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Keith_Richards [23]
3 years ago
7

The price of a four-star dinner was $100 two decades ago. A report says that a dinner is 250% of what it was 20 years ago, how m

uch would it cost today?
Mathematics
1 answer:
Nina [5.8K]3 years ago
7 0

The dinner will cost $250 today.

<u><em>Explanation</em></u>

The price of a four-star dinner was $100 two decades ago and now the cost of the dinner is 250% of what it was 20 years ago.

So for finding the current cost, we will find 250% of $100 .

First we need to make 250% as decimal form after dividing it by 100. So....

250% = \frac{250}{100}= 2.50

Now we will multiply this 2.50 with $100 for getting the current cost of the dinner.

Thus, the current cost =(100* 2.50)dollar = 250 dollar



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Solve by using related function 1/2x^2 -2=0
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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
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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".

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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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