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victus00 [196]
3 years ago
13

GIVING BRAINLIEST TO THE CORRECT ANSWER!!

Mathematics
1 answer:
arlik [135]3 years ago
3 0

Answer:

Step-by-step explanation:

No of total brown = 2 + 1 = 3 .

probability = 3 / 6

= 1 /2 .

no of males = 2

probability

= 2 / 6

= 1 /3

no of person who is both male and brown = 1

probability = 1 /6

Required probability

= probability of being brown + probability of being male - probability of being both

= 1 / 2 + 1 / 3 - 1 /6

= 4 / 6

= 2 / 3 .

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Describe the transformation.
Lesechka [4]

Answer:

g(x) = 1/3 (x - 12) .. vertical compression with factor 1/3 from f(x-4)

Step-by-step explanation:

*f(x) = x - 8  ... horizontal translation right 8 units from parent function f(x)=x

f(x-4) = (x-4) - 8 = x - 12 .... horizontal translation right 4 units from f(x)=x-8

g(x) = 1/3 (x - 12) .. vertical compression with factor 1/3 from f(x-4)

4 0
3 years ago
Help with homework Algebra 2:
Darina [25.2K]
The answer is 12y^2 + 2y - 5
The solution:
(72y^3 + 12y^2 - 30y) / 6y =
(6y)*(12y^2 + 2y - 5) / 6y = 12y^2 + 2y - 5
6 0
3 years ago
Read 2 more answers
Easy Question<br><br> Topic: Volume<br><br> Focus on question 10
vovangra [49]

Answer:

a 120 cm^3

b 70 cm^3

Step-by-step explanation:

The volume of a right rectangular prism is given by

V = l*w*h

10 a  V = 12*2*5

          =120 cm^3

10 b V = 2.5 *8 *3.5

           =70 cm^3

4 0
3 years ago
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Please help me with this question:)
Lunna [17]

Answer:

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Step-by-step explanation:

7 0
4 years ago
What are the orders of 3,7,9,11,13,17 and 19(mod20)?does 20 have primitive roots?
bezimeni [28]
3\equiv3\mod{20}
3^2\equiv9\mod{20}
3^3\equiv27\equiv7\mod{20}
3^4\equiv3\cdot3^3\equiv3\cdot7\equiv21\equiv1\mod{20}

7\equiv7\mod{20}
7^2\equiv49\equiv9\mod{20}
7^3\equiv7\cdot7^2\equiv63\equiv3\mod{20}
7^4\equiv7\cdot7^3\equiv21\equiv1\mod{20}

9\equiv9\mod{20}
9^2\equiv3^4\equiv1\mod{20}

11\equiv11\mod{20}
11^2\equiv121\equiv1\mod{20}

13\equiv-7\equiv13\mod{20}
13^2\equiv169\equiv9\mod{20}
13^3\equiv13\cdot13^2\equiv(-7)9\equiv-63\equiv-3\mod{20}
13^4\equiv13\cdot13^3\equiv(-7)(-3)\equiv21\equiv1\mod{20}

17\equiv-3\equiv17\mod{20}
17^2\equiv(-3)^2\equiv9\mod{20}
17^3\equiv(-3)^3\equiv-27\equiv3\mod{20}
17^4\equiv(-3)^4\equiv81\equiv1\mod{20}

19\equiv-1\equiv19\mod{20}
19^2\equiv19(-1)\equiv-19\equiv1\mod{20}

Generally speaking, a number x coprime to n will be a primitive root of n if we have x^n\equiv x\mod{n}, or x^{n-1}\equiv1\mod{n}. In other words, if x is of order n-1 modulo n, then x is a primitive root of n.

Since none of these numbers has order 19, it follows that 20 does not have any primitive roots.
6 0
4 years ago
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