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butalik [34]
3 years ago
15

Express the given percent as a decimal.1/2% (Round to five decimal places as needed.)

Mathematics
1 answer:
grin007 [14]3 years ago
6 0

Answer:

The required expression is \frac{1}{2}\%=0.00500

Step-by-step explanation:

Given : Percentage \frac{1}{2}\%

To find : Express the given percent as a decimal ?

Solution :

To convert percentage into decimal,

To remove percentage sign we divide the given number by 100.

So, \frac{1}{2}\%=0.5\%

Remove percentage,

\frac{1}{2}\%=\frac{0.5}{100}

Now, Put decimal after three digit

\frac{1}{2}\%=0.005

As we have to round it to five decimal places,

\frac{1}{2}\%=0.00500

Therefore, The required expression is \frac{1}{2}\%=0.00500

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The coordinates of the vertices of the image are L' = (-4, 6), M' = (-5, 1) and  N' = (-7, 3)

<h3>What are the coordinates of the vertices of the image?</h3>

The vertices of the preimage of the triangle are given as:

L = (4, -6)

M = (5, -1)

N = (7, -3)

The rotation is given as: 180 degrees counterclockwise

<h3 />

The rule of this rotation is

(x, y) => (-x, -y)

So, we have:

L' = (-4, 6)

M' = (-5, 1)

N' = (-7, 3)

Hence, the coordinates of the vertices of the image are L' = (-4, 6), M' = (-5, 1) and  N' = (-7, 3)

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Cual es la temperatuta del jueves 9 de julio a las 10 hs en el calafate?<br>​
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Step-by-step explanation:

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3 years ago
Which term is any number that can be shown as the quotient of two integers
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This is the definition of a rational number.
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3 years ago
Select TWO values of x that are roots
Sauron [17]

x_1=\dfrac{-b-\sqrt{b^2-4ac}}{2a};\ x_2=\dfrac{-b+\sqrt{b^2-4ac}}{2a}\\\\y=x^2+3x+5\\\\a=1;\ b=3;\ c=5\\\\b^2-4ac=3^2-4(1)(5)=9-20=-11\\\\x_1=\dfrac{-3-\sqrt{-11}}{2\cdot1}=\dfrac{-3-\sqrt{-11}}{2}\\\\x_2=\dfrac{-3+\sqrt{-11}}{2}

Answer: A and B.

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An insurance company has 25,000 automobile policy holders. If the yearly claim of a policy holder is a random variable with mean
telo118 [61]

Answer:

P(T>8300000)=1-P(T

Step-by-step explanation:

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".  

The central limit theorem states that "if we have a population with mean μ and standard deviation σ and take sufficiently large random samples from the population with replacement, then the distribution of the sample means will be approximately normally distributed. This will hold true regardless of whether the source population is normal or skewed, provided the sample size is sufficiently large".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".  

Data given

n = 25000 represent the automobile policy holders

\mu= 320 represent the population mean

\sigma =540 represent the population standard deviation

Let T the variable that represent the total of interest on this case. We can assume that the random variable for an individual policy holder is given by:

X\sim N(\mu = 540, \sigma=540)

From the central limit theorem we know that the distribution for the sample mean \bar X is given by:

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

Solution to the problem

First we need to find the distribution for the random variable T like this:

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

And the total T is given by:

T=\sum_{i=1}^n X_i =n \bar X

We can find the expected value, variance and deviation for this random variable like this:

E(T)= n E(\bar X) = n \mu = 25000*320=8000000

Var(T)= Var(n\bar X)= n^2 Var(\bar X) = n^2 \frac{\sigma^2}{n}=n \sigma^2 =25000*(540^2)=7290000000

Sd(T)=\sqrt{7290000000}=85381.497

And we are interested on this probability:

P(T>8300000)

And we can use the Z score formula given by:

Z=\frac{T-E(T)}{\sigma_T}

P(T>8300000)=1-P(T

6 0
3 years ago
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