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12345 [234]
3 years ago
5

Before sending track and field athletes to the Olympics, the U.S. holds a qualifying meet. The upper dot plot shows the distance

s (in meters) of the top 888 shot putters in the preliminary round of the 201220122012 Olympic games. The lower dot plot shows the top 888 throwers in the shot put at the U.S. qualifying meet.
Mathematics
2 answers:
tiny-mole [99]3 years ago
4 0

Answer:

A & C

Step-by-step explanation:

seraphim [82]3 years ago
3 0

Answer:

C

Step-by-step explanation:

Let's consider the first answer choice:

The times in the semifinals were faster on average than the times in the finals.

We don't have to calculate the mean values of each distribution because we can see visually that the center of the final round distribution is lower than the center of the semifinal round distribution.

Since lower times are faster, we can't say that the times in the semifinals were faster on average than the times in the finals.

Hint #22 / 3

Now, let's consider the second answer choice:

The times in the finals vary noticeably less than the times in the semifinals.

We can see visually that the final round distribution is more spread out than the semifinal round distribution. We could calculate MAD or IQR from these displays, but we can see that the range of the final round distribution is larger.

So we can't conclude that the times in the finals vary noticeably less than the times in the semifinals.

Hint #33 / 3

Select this answer:

None of the above

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Your teacher will report the mean and standard deviation of the sampling distribution created by the class.
sukhopar [10]

Answer:

\hat p = \frac{\sum_{i=1}^{40} \hat p_i}{40}

\hat p = 0.493

And the deviation is given by this formula:

s_{\hat p}= \frac{\sum_{i=1}^{40} (\hat p_i - \hat p)^2}{n-1}= 0.085

And as we can see the population proportion expected for the number of heads 0.5  is very close to the mean of the sampling distribution, the error is :

\% Error = \frac{0.5-0.493}{0.5}* 100 = 1.4\%

Step-by-step explanation:

Assuming the data on the figure attached. We ar assuming that this is a sampling distribution of sample proportions of heads in 40 flips of a coin.

As we can see we have the following values:

0.25, 0.35, 0.375,0.375, 0.40,0.40,0.40, 0.425,0.425,0.425, 0.45,0.45,0.45,0.45, 0.475,0.475,0.475, 0.475,0.475, 0.50,0.50,0.50, 0.525,0.525,0.525,0.525, 0.55,0.55,0.55,0.55,0.55, 0.575,0.575,0.575 0.575, 0.575, 0.60,0.60, 0.65,0.65

And we can calculate the sample proportion with the following formula:

\hat p = \frac{\sum_{i=1}^{40} \hat p_i}{40}

\hat p = 0.493

And the deviation is given by this formula:

s_{\hat p}= \frac{\sum_{i=1}^{40} (\hat p_i - \hat p)^2}{n-1}= 0.085

And as we can see the population proportion expected for the number of heads 0.5  is very close to the mean of the sampling distribution, the error is :

\% Error = \frac{0.5-0.493}{0.5}* 100 = 1.4\%

4 0
3 years ago
Whoever answers first and it is correct I will award Brainliest!!!
victus00 [196]

Answer:

below

Step-by-step explanation:

that is the procedure above

fraction of arc RS

ANGLE RcS = 60 °

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therefore the total fraction is 1/6

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3 years ago
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xz_007 [3.2K]
130 =

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The prime factors are 2, 5, and 13.

No prime factors are used more than once.

The only exponents to write might be the understood 1.

130 = 2^1  *  5^1  *  13^1
4 0
3 years ago
A middle school has 650 students. A survey was given to 65 randomly selected students,
arsen [322]

Answer:

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Step-by-step explanation:

assuming the sample is representative

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