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2 years ago

Solve for x:0.75x=22.50

Mathematics

2 answers:

Natasha2012 [34]2 years ago

8 0

Answer:

Step-by-step explanation:

IgorC [24]2 years ago

4 0

divide by 0.75 on both sides.

so,

$x = \frac{22.50}{0.75}$

the answer is x=30

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kompoz [17]

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3 years ago

Assume that the amount of beverage in a randomly selected 16-ounce beverage can has a normal distribution. Compute a 99% confide

ioda

Answer:

The question is incomplete, but the step-by-step procedures are given to solve the question.

Step-by-step explanation:

We have that to find our $\alpha$ level, that is the subtraction of 1 by the confidence interval divided by 2. So:

$\alpha = \frac{1 - 0.99}{2} = 0.005$

Now, we have to find z in the Ztable as such z has a pvalue of $1 - \alpha$ .

That is z with a pvalue of $1 - 0.005 = 0.995$ , so Z = 2.575.

Now, find the margin of error M as such

$M = z\frac{\sigma}{\sqrt{n}}$

In which $\sigma$ is the standard deviation of the population and n is the size of the sample.

$M = 2.575\frac{\sigma}{\sqrt{n}}$

The lower end of the interval is the sample mean subtracted by M.

The upper end of the interval is the sample mean added to M.

The 99% confidence interval for the population mean amount of beverage in 16-ounce beverage cans is (lower end, upper end).

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2 years ago

Help please photo is above

Zina [86]

Answer:

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Step-by-step explanation:

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2 years ago

What is 1.7¯ expressed as a fraction in simplest form? 13/7 25/9 8/7 16/9

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2 years ago

Read 2 more answers

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