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Maksim231197 [3]
2 years ago
15

Over the first five years of owning her car, Gina drove about 12,200 miles the first year, 16,211 miles the second year, 12,050

the third year, 11,350 the fourth year, and 13,325 the fifth year. Find the mean, median, and mode of this data.
mean = 12,200; median = 12,200; no mode
mean = 13,027; median = 12,200; mode = 4,861
mean = 12,200; median = 13,027; no mode
mean = 13,027; median = 12,200; no mode
Mathematics
2 answers:
vampirchik [111]2 years ago
8 0
(12,200 + 16,211 + 12,050 + 11,350 + 13,325) / 5 = 65136/5 = 
13027.2 <== this is the mean (average)

11,350 , 12,050 , 12,200 , 13,325, 16,211
median (middle number) = 12,200

there is no mode...a mode is a number that appears most often...all the numbers appear once in this data.

attashe74 [19]2 years ago
7 0
The answer is <span>mean = 13,027; median = 12,200; no mode
</span>
Let's rearrange values from the lowest to the highest:
11350, 12050, 12200, 13325, 16211

<span>The mean is the sum of all values divided by the number of values:
</span>(11350 + 12050 + 12200 + 13325 + 16211)/5 ≈ 13027

The median is the middle value. If there is an odd number of data, then the median is the value in the middle. In the data set 11350, 12050, 12200, 13325, 16211, the median (the middle value) is 12200

<span>The mode is the value that occurs most frequently. Since none of the number does not occur most frequently, there is no mode.
</span>
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Answer:

It has millions of tickets. On each ticket is written a number a dollar amount. The exact average and SD are unknown but are estimated from the sample to be $20,000 and $5,000 respectively.

Step-by-step explanation:

Given that:

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The objective is to choose from the given option about what most closely resembles the relevant box model.

The correct answer is:

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Find the solution of the given initial value problem:<br><br> y''- y = 0, y(0) = 2, y'(0) = -1/2
igor_vitrenko [27]

Answer:  The required solution of the given IVP is

y(x)=\dfrac{3}{4}e^x+\dfrac{5}{4}e^{-x}.

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y^{\prime\prime}-y=0,~~~y(0)=2,~~y^\prime(0)=-\dfrac{1}{2}.

Let y=e^{mx} be an auxiliary solution of the given differential equation.

Then, we have

y^\prime=me^{mx},~~~~~y^{\prime\prime}=m^2e^{mx}.

Substituting these values in the given differential equation, we have

m^2e^{mx}-e^{mx}=0\\\\\Rightarrow (m^2-1)e^{mx}=0\\\\\Rightarrow m^2-1=0~~~~~~~~~~~~~~~~~~~~~~~~~~[\textup{since }e^{mx}\neq0]\\\\\Rightarrow m^2=1\\\\\Rightarrow m=\pm1.

So, the general solution of the given equation is

y(x)=Ae^x+Be^{-x}, where A and B are constants.

This gives, after differentiating with respect to x that

y^\prime(x)=Ae^x-Be^{-x}.

The given conditions implies that

y(0)=2\\\\\Rightarrow A+B=2~~~~~~~~~~~~~~~~~~~~~~~~~~~(i)

and

y^\prime(0)=-\dfrac{1}{2}\\\\\\\Rightarrow A-B=-\dfrac{1}{2}~~~~~~~~~~~~~~~~~~~~~~~~(ii)

Adding equations (i) and (ii), we get

2A=2-\dfrac{1}{2}\\\\\\\Rightarrow 2A=\dfrac{3}{2}\\\\\\\Rightarrow A=\dfrac{3}{4}.

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Substituting the values of A and B in the general solution, we get

y(x)=\dfrac{3}{4}e^x+\dfrac{5}{4}e^{-x}.

Thus, the required solution of the given IVP is

y(x)=\dfrac{3}{4}e^x+\dfrac{5}{4}e^{-x}.

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