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lisabon 2012 [21]
2 years ago
7

How to write 1 2/5 as a percent

Mathematics
2 answers:
Tems11 [23]2 years ago
4 0

140%

Step-by-step explanation:

ANTONII [103]2 years ago
4 0

Answer:

140%

Step-by-step explanation:

First, we change 1 2/5 to a decimal number.

1 2/5 = 1 + 2/5 = 1 + 0.4 = 1.4

To change a decimal number to a percent, multiply the number by 100, which means you move the decimal point two places to the right.

1.4 = 1.40 = 140%

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STatiana [176]

your answer is -1/100 (a fraction)

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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

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3 years ago
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Mkey [24]

Answer:

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Step-by-step explanation:

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horrorfan [7]

Answer:

h= 12/pi OR 3.81972

Step-by-step explanation:

100= \frac{1}{3}5^{2}h

100= \frac{1}{3}\pi 25h

100= \frac{25\pi h}{3}

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qwelly [4]

Answer:

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Step-by-step explanation:

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