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sattari [20]
2 years ago
12

An insurance agent says the standard deviation of the total hospital charges for patients involved in a crash in which the vehic

le struck a construction barricade is less than ​$4000. A random sample of 25total hospital charges for patients involved in this type of crash has a standard deviation of ​$4400. At α=0.01
can you support the​ agent's claim? Use the​ P-value method to test the claim
Mathematics
1 answer:
lisov135 [29]2 years ago
7 0
Let the null hypothesis be that the agent's claim is correct. Using a standard normal probability table we find:
p-value=P(Z \ \textgreater \ 1.1)=0.14
The p-value 0.14 is much larger than alpha = 0.01, so we cannot reject the agent's claim.
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How many feet are in 5 yards ?? Explain how you calculated your answer.
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Step-by-step explanation:

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The table shows transactions from five different bank accounts.
lisabon 2012 [21]

Answer:

 New Account Balance ($)

Acc #         Old Acc Balance ($)      Trans Amount ($)      New Acc Balance ($)

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4                       40                                        52                       12

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5 0
2 years ago
Integral of 17/(x^3-125)
daser333 [38]

Answer:

17/75 ln│x − 5│− 17/150 ln(x² + 5x + 25) − 17/(5√75) tan⁻¹((2x + 5) / √75) + C

Step-by-step explanation:

∫ 17 / (x³ − 125) dx

= 17 ∫ 1 / (x³ − 125) dx

= 17 ∫ 1 / ((x − 5) (x² + 5x + 25)) dx

Use partial fraction decomposition:

= 17 ∫ [ A / (x − 5) + (Bx + C) / (x² + 5x + 25) ] dx

Use common denominator to find the missing coefficients.

A (x² + 5x + 25) + (Bx + C) (x − 5) = 1

Ax² + 5Ax + 25A + Bx² − 5Bx + Cx − 5C = 1

(A + B) x² + (5A − 5B + C) x + 25A − 5C = 1

Match the coefficients and solve the system of equations.

A + B = 0

5A − 5B + C = 0

25A − 5C = 1

A = 1/75

B = -1/75

C = -2/15

So the integral is:

= 17 ∫ [ 1/75 / (x − 5) + (-1/75 x − 2/15) / (x² + 5x + 25) ] dx

Simplify:

= 17/75 ∫ [ 1 / (x − 5) − (x + 10) / (x² + 5x + 25) ] dx

Factor ½ from the numerator of the second fraction:

= 17/75 ∫ [ 1 / (x − 5) − ½ (2x + 20) / (x² + 5x + 25) ] dx

Split the fraction:

= 17/75 ∫ [ 1 / (x − 5) − ½ (2x + 5) / (x² + 5x + 25) − ½ (15) / (x² + 5x + 25) ] dx

Multiply the last fraction by 4/4:

= 17/75 ∫ [ 1 / (x − 5) − ½ (2x + 5) / (x² + 5x + 25) − 30 / (4x² + 20x + 100) ] dx

Complete the square:

= 17/75 ∫ [ 1 / (x − 5) − ½ (2x + 5) / (x² + 5x + 25) − 15 / ((2x + 5)² + 75) ] dx

Split the integral:

= 17/75 ∫ 1 / (x − 5) dx − 17/150 ∫ (2x + 5) / (x² + 5x + 25) dx − 17/5 ∫ 1 / ((2x + 5)² + 75) dx

The first integral is:

∫ 1 / (x − 5) dx = ln│x − 5│

The second integral is:

∫ (2x + 5) / (x² + 5x + 25) dx = ln(x² + 5x + 25)

The third integral is:

∫ 1 / ((2x + 5)² + 75) dx = 1/√75 tan⁻¹((2x + 5) / √75)

Plug in:

= 17/75 ln│x − 5│− 17/150 ln(x² + 5x + 25) − 17/(5√75) tan⁻¹((2x + 5) / √75) + C

4 0
2 years ago
Bob has two weekend jobs last weekend he made a total of $77 after working as a cashier for 5 hours and delivering newspapers fo
OlgaM077 [116]

Answer: $9 per hour at his job as a cashier and $8 per hour at his job delivering newspapers.

Step-by-step explanation:

1. Let's call the amount he got paid per hour at his job as a cashier: x.

Let's call the amount he got paid per hour at his job delivering newspapers: y.

2. Keeping on mind the information given in the problem above, you can make the following system of equations:

\left \{ {{5x+4y=77 \atop {6x+3y=78}} \right.

3. You can solve it by applying the Substitution method, as following:

- Solve for one of the variables from one of the equations and substitute it into the other equation to solve for the other variable and calculate its value.

- Substitute the value obtained into one of the original equations to solve for the other variable and calculate its value.

 4. Therefore, you have:

5x+4y=77\\5x=77-4y\\x=\frac{77-4y}{5}=15.4-0.8y

Then:

6(15.4-0.8y)+3y=78\\92.4-4.8y+3=78\\-1.8y=-14.4\\y=8

Finally:

5x+4(8)=77\\5x+32=77\\5x=45\\x=9

Therefore he got paid $9 per hour at his job as a cashier and $8 per hour at his job delivering newspapers.

5 0
3 years ago
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