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Nonamiya [84]
3 years ago
10

Evaluate the series 1 + 0.1 + 0.01 + . . . 21.111.120.33

Mathematics
2 answers:
Sonbull [250]3 years ago
4 0
The correct answer is 1.11
AURORKA [14]3 years ago
3 0
The answer to this is 1.11

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1944 mm Please mark as brainliest if im correct
3 0
3 years ago
Please help I'm really stuck
Anit [1.1K]
First factor the numerator
b^2-6b+8/b-4  * b+8/b-2

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7 0
3 years ago
The sum of the lengths of two opposite sides of the circumscribed quadrilateral is 12 cm, the length of a radius of the circle i
nasty-shy [4]

Answer:

60cm^2

Step-by-step explanation:

We assume that is a circumscribing quadrilateral, rather than one that is circumscribed. It is also called a "tangential quadrilateral" and its area is ...

 K = sr

where s is the semi-perimeter, the sum of opposite sides, and r is the radius of the incircle.

 K = (12 cm) (5cm) = 60 cm²

_____

A quadrilateral can only be tangential if pairs of opposite sides add to the same length. Hence the given sum is the semiperimeter.

8 0
3 years ago
HELP ASAP! WILL MARK BRAINLIEST!!​
malfutka [58]

Answer:

r = 6

Step-by-step explanation:

V(cyl) = π · r² · h

144π =  π · r² · 4

144 = r² · 4

<u>144</u> = r²

 4

r = \sqrt{144} ÷ \sqrt{4}

r = 12/2

r = 6

5 0
2 years ago
Susie has a bag of marbles containing 3 Red, 7 Green, and 10 Blue marbles.
morpeh [17]

Answer:

1) a) 0.0945

b) 0.1062

2) a) 0.0138

b) 0.00081

3) a) 0.00094

b) 0.00083

4) 301.68 cents

5) 0.0056

Step-by-step explanation:

Since there are 3 Red, 7 Green, and 10 Blue marbles.

Total number of marbles N = 20

Probability (Red) = 3/20

Probability (Green) = 7/20

Probability (Blue) = 10/20

1) the probability of picking 5 marbles and getting at least one red marble

A) with replacement

P(at least 1 red of 5)= (3/20 * 17/20 * 17/20 * 17/20 * 17/20) + (3/20 * 3/20 * 17/20 * 17/20 * 17/20) + (3/20 * 3/20 * 3/20 * 17/20 * 17/20)

P(at least 1 red of 5) = 0.0783 +0.0138 + 0.0024 = 0.0945

B) without replacement

P(at least 1 red of 5) = ( 3/20 * 17/19* 16/18 * 15/17 * 14/16) + (3/20 * 2/19 * 17/18 * 16/17 * 15/16) + (3/20 * 2/19 * 1/18 * 17/17 * 16/16)

P(at least 1 red of 5) = 0.0921 + 0.01316 + 0.0009 = 0.1062

2) the probability of picking 6 marbles having 2 of each color

A) with replacement

P( 6, 2 of each) = 3/20 * 3/20 * 7/20 * 7/20 * 10/20 * 10/20

P( 6, 2 of each) = 0.0138

B) without replacement

P( 6, 2 of each) = 3/20 * 2/19 * 7/18 * 6/17 * 10/16 * 9/15

P( 6, 2 of each) = 0.00081

3) Pick 8 marbles: 4 green and 4 blue

A) with replacement

P(8, 4G and 4 B) = 7/20*7/20*7/20*7/20*10/20*10/20*10/20

P(8, 4G and 4 B) = 0.00094

B) without replacement

P(8, 4G and 4 B) = 7/20*6/19*5/18*4/17*10/16*9/15*8/14*7/13 = 0.00083

4) getting at least 6 marbles of the same color

Only Green and Blue marbles are more than 6

P(at least 6 marbles of the same color) = (7/20*6/19*5/18*4/17*3/16*2/15) + (10/20*9/19*8/18*7/17*6/16*5/15)

= 0.00018 + 0.00542

P(at least 6 marbles of the same color) = 0.0056

Cost of 6 .marbles= 6* 50 cents

C = 300 cents

Therefore, You will have to pay

(1 + 0.0056) 300 cent = 301.68 cents to be sure of getting at least 6 marbles of the same color

5) getting at least 6 marbles of the same color

Only Green and Blue marbles are more than 6

P(at least 6 marbles of the same color) = (7/20*6/19*5/18*4/17*3/16*2/15) + (10/20*9/19*8/18*7/17*6/16*5/15)

= 0.00018 + 0.00542

P(at least 6 marbles of the same color) = 0.0056

6 0
2 years ago
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