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2 years ago

Answer this question please ASAP!!!

Mathematics

1 answer:

Ymorist [56]2 years ago

7 0

17.4%

If helped mark me the brainiest!!

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Log, 7 1/2<br> Solve in exponential form

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Place value of 0 1 4 3 6 9 in 987,164.302

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3 years ago

A triangle has a 90° angle and two sides that measure 5 cm in length. Which statements are true about this triangle?

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Answer:

what are the statements

Step-by-step explanation:

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two positive integers are 3 units apart on a number line. their product is 180. wich equation can be used to solve for m, the gr

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Answer:

m^2-3m-180=0

m=15

Step-by-step explanation:

greater integer = m

small integer = n

we know that m = n+3 or n = m-3

we also know that n*m = 180

replace n for m-3

(m-3)*m=180

m^2-3m-180 = 0

(m-15)(m+12)=0

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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