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2 years ago

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An aircraft carrier can travel 450 km in the same time a container ship travels 300 km. If the aircraft carrier was travelling 5

km/h faster what was the speed of the container ship? (answer is a rational expression)

1 answer:

Leto [7]2 years ago

6 0

Answer:

10 km/hr

Step-by-step explanation:

The aircraft carrier and the container ship both took 30 hours to travel their respective distances. If you divide both numbers by 30, you find that their differences is 5. We can tell that the aircraft carrier traveled 450 ÷ 30 = 15 km/hr and the container ship traveled at a speed of 300 ÷ 30 = 10 km/hr. The aircraft carrier traveled 5 km/hr faster. So, we can tell the container ship traveled at 10 km/hr.

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Step-by-step explanation:

Given the following :

Inventory (I) = 180

Lead time (L) = 7 days

Review time (T) = 2 weeks = 14 days

Demand (D) = 20

Standard deviation (σ) = 5

Zscore for 95% probability = 1.645

Units to be ordered :

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(σT+L) = √(T + L)σ²

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During April of 2013, Gallup randomly surveyed 500 adults in the US, and 47% said that they were happy, and without a lot of str

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Answer:

number of successes

$k = 235$

number of failure

$y = 265$

The criteria are met

A

The sample proportion is $\r p = 0.47$

B

$E =4.4 \%$

C

What this mean is that for N number of times the survey is carried out that the which sample proportion obtain will differ from the true population proportion will not more than 4.4%

Ci

$r = 0.514 = 51.4 \%$

$v = 0.426 = 42.6 \%$

D

This 95% confidence interval mean that the the chance of the true population proportion of those that are happy to be exist within the upper and the lower limit is 95%

E

Given that 50% of the population proportion lie with the 95% confidence interval the it correct to say that it is reasonably likely that a majority of U.S. adults were happy at that time

F

Yes our result would support the claim because

$\frac{1}{3 } \ of N < \frac{1}{2} (50\%) \ of \ N , \ Where\ N \ is \ the \ population\ size$

Step-by-step explanation:

From the question we are told that

The sample size is $n = 500$

The sample proportion is $\r p = 0.47$

Generally the number of successes is mathematical represented as

$k = n * \r p$

substituting values

$k = 500 * 0.47$

$k = 235$

Generally the number of failure is mathematical represented as

$y = n * (1 -\r p )$

substituting values

$y = 500 * (1 - 0.47 )$

$y = 265$

for approximate normality for a confidence interval criteria to be satisfied

$np > 5 \ and \ n(1- p ) \ >5$

Given that the above is true for this survey then we can say that the criteria are met

Given that the confidence level is 95% then the level of confidence is mathematically evaluated as

$\alpha = 100 - 95$

$\alpha = 5 \%$

$\alpha =0.05$

Next we obtain the critical value of $\frac{\alpha }{2}$ from the normal distribution table, the value is

$Z_{\frac{ \alpha }{2} } = 1.96$

Generally the margin of error is mathematically represented as

$E = Z_{\frac{\alpha }{2} } * \sqrt{ \frac{\r p (1- \r p}{n} }$

substituting values

$E = 1.96 * \sqrt{ \frac{0.47 (1- 0.47}{500} }$

$E = 0.044$

=> $E =4.4 \%$

What this mean is that for N number of times the survey is carried out that the proportion obtain will differ from the true population proportion of those that are happy by more than 4.4%

The 95% confidence interval is mathematically represented as

$\r p - E < p < \r p + E$

substituting values

$0.47 - 0.044 < p < 0.47 + 0.044$

$0.426 < p < 0.514$

The upper limit of the 95% confidence interval is $r = 0.514 = 51.4 \%$

The lower limit of the 95% confidence interval is $v = 0.426 = 42.6 \%$

This 95% confidence interval mean that the the chance of the true population proportion of those that are happy to be exist within the upper and the lower limit is 95%

Given that 50% of the population proportion lie with the 95% confidence interval the it correct to say that it is reasonably likely that a majority of U.S. adults were happy at that time

Yes our result would support the claim because

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