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Mnenie [13.5K]
3 years ago
7

How many paychecks do you get a year if you're paid semimonthly?

Mathematics
1 answer:
hjlf3 years ago
6 0
Semi --> twice
semimonthly --> occuring twice a month

Thus:
You are paid twice a month

12 months in a year

12 x 2 = 24 paychecks an year
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Of the 120 rooms in a hotel, 55% are single bed rooms, 30% are double bed rooms and the rest are deluxe rooms. How many deluxe r
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3 years ago
Reasoning triangle $abc$ has a perimeter of $12$ units. the vertices of the triangle are $a(x,\ 2),\ b(2,-2)$ , and $c(-1,\ 2)$
OlgaM077 [116]

hmmm let's find the distance between B and C first off.

~~~~~~~~~~~~\textit{distance between 2 points} \\\\ B(\stackrel{x_1}{2}~,~\stackrel{y_1}{-2})\qquad C(\stackrel{x_2}{-1}~,~\stackrel{y_2}{2})\qquad \qquad d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2} \\\\\\ BC=\sqrt{(~~-1 - 2~~)^2 + (~~2 - (-2)~~)^2} \implies BC=\sqrt{(-1 -2)^2 + (2 +2)^2} \\\\\\ BC=\sqrt{( -3 )^2 + ( 4 )^2} \implies BC=\sqrt{ 9 + 16 } \implies BC=\sqrt{ 25 }\implies BC=5

hmmm if BC is that much, then AB + AC = 7, since AB + AC + BC = 12 which is the perimeter, Check the picture below.

6 0
2 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:

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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
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