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Dominik [7]
2 years ago
8

Four friends had dinner togethere.

Mathematics
1 answer:
nataly862011 [7]2 years ago
8 0

Answer:

Your answer is $16.40

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Combine the like terms to create an equivalent expression: 8n+12+(-9)-(-6n)
Scilla [17]

Work done below:

8n+12+(-9)-(-6n)=

8n+12-9+6n=

14n+3

6 0
3 years ago
I Will give you 98 points if you will answer this.
Alex777 [14]
Answer: -1

Explanation: The subscript (3) under 6 has no value so 6 subscript 3 is just 6. Use BEDMAS to solve for the equation (Brackets, Exponents, Division, Multiplication, Addition, Subtraction)

1. 6 x 2 - 24 / square root 16 + (8)
2. 6 x 2 - 24 / 4 + 8
3. 6 x 2 - 24 / 12
4. 12 - 24 / 12
5. -12 / 12
6. -1

Therefore the answer is -1.
3 0
3 years ago
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Need help taking test now !! thank you
nlexa [21]

Answer:

81/15

Step-by-step explanation:

14/3 would become 70/15 and 11/5 would become 33/15 and 70+33 is 103, so 103/5. 103/15 - 22/15 = 81/15.

6 0
3 years ago
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Two student council representatives are chosen at random from a group of 7 girls and 3 boys. Let G be the random variable denoti
Archy [21]

This question is incomplete, the complete question is;

Two student council representatives are chosen at random from a group of 7 girls and 3 boys. Let G be the random variable denoting the number of girls chosen.

a) What is E[G] or range of G

b) Give the distribution over the random variable G

Answer:

a) E[G] is [ 0, 1 , 2 ]

b)

the distribution over the random variable G.

x     P(x)

0     0.0667

1      0.4667

2     0.4667

Step-by-step explanation:

Given the data in the question;

Number to be chosen is 2

from 7 girls and 3 boys

a) range of G

number of girls can be; [ 0, 1 , 2 ]

Therefore, E[G] is [ 0, 1 , 2 ]

b) he distribution over the random variable G.

Distribution of the random variable G is Hypergeometric;

so

with P( G=g ) = P( getting g girls from 7 and 2-g boys from 3)

= ((^7C_g) × (^3C_{2-g)) / ^{10}C_2

now since, the range of G is [ 0, 1 , 2 ]

P( G=0 ) = ((^7C_0) × (^3C_{2)) / ^{10}C_2

= [(7!/(0!(7-0)!)) × (3!/(2!(3-2)!))] / (10!/(2!(10-2)!))

= [1 × 3] / 45 = 3/45 = 0.0667

P( G=1 ) = ((^7C_1) × (^3C_{1)) / ^{10}C_2

= [(7!/(1!(7-1)!)) × (3!/(1!(3-1)!))] / (10!/(2!(10-2)!))

= [ 7 × 3 ] / 45 = 21/45 = 0.4667

P( G=2 ) = ((^7C_2) × (^3C_{0)) / ^{10}C_2

= [(7!/(2!(7-2)!)) × (3!/(0!(3-0)!))] / (10!/(2!(10-2)!))

= [ 21 × 1 ] / 45 = 21/45 = 0.4667

Therefore, the distribution over the random variable G.

x     P(x)

0     0.0667

1      0.4667

2     0.4667

3 0
3 years ago
-2x^4 +50x^2+0x-3/x-5
Gemiola [76]
1) "x2" was replaced by "x^2". 1 more similar replacement(s).
Raising zero to a power is not allowed
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4 years ago
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