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adoni [48]
3 years ago
15

Find the area of a sector

Mathematics
1 answer:
Annette [7]3 years ago
3 0

Answer:

990 cm²

Step-by-step explanation:

Given ratio of radii = a : b , then

ratio of areas = a² : b²

Here

ratio of radii = 50 : 30 = 5 : 3 , then

ratio of areas = 5² : 3² = 25 : 9

let x be the area of sector DEF , then using proportion

\frac{25}{2750} = \frac{9}{x} ← \frac{ratio}{area}

cross- multiply

25x = 24750 ( divide both sides by 25 )

x = 990

Area of sector DEF = 990 cm²

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3 years ago
According to Masterfoods, the company that manufactures M&M's, 12% of peanut M&M's are brown, 15% are yellow, 12% are re
Luda [366]

Answer:

The questions asked are

If you randomly select 4 peanuts

1. Compute the probability that exactly three of the four M&M’s are brown

2. Compute the probability that two or three of the four M&M’s are brown.

3. Compute the probability that at most three of the four M&M’s are brown.

4. Compute the probability that at least three of the four M&M’s are brown.

Step-by-step explanation:

Given the following information

Brown=12%. P(B)=0.12

Yellow=15%. P(Y)=0.15

Red=12%. P(R), =0.12

Blue=23%. P(B) =0.23

Orange, =23%. P(O) =0.23

Green=15%. P(G)=0.15

Question 1.

They are independent events

If there are exactly three brown and the last is not brown

P(B n B n B n B')

P(B)×P(B)×P(B)×P(B')

0.12×0.12×0.12×(1-P(B))

0.001728×(1-0.12)

0.001728×0.88

0.00152.

0.152%

2. If two or three are brown

I.e we are going to two brown and two none brown or three brown and one not brown. (P(B)×P(B)×P(B')×P(B'))+ (P(B)×P(B)×P(B'))

(0.12×0.12×0.88×0.88)+(0.12×0.12×0.12×0.88)

0.0112+0.00152

0.0127

1.27%

3. At most three brown out of four then we are going to have

BBBB', BBB'B', BB'B'B', B'B'B'B'

These are the cases of at most three brown.

P(B)×P(B)×P(B)×P(B') + P(B)×P(B)×P(B')×P(B') + P(B)×P(B')×P(B')×PB')+ P(B')×P(B')×P(B')×P(B')=

0.12×0.12×0.12×0.88+ 0.12×0.12×0.88×0.88+ 0.12×0.88×0.88×0.88+ 0.88×0.88×0.88×0.88=0.694

0.694

69.4%

4. At least 3 brown out of four selection

I.e BBBB', BBBB

These are the two options

P(B)×P(B)×P(B)×P(B') + P(B)×P(B)×P(B)×P(B)=

0.12×0.12×0.12×0.88 + 0.12×0.12×0.12×0.12

0.001728

0.173%

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Answer:

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Step-by-step explanation:

21*.12=

2.52

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