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rosijanka [135]
3 years ago
12

Select the three statements that are true. A. -11 > -6 B. -5 > -7 C.-4 > -6 D. -9 > 7 E. 8 (-1 F. -5 -4​

Mathematics
1 answer:
OLEGan [10]3 years ago
4 0

Answer: B, C, And F

Step-by-step explanation:

My answer is correct because, for the negative numbber whatever is close to 1 like -1 is greater then the other negative number.

#AnimePower

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Weary of the low turnout in student elections, a college administration decides to choose an SRS of three students to form an ad
-Dominant- [34]

Answer:

P(ABC) = 0.110592

P(ABC^c) = 0.119808

P(AB^cC) = 0.119808

P(A^cBC) = 0.119808

P(AB^cC^c)  = 0.129792

P(A^cBC^c)  = 0.129792

P(A^cB^cC)  = 0.129792

P(A^cB^cC^c)  = 0.140608

Step-by-step explanation:

Given

P(A) = P(B) = P(C) = 48\%

Convert the probability to decimal

P(A) = P(B) = P(C) = 0.48

Solving (a): P(ABC)

This is calculated as:

P(ABC) = P(A) * P(B) * P(C)

This gives:

P(ABC) = 0.48*0.48*0.48

P(ABC) = 0.110592

Solving (b): P(ABC^c)

This is calculated as:

P(ABC^c) = P(A) * P(B) * P(C^c)

In probability:

P(C^c) = 1 - P(C)

So, we have:

P(ABC^c) = P(A) * P(B) * (1 - P(C))

P(ABC^c) = 0.48 * 0.48 * (1 - 0.48)

P(ABC^c) = 0.48 * 0.48 * 0.52

P(ABC^c) = 0.119808

Solving (c): P(AB^cC)

This is calculated as:

P(AB^cC) = P(A) * P(B^c) * P(C)

P(AB^cC) = P(A) * [1 - P(B)] * P(C)

P(AB^cC) = 0.48 * (1 - 0.48)* 0.48

P(AB^cC) = 0.48 * 0.52* 0.48

P(AB^cC) = 0.119808

Solving (d): P(A^cBC)

This is calculated as:

P(A^cBC) = P(A^c) * P(B) * P(C)

P(A^cBC) = [1-P(A)] *P(B) * P(C)

P(A^cBC) = (1 - 0.48)* 0.48 * 0.48

P(A^cBC) = 0.52* 0.48 * 0.48

P(A^cBC) = 0.119808

Solving (e): P(AB^cC^c)

This is calculated as:

P(AB^cC^c)  = P(A) * P(B^c) * P(C^c)

P(AB^cC^c)  = P(A) * [1-P(B)] * [1-P(C)]

P(AB^cC^c)  = 0.48 * [1-0.48] * [1-0.48]

P(AB^cC^c)  = 0.48 * 0.52*0.52

P(AB^cC^c)  = 0.129792

Solving (f): P(A^cBC^c)

This is calculated as:

P(A^cBC^c)   = P(A^c) * P(B) * P(C^c)

P(A^cBC^c)   = [1-P(A)] * P(B) * [1-P(C)]

P(A^cBC^c)   = [1-0.48] * 0.48 * [1-0.48]

P(A^cBC^c)   = 0.52 * 0.48 * 0.52

P(A^cBC^c)  = 0.129792

Solving (g): P(A^cB^cC)

This is calculated as:

P(A^cB^cC)  = P(A^c) * P(B^c) * P(C)

P(A^cB^cC)  = [1-P(A)] * [1-P(B)] * P(C)

P(A^cB^cC)  = [1-0.48] * [1-0.48] * 0.48

P(A^cB^cC)  = 0.52 * 0.52 * 0.48

P(A^cB^cC)  = 0.129792

Solving (h): P(A^cB^cC^c)

This is calculated as:

P(A^cB^cC^c)  = P(A^c) * P(B^c) * P(C^c)

P(A^cB^cC^c)  = [1-P(A)] * [1-P(B)] * [1-P(C)]

P(A^cB^cC^c)  = [1-0.48] * [1-0.48] * [1-0.48]

P(A^cB^cC^c)  = 0.52*0.52*0.52

P(A^cB^cC^c)  = 0.140608

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Step-by-step explanation:

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