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4 years ago

At marios italian restaurant, 118 of the 200 plates are dirty, what percent if the plates are clean

Mathematics

1 answer:

JulijaS [17]4 years ago

8 0

Where is this restaurant? It sounds good right now.<span />

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3 years ago

What is the value of -2/5 ÷ 3/4

ale4655 [162]

-8/15 hope it helps!

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3 years ago

Given f(x)=-x+7 and g(x)=f(x+4) write a function for function g

finlep [7]

$\begin{array}{llrlll}f(x) = -x +7&~\hspace{10em} &f\left( \boxed{x+4} \right) &=& \boxed{-x+4}+7\\\\&&g(x) &=& -x+11\end{array}$

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3 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

What is the integral of ∫4x^2 dx

Nataly_w [17]

Answer:

$\boxed{\frac{4}{3} x^3 + c}$

Step-by-step explanation:

$\int\limits {4x^2} \ dx$

$= \frac{4}{3} x^3 + c$

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3 years ago

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