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ziro4ka
2 months ago
11

Part B

Mathematics
1 answer:
AnnZ [12.3K]2 months ago
7 0

Answer:

Part A) The straight-line distance between the school and the Fire Station is 4.6\ miles

Part B) The straight-line distance between the school and the hospital is 2.1\ miles

Step-by-step explanation:

The full question can be found in the attached image.

Let

The positive x-coordinates represent the East

The positive y-coordinates represent the North

The negative x-coordinates indicate the West

The negative y-coordinates indicate the South

Treat point A (0,0) as the Middle School (reference point).

Part A) The Town Hall is situated 4.3 miles directly east of the Middle School,

thus the coordinates for Town Hall are B(4.3,0).

The Fire Station is positioned 1.7 miles directly north of Town Hall, therefore

the coordinates for the Fire Station are C(4.3,1.7).

What is the straight-line distance between the school and the Fire Station?

Keep in mind that

the distance formula between two points is indicated as

d=\sqrt{(y2-y1)^{2}+(x2-x1)^{2}}

We have

A(0,0) and C(4.3,1.7).

Now substitute

d=\sqrt{(1.7-0)^{2}+(4.3-0)^{2}}

d=\sqrt{(1.7)^{2}+(4.3)^{2}}

d=4.6\ miles

Part B) Now considering

The hospital is 3.1 miles west of the fire station.

The coordinates for the hospital are D(4.3-3.1,1.7).

D(1.2,1.7)

What is the straight-line distance between the school and the hospital?

Remember that

the distance formula applies here as well,

d=\sqrt{(y2-y1)^{2}+(x2-x1)^{2}}

with points A(0,0) and D(1.2,1.7).

Substituting the values gives us

d=\sqrt{(1.7-0)^{2}+(1.2-0)^{2}}

d=\sqrt{(1.7)^{2}+(1.2)^{2}}

d=2.1\ miles

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Step-by-step explanation:

1) Provided data and references

\bar X represents the average of the samples

s denotes the standard deviation of the samples

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t represents the specific statistic of interest

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To assess if the true mean is at least 10 hours, we must set up a hypothesis:

Part a

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If we consider the sample size being less than 30 and the population deviation unknown, it’s more appropriate to use a t-test to compare the actual mean with the reference value, calculated as:

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In this instance, with the p-value being less than the significance level, we have sufficient evidence to reject the null hypothesis.

Part c

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Part d

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Check the attached figure for further clarification

Therefore

The true statement is

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