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AleksAgata [21]
3 years ago
6

Michael threw for 1,654 yards in his first five games. At this rate, how many yards will he have thrown for in fifteen games? ✗

Mathematics
2 answers:
vodka [1.7K]3 years ago
7 0

Answer:

4,962 yards.

Step-by-step explanation:

We are given that,

For the first 5 games, Michael threw 1,654 yards.

Since, we see that, the relationship between the yards and the games are directly proportional.

Then for 15 games,

Michael will have thrown for \frac{1654}{5}\times 3 yards i.e. 1654 × 3 = 4,962 yards

Hence, for 15 games, Michael have thrown for 4,962 yards

suter [353]3 years ago
6 0

Answer: 4,962 yards

Step-by-step explanation:

Given: Michael threw for 1,654 yards in his first five games.

The rate of threw = \dfrac{\text{Distance}}{\text{Time}}

Then , The rate of threw = \dfrac{1654}{\text{5}} yard/ minute.

i.e. the distance of yards thrown in 1 minute =  \dfrac{1654}{\text{5}} yard

If the rate remains same , then the distance of yards thrown for in fifteen games is given by :-

\dfrac{1654}{5}\times15=4,962\text{ yards}

Hence, At this rate, 4,962 yards will he have thrown for in fifteen games.

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Answer:

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Step-by-step explanation:

Assuming the following question: Because of staffing decisions, managers of the Gibson-Marimont Hotel are interested in  the variability in the number of rooms occupied per day during a particular season of the  year. A sample of 20 days of operation shows a sample mean of 290 rooms occupied per  day and a sample standard deviation of 30 rooms

Part a

For this case the best point of estimate for the population variance would be:

s^2 =30^2 =900

Part b

The confidence interval for the population variance is given by the following formula:

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df=n-1=20-1=19

Since the Confidence is 0.90 or 90%, the significance \alpha=0.1 and \alpha/2 =0.05, the critical values for this case are:

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567.28 \leq \sigma^2 \leq 1690.224

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23.818 \leq \sigma \leq 41.112

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Arc measure = 58.24°

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Calculate the measure of the angle KLN (as this equals m∠KLM which is the measure of arc MK)

ΔKNL is a right triangle, so we can use the cos trig ratio to find ∠KLM:

\sf \cos(\theta)=\dfrac{A}{H}

where:

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Given:

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\implies \textsf{Arc length MK}=2 \pi (15.2)\left(\dfrac{\sf \angle KLM}{360^{\circ}}\right)

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3 years ago
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