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3 years ago

Mike, who has 6 video games, has half as many games as Paul. Turn this into an algebraic equation.

Mathematics

1 answer:

cricket20 [7]3 years ago

4 0

$x-games \ of \ Paul \\ \\ 6=\frac{x}{2} \\ \\ \boxed{x=2*6=12}$

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What's the answer of 75% of what number is 60

elena55 [62]

75% of 60
OF MEANS times
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3 years ago

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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

Does anyone have any idea how to do this?

Mashutka [201]

Answer: D and A, B and E, C and F

Step-by-step explanation:

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2 years ago

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15/45 and 45/135 simplified

iren2701 [21]

"15/45" = "1/3"
"45/135" = "1/3"

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3 years ago

Factorize 2^x+7 + 2^x+6

joja [24]

Answer:

Step-by-step explanation:

$2^{(x+ 7)} + 2^{(x+ 6 )} = 2^{x} * 2^{7} + 2^{x}*2^{6}\\\\=2^{x} *(2^{7} + 2^{6})$

$= 2^{x} * 2^{6} ( 2+ 1)\\\\= 2^{x} * 2^{6} * 3$

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2 years ago