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3 years ago

Please help I need answer to this problem.

Mathematics

2 answers:

Svet_ta [14]3 years ago

5 0

Answer:

d 3.38

Step-by-step explanation:

seropon [69]3 years ago

4 0

Are the first numbers ratios?

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Furkat [3]

$\huge\bold{\purple{\bold{⚡133°⚡}}}$

Step-by-step explanation:

Co interior angles' sum is 180°

So, 180-67=133
133°

5 0

2 years ago

Solve the division problem. Round answer to the nearest hundredth. 9.252.063

Sergeu [11.5K]

9.252.063 rounded to the nearest hundredth is 9.252.060

6 0

3 years ago

Read 2 more answers

Please help me with this statistical question if you can, it would mean alot im really struggling right now

MA_775_DIABLO [31]

Answer:

vcc bc cvfdf

Step-by-step explanation:

3 0

3 years ago

I have the answer but can someone explain how to solve this problem? Thanks!

uranmaximum [27]

Answer:

$32

Step-by-step explanation:

27 was how much money Taylor had after buying a salad that was worth 5 dollars.

Money before buying the salad is 27+5 = 32.

4 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