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e-lub [12.9K]
3 years ago
9

Помогите математика !!!!!!

Mathematics
1 answer:
ohaa [14]3 years ago
4 0

Answer:

sorry say nice its out of understand

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Combine like terms. 5y - 8 - y
Svetlanka [38]
<h3>Question:-</h3>
  • Combine like terms. 5y - 8 - y

<h3>Answer:-</h3>
  • here like terms are 5y,-y
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\bold {4y - 8}

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h Tu to be a great day of school and

Step-by-step explanation:

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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

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3 years ago
Perry pays $340 per month to lease her car. How much will she pay to lease her car for two years?
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