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damaskus [11]
2 years ago
15

How much oil wells in a given field will ultimately produce is key information in deciding whether to drill more wells. Data was

collected estimating the total amount of oil recovered (in thousands of barrels) from 64 wells in the Devonian Richmond Dolomite area of the Michigan Basin. The data is in an Excel file on Canvas. Give a 95% confidence interval for the mean amount of oil recovered from all wells in this area.
DATA:

Oil

21.7

53.2

46.4

42.7

50.4

97.7

103.1

51.9

43.4

69.5

156.5

34.6

37.9

12.9

2.5

31.4

79.5

26.9

18.5

14.7

32.9

196

24.9

118.2

82.2

35.1

47.6

54.2

63.1

69.8

57.4

65.6

56.4

49.4

44.9

34.6

92.2

37

58.8

21.3

36.6

64.9

14.8

17.6

29.1

61.4

38.6

32.5

12

28.3

204.9

44.5

10.3

37.7

33.7

81.1

12.1

20.1

30.5

7.1

10.1

18

3

2

Mathematics
1 answer:
Ugo [173]2 years ago
3 0

Answer:

Check the explanation

Step-by-step explanation:

Kindly check the attached image below to see the step by step explanation to the question above.

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Alicia borrowed $15,000 to buy a car. She borrowed the money at 8% for 6 years. What is the interest she will pay for the loan ?
kykrilka [37]
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2 years ago
You earn $11 per hour delivering pizzas. you also work part-time at a convenience store where you earn $9 per hour. you want to
scoundrel [369]
We will traduce the sentences into equations. 
Let x be the number of hours in the first job, and y be the number of hours in the second job. 
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x+y\leq25\\11x+9y\geq 150

The above system has many solution, we can select for example:
x = 1, y = 16. So we can work one hour at the first job and 16 hour at the second job. 
7 0
2 years ago
The volume of wine in liters produced by a parcel of vineyard every year is modeled by a Gaussian distribution with an average o
saul85 [17]

Answer:

0.99865

Step-by-step explanation:

The question above is modelled by gaussian distribution. Gaussian distribution is also known as Normal distribution.

To solve the above question, we would be using the z score formula

The formula for calculating a z-score

z = (x-μ)/σ,

where x is the raw score

μ is the population mean

σ is the population standard deviation.

In the above question,

x is 115 liters

μ is 100

σ is the population standard deviation is unknown. But we were given variance in the question.

Standard deviation = √Variance

Variance = 9

Hence, Standard deviation = √9 = 3

We go ahead to calculate our z score

z = (x-μ)/σ

z = (115 - 100) / 3

z = 15/ 3

z score = 5

Using the z score table of normal distribution to find the Probability of having a z score of 5

P(x = 115) = P(z = 5) =

0.99865

Therefore the probability that this year it will produce 115 liters of wine = 0.99865

6 0
2 years ago
Solve the recurrence relation: hn = 5hn−1 − 6hn−2 − 4hn−3 + 8hn−4 with initial values h0 = 0, h1 = 1, h2 = 1, and h3 = 2 using (
musickatia [10]
(a) Suppose h_n=r^n is a solution for this recurrence, with r\neq0. Then

r^n=5r^{n-1}-6r^{n-2}-4r^{n-3}+8r^{n-4}
\implies1=\dfrac5r-\dfrac6{r^2}-\dfrac4{r^3}+\dfrac8{r^4}
\implies r^4-5r^3+6r^2+4r-8=0
\implies (r-2)^3(r+1)=0\implies r=2,r=-1

So we expect a general solution of the form

h_n=c_1(-1)^n+(c_2+c_3n+c_4n^2)2^n

With h_0=0,h_1=1,h_2=1,h_3=2, we get four equations in four unknowns:

\begin{cases}c_1+c_2=0\\-c_1+2c_2+2c_3+2c_4=1\\c_1+4c_2+8c_3+16c_4=1\\-c_1+8c_2+24c_3+72c_4=2\end{cases}\implies c_1=-\dfrac8{27},c_2=\dfrac8{27},c_3=\dfrac7{72},c_4=-\dfrac1{24}

So the particular solution to the recurrence is

h_n=-\dfrac8{27}(-1)^n+\left(\dfrac8{27}+\dfrac{7n}{72}-\dfrac{n^2}{24}\right)2^n

(b) Let G(x)=\displaystyle\sum_{n\ge0}h_nx^n be the generating function for h_n. Multiply both sides of the recurrence by x^n and sum over all n\ge4.

\displaystyle\sum_{n\ge4}h_nx^n=5\sum_{n\ge4}h_{n-1}x^n-6\sum_{n\ge4}h_{n-2}x^n-4\sum_{n\ge4}h_{n-3}x^n+8\sum_{n\ge4}h_{n-4}x^n
\displaystyle\sum_{n\ge4}h_nx^n=5x\sum_{n\ge3}h_nx^n-6x^2\sum_{n\ge2}h_nx^n-4x^3\sum_{n\ge1}h_nx^n+8x^4\sum_{n\ge0}h_nx^n
G(x)-h_0-h_1x-h_2x^2-h_3x^3=5x(G(x)-h_0-h_1x-h_2x^2)-6x^2(G(x)-h_0-h_1x)-4x^3(G(x)-h_0)+8x^4G(x)
G(x)-x-x^2-2x^3=5x(G(x)-x-x^2)-6x^2(G(x)-x)-4x^3G(x)+8x^4G(x)
(1-5x+6x^2+4x^3-8x^4)G(x)=x-4x^2+3x^3
G(x)=\dfrac{x-4x^2+3x^3}{1-5x+6x^2+4x^3-8x^4}
G(x)=\dfrac{17}{108}\dfrac1{1-2x}+\dfrac29\dfrac1{(1-2x)^2}-\dfrac1{12}\dfrac1{(1-2x)^3}-\dfrac8{27}\dfrac1{1+x}

From here you would write each term as a power series (easy enough, since they're all geometric or derived from a geometric series), combine the series into one, and the solution to the recurrence will be the coefficient of x^n, ideally matching the solution found in part (a).
3 0
2 years ago
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