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satela [25.4K]
3 years ago
14

uppose you are performing a hypothesis test with H subscript 0 superscript blank space colon space mu equals 10 and H subscript

A colon mu greater than 10, using a sample size of 50 observations. You sample data has a mean of 10.7, with a standard deviation of 3.1. Compute the test statistic for this hypothesis test
Mathematics
1 answer:
Tems11 [23]3 years ago
7 0

Answer: The test statistic for this hypothesis test =1.597

Step-by-step explanation:

When population standard deviation is unknown, we use t-test statistic:

t=\dfrac{\overline{x}-\mu}{\dfrac{s}{\sqrt{n}}}

\overline{x} = sample mean, \mu = population mean , s= sample standard deviation, n = sample size.

As per given,

t=\dfrac{10.7-10}{\dfrac{3.1}{\sqrt{50}}}\\\\=1.59669273171\approx1.597

Hence, the test statistic for this hypothesis test =1.597

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A cruise ship needs to book at least 2,052 passengers to be profitable, but the most passengers the ship can accommodate is 2,46
mart [117]

The numbers of passengers that need to be booked to ensure the cruise line makes a profit, using a compound inequality is x ≥ 2,052 and x ≤ 2,462.

<h3>How to illustrate the inequality?</h3>

From the information given, we are told that the cruise ship needs to book at least 2,052 passengers to be profitable, but the most passengers the ship can accommodate is 2,462.

Learn x be the number of people that can be accommodated. Therefore, the model is x ≥ 2,052 and x ≤ 2,462.

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8 0
2 years ago
Two airplanes leave an airport at the same time, the first headed due north and the second at a bearing of N42^ E . At 2:00PM, t
Leno4ka [110]

Answer:

The two airplanes are about 330miles apart.

Step-by-step explanation:

The diagram interpreting the question has been attached to this response.

As shown in the diagram,

i. the airplanes leave at point C.

ii. at 2.00pm the first and second airplanes are at points A and B respectively, where they are 312miles and 487miles away from the starting point C in directions due north and N42E from the point C.

iii. the points A, B and C form a triangle with sides a, b and c.

To solve for the value of c which is the distance between the two planes at 2.00pm, the cosine rule is used.

c² = a² + b² - 2abcosC            --------------(i)

where;

b  = 312miles

a = 487miles

C = 42°

Substitute these values into equation (i) and solve as follows;

c² = (487)² + (312)² - 2(312)(487)cos(42)

c² = (237169) + (97344) - 303888cos(42)

c² = (237169) + (97344) - 303888(0.7431)

c² = 334513 - 225819.1728

c² = 108693.8272

<em>Take the square root of both sides</em>

√c² = √108693.8272

c = 329.69

c ≅ 330miles

Therefore, the two airplanes are far apart by 330miles

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