Please help with this question due asap

Roger Hunt intends to purchase one of two car dealerships currently for sale in a certain city. Records obtained from each of th

ValentinkaMS [17]

Answer:

(a) Dahl Motors = 8 Cars

Farthington Auto Sales = 7 Cars

(b) Roger will purchase from Farthington Auto Sales with higher weekly profit

Step-by-step explanation:

Dahl Motors / Week

(1) Car Sold 5 6 7 8 9 10 11 12

(2) Probability 5% 9% 14% 24% 18% 14% 11% 5%

(3) (2 x 3 ) 0.25 0.54 0.98 1.92 1.62 1.40 1.21 0.60

(4) Average sales = 0.25 + 0.54 + 0.98 + 1.92 + 1.62 + 1.40 + 1.21+0.60

= 8.52‬ cars/ week

= 8 cars /week

Profit/ Car = $544

Total weekly profit = 544 x 8

= $4,352

Farthington Auto Sales/ Week

(1) Car Sold 5 6 7 8 9 10

(2) Probability 8% 21% 31% 24% 10% 6%

(3) (2 x 3 ) 0.40 1.26 2.17 1.92 0.9 0.6

(4) Average sales = 0.40 + 1.26 + 2.17 + 1.92 + 0.90 + 0.60

= 7.25‬ cars/ week

= 7 cars/week

Profit/ Car = $651

Total weekly profit = 651 x 7

= $4,557‬

From the calculations above , Roger will purchase from Farthington Auto Sales

3 0

3 years ago

Find the nth term of this quadratic sequence 2,8,18,32,50

olga_2 [115]

Answer:

= 2n²

Step-by-step explanation:

From the sequence 2,8,18,32,50

The difference between each numbers are

: 6, 10, 14, 18

The difference between the second sequence is 4.

Therefore the nth term of the sequence is 2n²

6 0

3 years ago

Read 2 more answers

Francine uses 2/3 cups of pineapple juice for every 1/3 cup of orange juice to make a smoothie.Enter the number of cups of pinea

blondinia [14]

Answer:

1 cup i think

Step-by-step explanation:

Sorry im an idiot but i think its right

4 0

3 years ago

Read 2 more answers

2-31. Write an equation and solve it to find the answer to the question below. Use the 5-D Process to help you write the equati

raketka [301]

r = 25+b

n = 3b

n+r+b=435

3b+25+b+b=435

5b+25=435

5b=210

b=42

r = 25+42 = 67

n = 3(42) = 126

6 0

4 years ago

Find the Taylor series for f(x) centered at the given value of a. [Assume that f has a power series expansion. Do not show that

FromTheMoon [43]

Answer:

The Taylor series is $\ln(x) = \ln 3 + \sum_{n=1}^{\infty} (-1)^{n+1} \frac{(x-3)^n}{3^n n}$ .

The radius of convergence is $R=3$ .

Step-by-step explanation:

<em>The Taylor expansion.</em>

Recall that as we want the Taylor series centered at $a=3$ its expression is given in powers of $(x-3)$ . With this in mind we need to do some transformations with the goal to obtain the asked Taylor series from the Taylor expansion of $\ln(1+x)$ .

Then,

$\ln(x) = \ln(x-3+3) = \ln(3(\frac{x-3}{3} + 1 )) = \ln 3 + \ln(1 + \frac{x-3}{3}).$

Now, in order to make a more compact notation write $\frac{x-3}{3}=y$ . Thus, the above expression becomes

$\ln(x) = \ln 3 + \ln(1+y).$

Notice that, if x is very close from 3, then y is very close from 0. Then, we can use the Taylor expansion of the logarithm. Hence,

$\ln(x) = \ln 3 + \ln(1+y) = \ln 3 + \sum_{n=1}^{\infty} (-1)^{n+1} \frac{y^n}{n}.$

Now, substitute $\frac{x-3}{3}=y$ in the previous equality. Thus,

$\ln(x) = \ln 3 + \sum_{n=1}^{\infty} (-1)^{n+1} \frac{(x-3)^n}{3^n n}.$

<em>Radius of convergence.</em>

We find the radius of convergence with the Cauchy-Hadamard formula:

$R^{-1} = \lim_{n\rightarrow\infty} \sqrt[n]{|a_n|},$

Where $a_n$ stands for the coefficients of the Taylor series and $R$ for the radius of convergence.

In this case the coefficients of the Taylor series are

$a_n = \frac{(-1)^{n+1}}{ n3^n}$

and in consequence $|a_n| = \frac{1}{3^nn}$ . Then,

$\sqrt[n]{|a_n|} = \sqrt[n]{\frac{1}{3^nn}}$

Applying the properties of roots

$\sqrt[n]{|a_n|} = \frac{1}{3\sqrt[n]{n}}.$

Hence,

$R^{-1} = \lim_{n\rightarrow\infty} \frac{1}{3\sqrt[n]{n}} =\frac{1}{3}$

Recall that

$\lim_{n\rightarrow\infty} \sqrt[n]{n}=1.$

So, as $R^{-1}=\frac{1}{3}$ we get that $R=3$ .

8 0

4 years ago