Which of the following rates converts to a unit rate of $12.50 per hour

How many megabytes of data can a 4.7 gigabyte dvd store?

dedylja [7]

The answer is 4700 megabytes.

1 giga = 1000 mega

1 gigabyte = 1000 megabyte
4.7 gigabyte = x megabyte

1 gb : 1000 mb = 4.7 gb : x
x = 1000 mb * 4.7 gb : 1 gb
x = 4700 mb

3 0

3 years ago

Please solve this...question..it may come in my exam.#math problem fast solve it and help me

ehidna [41]

Answer:

n + y

Step-by-step explanation:

a² - b² = (a +b)(a - b)

$\frac{n^{2}-y^{2}}{n+y}$ ÷ $\frac{n - y }{n + y } = \frac{(n + y)(n - y)}{(n + y)}*\frac{(n + y)}{(n - y)}\\\\$

{ (n + y) in the denominator and numerator will be cancelled ;

(n - y) in the denominator and numerator will be cancelled

= n + y

3 0

3 years ago

Catherine has $54. She plans to spend more than $20 of the money for a painting canvas. The rest will go toward paints. Each tub

erastova [34]

She can buy 3 tubes be cause 8.50 time 2 is less then 20$ but 8.50 times three is more than 20$ its 25.5

3 0

3 years ago

A 35 foot rope is cut into two pieces with one section 6 times as long as the other section, how is the shorter piece?

guajiro [1.7K]

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Define x:
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Let the shorter piece be x.
Shorter piece = x
Longer piece = 6x

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Construct equation:
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x +6x =35
7x = 35
x =5

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Find the length:
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Shorter length = x = 5 feet
Larger length = 6x = 6(5) = 30 feet

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Answer: The shorter piece is 5 feet.
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4 0

3 years ago

A simple random sample of 100 observations was taken from a large population. The sample mean and the standard deviation were de

olga2289 [7]

Answer:

Se=1.2

Step-by-step explanation:

The standard error is the standard deviation of a sample population. "It measures the accuracy with which a sample represents a population".

The central limit theorem (CLT) states "that the distribution of sample means approximates a normal distribution, as the sample size becomes larger, assuming that all samples are identical in size, and regardless of the population distribution shape"

The sample mean is defined as:

$\bar X =\frac{\sum_{i=1}^n x_i}{n}$

And the distribution for the sample mean is given by:

$\bar X \sim N(\mu,\frac{\sigma}{\sqrt{n}})$

Let X denotes the random variable that measures the particular characteristic of interest. Let, X1, X2, …, Xn be the values of the random variable for the n units of the sample.

As the sample size is large,(>30) it can be assumed that the distribution is normal. The standard error of the sample mean X bar is given by:

$Se=\frac{\sigma}{\sqrt{n}}$

If we replace the values given we have:

$Se=\frac{12}{\sqrt{100}}=1.2$

So then the distribution for the sample mean $\bar X$ is:

$\bar X \sim N(\mu=80,Se=1.2)$

5 0

3 years ago