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Vesna [10]
3 years ago
13

Please answer with the correct letter for brainliest

Mathematics
2 answers:
Sidana [21]3 years ago
8 0

Answer:

Given~inequality~is\\2(x-3) -5x

So, the required answer is C. [Look at the screenshot]

Andru [333]3 years ago
5 0
I really think that it’s C
Hopefully this helps
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Five pounds is 2.268 times as heavy as one kilogram. How many times as heavy is two pounds than one kilogram?
BlackZzzverrR [31]

Answer:

4.536

Step-by-step explanation:

2.268*2=4.536

8 0
3 years ago
Read 2 more answers
The diameter of a ball is 8 in.
Cerrena [4.2K]
267. 95cm^3

hope this helped
5 0
2 years ago
Solve the equation 1.5y < -6.75
finlep [7]
1.5y < -6.75

First, divide both sides by '1.5'.
\frac{-6.5}{1.5}
Second, since 4.5 × 1.5 = 6.5, simplify the fraction to '-4.5'
y \ \textless \  -4.5

Answer: y < -4.5

7 0
2 years ago
Suppose that an airline uses a seat width of 16.5 in. Assume men have hip breadths that are normally distributed with a mean of
Alexxx [7]

Answer:

a) 0.018

b) 0            

Step-by-step explanation:

We are given the following information in the question:

Mean, μ =  14.4 in

Standard Deviation, σ = 1 in

We are given that the distribution of breadths is a bell shaped distribution that is a normal distribution.

Formula:

z_{score} = \displaystyle\frac{x-\mu}{\sigma}

a) P(breadth will be greater than 16.5 in)

P(x > 16.5)

P( x > 16.5) = P( z > \displaystyle\frac{16.5 - 14.4}{1}) = P(z > 2.1)

= 1 - P(z \leq 2.1)

Calculation the value from standard normal z table, we have,  

P(x > 16.5) = 1 - 0.982 = 0.018 = 1.8\%

0.018 is the probability that if an individual man is randomly​ selected, his hip breadth will be greater than 16.5 in.

b) P( with 123 randomly selected​ men, these men have a mean hip breadth greater than 16.5 in)

Formula:

z_{score} = \displaystyle\frac{x-\mu}{\frac{\sigma}{\sqrt{n}}}  

P(x > 16.5)  

P( x > 16.5) = P( z > \displaystyle\frac{16.5-14.4}{\frac{1}{\sqrt{123}}}) = P(z > 23.29)  

= 1 - P(z \leq 23.29)

Calculation the value from standard normal z table, we have,  

P(x > 16.5) = 1 - 1 = 0

There is 0 probability that 123 randomly selected men have a mean hip breadth greater than 16.5 in

4 0
3 years ago
Fosmine lounches a model rocket from a elift. The rocket goes up 16.8 feet above the cliff,
7nadin3 [17]

Answer: The final answer I got is -44.9

Step-by-step explanation:  At the first chose I put Positive, then at the second choise I put negative, at the third choose I put 16.8-(-28.1), and at the last choice I put -44.9.

3 0
3 years ago
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