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3 years ago

A car salesman has 8 used cars for sale. They have a mean price of $4500. What is the total price of all 8 cars?

Mathematics

1 answer:

muminat3 years ago

7 0

Answer:

$36000

Step-by-step explanation:

We have a mean price of $4500 this means that in the 8 cars we have cars with higer and lower prices than $4500 but that if we have to give the same price to 8 cars that would be $4500. Then to know the total price of the all 8 cars we just have to multiply:

$\$4500\cdot 8=$36000$

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Step-by-step explanation:

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3 years ago

What is the sum of the rational expression below? 3x/x+2 + x/x+4

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Step-by-step explanation:

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2 years ago

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1) Use power series to find the series solution to the differential equation y'+2y = 0 PLEASE SHOW ALL YOUR WORK, OR RISK LOSING

iogann1982 [59]

$y=\displaystyle\sum_{n=0}^\infty a_nx^n$

then

$y'=\displaystyle\sum_{n=1}^\infty na_nx^{n-1}=\sum_{n=0}^\infty(n+1)a_{n+1}x^n$

The ODE in terms of these series is

$\displaystyle\sum_{n=0}^\infty(n+1)a_{n+1}x^n+2\sum_{n=0}^\infty a_nx^n=0$

$\displaystyle\sum_{n=0}^\infty\bigg(a_{n+1}+2a_n\bigg)x^n=0$

$\implies\begin{cases}a_0=y(0)\\(n+1)a_{n+1}=-2a_n&\text{for }n\ge0\end{cases}$

We can solve the recurrence exactly by substitution:

$a_{n+1}=-\dfrac2{n+1}a_n=\dfrac{2^2}{(n+1)n}a_{n-1}=-\dfrac{2^3}{(n+1)n(n-1)}a_{n-2}=\cdots=\dfrac{(-2)^{n+1}}{(n+1)!}a_0$

$\implies a_n=\dfrac{(-2)^n}{n!}a_0$

So the ODE has solution

$y(x)=\displaystyle a_0\sum_{n=0}^\infty\frac{(-2x)^n}{n!}$

which you may recognize as the power series of the exponential function. Then

$\boxed{y(x)=a_0e^{-2x}}$

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3 years ago

The following data represent highway fuel consumption in miles per gallon (mpg) for a random sample of 55 models of passenger ca

Varvara68 [4.7K]

Answer:

11 ;

Step-by-step explanation:

Given the data:

30 27 22 25 24 25 24 15 35 35 33 52 49 10 27 18 20 23 24 25 30 24 24 24 18 20 25 27 24 32 13 13 21 2 37 35 32 33 29 3 28 28 25 29 31

Number of classes = 5

Class width : Range / number of classes

Class width = (maximum - minimum) / 5

Class width = (52 - 2) / 5 = 50/5 = 10 + 1 = 11

For the frequency table showing class limits, class boundaries, midpoints, frequencies, relative frequencies, and cumulative frequencies and histogram

Kindly check attached picture

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2 years ago

What is the least common denominator of the fractions 4/15 1/6 and 3/4

Alexus [3.1K]

Answer:

Step-by-step explanation:

We have to find least common denominator for these fractions or we just have to find the least common multiple for 15, 6 and 4

In order to find the least common multiple, we have to represent these numbers as factors of prime numbers:

15 = 3 * 5

6 = 2 * 3

4 = 2 * 2

Least common multiples will be common factors (considering that there are no more than of them in other numbers) and the uncommon factors. Then, we can write that:

2 * 2 * 3 * 5 = 60 (there are two 2's, one common 3 and an uncommon 5)

The denominator will be 60 for the least common multiples.

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3 years ago

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