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KATRIN_1 [288]
3 years ago
11

I sailed west to Hawaii against the wind at the speed of 12 knots/hour. When I returned, I sailed the same route with the wind a

t the speed of 18 knots/hour. What was my average speed for the entire trip?
Mathematics
1 answer:
expeople1 [14]3 years ago
7 0

Answer:

Step-by-step explanation:

total time is 5 hr and total distance is 72 nautical miles

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A package weighing 5.4 lb is to be shipped to Europe. How many kilograms does this package weigh
Softa [21]
1kg = 2.205 lbs

Therefore, 1 lb = 1/2.205 kg
Or 5.4lb = 5.4/2.205
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Thus, this package will weigh 2.44 kgs
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Shannon saves 1235 in november. In november, she saved $250 less than she did in december. How much money did she save in the tw
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In December she would have saved $1585 because 1235 + 250 = 1485. You can find how much money she raised in those 2 months by adding the totals of both months together, so 1485 + 1235 = 2720. She saved $2720 in the 2 months.
4 0
3 years ago
Question 7 of 10
andriy [413]

Answer:

C. n = 90; p = 0.8

Step-by-step explanation:

According to the Central Limit Theorem, the distribution of the sample means will be approximately normally distributed when the sample size, 'n', is equal to or larger than 30, and the shape of sample distribution of sample proportions with a population proportion, 'p' is normal IF n·p ≥ 10 and n·(1 - p) ≥ 10

Analyzing  the given options, we have;

A. n = 45, p = 0.8

∴ n·p = 45 × 0.8 = 36 > 10

n·(1 - p) = 45 × (1 - 0.8) = 9 < 10

Given that for n = 45, p = 0.8, n·(1 - p) = 9 < 10, a normal distribution can not be used to approximate the sampling distribution

B. n = 90, p = 0.9

∴ n·p = 90 × 0.9 = 81 > 10

n·(1 - p) = 90 × (1 - 0.9) = 9 < 10

Given that for n = 90, p = 0.9, n·(1 - p) = 9  < 10, a normal distribution can not be used to approximate the sampling distribution

C. n = 90, p = 0.8

∴ n·p = 90 × 0.8 = 72 > 10

n·(1 - p) = 90 × (1 - 0.8) = 18 > 10

Given that for n = 90, p = 0.9, n·(1 - p) = 18 > 10, a normal distribution can be used to approximate the sampling distribution

D. n = 45, p = 0.9

∴ n·p = 45 × 0.9 = 40.5 > 10

n·(1 - p) = 45 × (1 - 0.9) = 4.5 < 10

Given that for n = 45, p = 0.9, n·(1 - p) = 4.5 < 10, a normal distribution can not be used to approximate the sampling distribution

A sampling distribution Normal Curve

45 × (1 - 0.8) = 9

90 × (1 - 0.9) = 9

90 × (1 - 0.8) = 18

45 × (1 - 0.9) = 4.5

Now we will investigate the shape of the sampling distribution of sample means. When we were discussing the sampling distribution of sample proportions, we said that this distribution is approximately normal if np ≥ 10 and n(1 – p) ≥ 10. In other words

Therefore;

A normal curve can be used to approximate the sampling distribution of only option C. n = 90; p = 0.8

3 0
3 years ago
Can someone help me pls
Stels [109]

Answer: This answer for this is A

Step-by-step explanation: Hope this help :D

4 0
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What is−1/2−(−5/9) as a fraction in simplest form
Rasek [7]

Answer:

1/18

Step-by-step explanation:

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7 0
3 years ago
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