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Savatey [412]
2 years ago
14

84, 87, 90, 93

Mathematics
1 answer:
damaskus [11]2 years ago
3 0

Even odd even odd, aka D.

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Need the answer asapp with the working out please thanks :)
patriot [66]
Hi,

The two numbers should be 12 and 30. 12=2x2x3 while 30=2x3x5.

Their HCF is 2x3=6 and their LCM is 2x3x2x5. Because of their HFC, we know that they are both multiple of 6. Also, the question says they both are GREATER than 6, so they can’t be 6 but are 6 times “something”. Thanks to the LCM, we know that “something” is equal to 2 for the first number and to 5 for the second one, the numbers hence being 12 and 30.

I hope this helps. If I was not clear enough or if you’d like further explanation please let me know. Also, English is not my first language, so I’m sorry for any mistakes.
7 0
2 years ago
#16 is really really hard thanks
Leokris [45]

7 million acres of Nevadas land is unfedarally owned!

Hope This Helps

7 0
3 years ago
Write a value-returning recursive function that computes the sum of the digits in a given positive int argument. for example, if
mars1129 [50]
244556
2+4+4+5+5+6=26
5 0
3 years ago
What is the greatest common factor of theses variables
DerKrebs [107]
B is the answer!

Explanation:
m^2 n^2 fits into both, you can take m^2 n^2 out of it both.

m is a common factor since you can take m out of both but it’s not the greatest common factor.

2n & 2m wouldn’t even be a common factor bc there is no 2n or 2m in neither.
8 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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