 footballers in the squad 32
DENIUS [597]

Answer:

15 / 18

Step-by-step explanation:

19-4= 15

50-32=18

So 15/18 is fraction for injured players that didn't stretch

4 0
3 years ago
(a) Consider a class with 30 students. Compute the probability that at least two of them have their birthdays on the same day. (
Galina-37 [17]

Answer:

a.) 0.7063

b.) 23

Step-by-step explanation:

a.)

Let X be an event in which at least 2 students have same birthday

     Y be an event in which no student have same birthday.

Now,

P(X) + P(Y) = 1

⇒P(X) = 1 - P(Y)

as we know that,

Probability of no one has birthday on same day = P(Y)

⇒P(Y) = \frac{365!}{(365)^{n} (365-n)! }      where there are n people in a group

As given,

n = 30

⇒P(Y) = \frac{365!}{(365)^{30} (365-30)! } = \frac{365!}{(365)^{30} (335)! } = 0.2937

∴ we get

P(X) = 1 - 0.2937 = 0.7063

So,

The probability that at least two of them have their birthdays on the same day  =  0.7063

b.)

Given, P(X) > 0.5

As

P(X) + P(Y) = 1

⇒P(Y) ≤ 0.5

As

P(Y) = \frac{365!}{(365)^{n} (365-n)! }

We use hit and trial method

If n = 1 , then

P(Y) = \frac{365!}{(365)^{1} (365-1)! } = \frac{365!}{(365)^{1} (364)! }  = 1 \nleq 0.5

If n = 5 , then

P(Y) = \frac{365!}{(365)^{5} (365-5)! } = \frac{365!}{(365)^{5} (360)! }  = 0.97 \nleq 0.5

If n = 10 , then

P(Y) = \frac{365!}{(365)^{10} (365-10)! } = \frac{365!}{(365)^{10} (354)! }  = 0.88 \nleq 0.5

If n = 15 , then

P(Y) = \frac{365!}{(365)^{15} (365-15)! } = \frac{365!}{(365)^{15} (350)! }  = 0.75 \nleq 0.5

If n = 20 , then

P(Y) = \frac{365!}{(365)^{20} (365-20)! } = \frac{365!}{(365)^{20} (345)! }  = 0.588 \nleq 0.5

If n = 22 , then

P(Y) = \frac{365!}{(365)^{22} (365-22)! } = \frac{365!}{(365)^{22} (343)! }  = 0.52 \nleq 0.5

If n = 23 , then

P(Y) = \frac{365!}{(365)^{23} (365-23)! } = \frac{365!}{(365)^{23} (342)! }  = 0.49 \nleq 0.5

∴ we get

Number of students should be in class in order to have this probability above 0.5 = 23

5 0
3 years ago
 Create a single polynomial of the factored form that meets the following criteria:
Dmitry_Shevchenko [17]

Answer:

ummmm m m

Step-by-step explanation:

,I think the answer would be 10

5 0
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Gala2k [10]

Answer: since I don’t understand this question I’m choosing D

Step-by-step explanation:

8 0
3 years ago
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