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2 years ago

One morning, eight buses arrive at a bus stop.

Mathematics

2 answers:

Semenov [28]2 years ago

6 0

Answer:

0 and 5 minutes

Step-by-step explanation:

Median is the middle observation of given data. It can be found by following steps:

Arranging data in ascending or descending order.

Taking the average of middle two value if the total number of observation is even, and this average is our median.

or, if we odd number of observation then the most middle value is our median.

The mode is the observation which has a high number of repetitions (frequency).

We have given 8 numbers and we have to find the other two numbers a minute late for each bus to arrive at the bus stop.

Also, we have given the median and mode of all 10 buses i.e. 3.5 and 0 respectively.

Since the mode is 0, so the frequency of 0 must be greater than 2 times and less than or equal to 4 times.

The frequency of 0 is 4 is not possible as then we didn't get a median 3.5 then.

So the frequency of 0 is 3.

Hence one of number is 0 in two unknown number of a minute late for each bus arrive at the bus stop.

Now, we have the number of minutes late for each bus after arranging them in ascending order is: 0 0 0 2 2 6 8 8 9

For getting a median 3.5 we must have a number between 2 and 6. Let it be x then:

$\frac{2+x}{2} = 3.5$

⇒x = 5

Hence the number of minutes late for each bus is: 0 0 0 2 2 5 6 8 8 9

Thus, 0 and 5 are minutes late for the two-afternoon buses.

svet-max [94.6K]2 years ago

3 0

Answer:

5 minutes

Step-by-step explanation:

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1 year ago

Javier writes an integer on a piece of paper. The absolute value of his number is less than 4. Which is true of the possible val

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2 years ago

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Answer:

The Field's length is 93ft

Step-by-step explanation:

P=2L+2w

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2 years ago

A juice drink filling machine, when in perfect adjustment, fills the bottles with 12 ounces of drink on an average. Any overfill

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Answer:

d. The average is equal to 12 ounces.

Step-by-step explanation:

In this problem, the drink filling machine must be perfectly calibrated at 12 ounces since it needs to be shut down in cases of overfilling (mean > 12 ounces) and underfilling (mean < 12 ounces). Therefore, the correct approach would be to test if the mean is 12 ounces and the correct set of hypothesis would be:

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2 years ago

A study was performed on green sea turtles inhabiting a certain area. Time-depth recorders were deployed on 6 of the 76 capture

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Answer:

One can be 99% confident the true mean shell length lies within the above interval.

The population has a relative frequency distribution that is approximately normal.

Step-by-step explanation:

We are given that Time-depth recorders were deployed on 6 of the 76 captured turtles. These 6 turtles had a mean shell length of 51.3 cm and a standard deviation of 6.6 cm.

The pivotal quantity for a 99% confidence interval for the true mean shell length is given by;

P.Q. = $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$ ~ $t_n_-_1$

where, $\bar X$ = sample mean shell length = 51.3 cm

s = sample standard deviation = 6.6 cm

n = sample of turtles = 6

$\mu$ = true mean shell length

Now, the 99% confidence interval for $\mu$ = $\bar X \pm t_(_\frac{\alpha}{2}_) \times \frac{s}{\sqrt{n} }$

Here, $\alpha$ = 1% so $(\frac{\alpha}{2})$ = 0.5%. So, the critical value of t at 0.5% significance level and 5 (6-1) degree of freedom is 4.032.

<u>So, 99% confidence interval for</u> $\mu$ = $51.3 \pm 4.032 \times \frac{6.6}{\sqrt{6} }$

= [51.3 - 10.864 , 51.3 + 10.864]

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The interpretation of the above result is that we are 99% confident that the true mean shell length lie within the above interval of [40.44 cm, 62.16 cm].

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C. The population has a relative frequency distribution that is approximately normal.

This assumption is reasonably satisfied as the data comes from the whole 76 turtles and also we don't know about population standard deviation.

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