The Poisson distribution with a mean of 6.0 is an appropriate model.
<h3>What is mean?</h3>
- In mathematics and statistics, the concept of mean is crucial. The most typical or average value among a group of numbers is called the mean.
- It is a statistical measure of a probability distribution's central tendency along the median and mode. It also goes by the name "expected value."
- There are different ways of measuring the central tendency of a set of values. There are multiple ways to calculate the mean. Here are the two most popular ones:
- Arithmetic mean is the total of the sum of all values in a collection of numbers divided by the number of numbers in a collection.
Hence, The Poisson distribution with a mean of 6.0 is an appropriate model.
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<h3>
Answer: No, this function is not linear</h3>
This function is a hyperbola. It graphs out two disjoint curves. A linear function produces a single straight line graph. All linear functions can be written in the form y = mx+b. The x being in the denominator is one indicator we cannot write the original equation in the form y = mx+b.
y = 3/x is the same as xy = 3; this tells us every point on y = 3/x has its (x,y) coordinate pair multiply to 3.
Answer:
70
Step-by-step explanation:
= First term = 
= Common difference = 
= Number of terms = 20
Sum of arithmetic progression is given by
![S=\dfrac{n}{2}[2a_1+(n-1)d]\\\Rightarrow S=\dfrac{20}{2}\times (2\times \dfrac{1}{3}+(20-1)\dfrac{1}{3})\\\Rightarrow S=70](https://tex.z-dn.net/?f=S%3D%5Cdfrac%7Bn%7D%7B2%7D%5B2a_1%2B%28n-1%29d%5D%5C%5C%5CRightarrow%20S%3D%5Cdfrac%7B20%7D%7B2%7D%5Ctimes%20%282%5Ctimes%20%5Cdfrac%7B1%7D%7B3%7D%2B%2820-1%29%5Cdfrac%7B1%7D%7B3%7D%29%5C%5C%5CRightarrow%20S%3D70)
The sum of the first 20 terms of the arithmetic sequence is 70.
Correct Choice is B.
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Answer:
They lose about 2.79% in purchasing power.
Step-by-step explanation:
Whenever you're dealing with purchasing power and inflation, you need to carefully define what the reference is for any changes you might be talking about. Here, we take <em>purchasing power at the beginning of the year</em> as the reference. Since we don't know when the 6% year occurred relative to the year in which the saving balance was $200,000, we choose to deal primarily with percentages, rather than dollar amounts.
Each day, the account value is multiplied by (1 + 0.03/365), so at the end of the year the value is multiplied by about
... (1 +0.03/365)^365 ≈ 1.03045326
Something that had a cost of 1 at the beginning of the year will have a cost of 1.06 at the end of the year. A savings account value of 1 at the beginning of the year would purchase one whole item. At the end of the year, the value of the savings account will purchase ...
... 1.03045326 / 1.06 ≈ 0.9721 . . . items
That is, the loss of purchasing power is about ...
... 1 - 0.9721 = 2.79%
_____
If the account value is $200,000 at the beginning of the year in question, then the purchasing power <em>normalized to what it was at the beginning of the year</em> is now $194,425.14, about $5,574.85 less.
