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2 years ago

Help me please! Did I do this right?

Mathematics

2 answers:

MrMuchimi2 years ago

5 0

Answer:

no its A and B

Step-by-step explanation:

snow_tiger [21]2 years ago

4 0

-9/36 Is equal to -1/4 and that’s all that is on there.
Hope that helps!!
:)

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A certain region of America with over 500,000 households has a mean household income of $70,000 with a standard deviation of $20

Llana [10]

Answer:

Mean = 70,000 dollars

SD = 800 dollars

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Step-by-step explanation:

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3 years ago

Raspberries are hannas favorite fruit. She can only buy one pins for 2.55$ how many pounds of raspberries can Hannah but for $12

jeyben [28]

Answer:

She can buy 36

Step-by-step explanation:

7 0

2 years ago

Find the percent cost of the total spent on each equipment $36, fees $158, transportation $59

Evgen [1.6K]

Percent cost of the total spent on

each equipment = 14.23%,fees=62.45% and transportation = 23.32%.

As given in the question,

Money spent on

Each equipment = $36

Fees = $158

Transportation = $59

Total money spent = $( 36 + 158 + 59)

= $253

Percent cost of the total spent on

Each equipment = (36/253) × 100

= 14.23%

Fees = (36/253) × 100

= 62.45%

Transportation =(36/253) × 100

= 23.32%

Therefore, percent cost of the total spent on

each equipment = 14.23%,fees=62.45% and transportation = 23.32%.

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6 0

1 year ago

A small pouch contains 8 black marbles, 5 white marbles, and 12 red marbles. What is the probability of picking a red then a bla

Alina [70]

Answer:

8/600

Step-by-step explanation:

4 0

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

4 0

3 years ago

Help me please! Did I do this right?​

Help me please! Did I do this right?