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trapecia [35]
3 years ago
5

A basketball coach keeps track of the points scored per game for all of her players. She gives a trophy to the player with the h

ighest mean score at the end of the season. Why does she uses the mean instead of the median?
Mathematics
2 answers:
VLD [36.1K]3 years ago
8 0

Answer:

The mean is the average of how many points a player usually scores while median is the number that is the middle number of all the points she scored in games. mean takes into count all the games while median only takes into account one game.

Step-by-step explanation:

pashok25 [27]3 years ago
5 0

Answer:

The mean is the average so whoever has the higher average gets the trophy instead of the median because that wouldn't be fair because a good player might have had a bad game.

Step-by-step explanation:

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Midway through the fundraiser, Mrs Little, the sponsor, discovers a container of buttermilk in the back of the fridge. It has 6
solong [7]

Answer: 6 recipes

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7 0
3 years ago
Which explicit formula gives the nth term of the sequence<br><br> 0, 3, 8, 15, 24,... ?
LiRa [457]

Answer: add over and over again

Step-by-step explanation:How do you get from 0 to 3? you add 3 to 0. Then, how do you get from 3 to 8? you add another 3 but then you a 2 too, so then 8. Lastly, my last example, how to get from 8 to 15. You take five, and then add 2 to it and add that number(then number you get when you add 5+1), 7 to 8.

5 0
3 years ago
Read 2 more answers
Simplify 3(2x+1) – 8.<br> A. 6х+5<br> В. 6х-5<br> с. 5x - 5<br> D. 5x - 7
victus00 [196]

Answer:

The answer to this question is B

Step-by-step explanation:

its 6x-5

3 0
3 years ago
When the polynomial in P(x) is divided by (x + a), the remainder equals P(a).
Effectus [21]

According to Remainder Theorem for the polynomials, for every polynomial P(x) there exist such polynomials G(x) and R(x), that P(x)=(x\pm a)G(x)+R(x).


When the polynomial in P(x) is divided by (x + a), then there exist such polynomials G(x) and R(x), that P(x)=(x+a)G(x)+R(x). Note that for x=-a:


P(-a)=(-a+a)G(-a)+R(-a)=0·G(a)+R(-a)=R(-a).
This means that P(-a) is the remainder, not P(a). <span>
</span><span>Answer:<span> correct choice is B.</span></span>




3 0
3 years ago
Do teachers find their work rewarding and satisfying? An article reports the results of a survey of 397 elementary school teache
Tju [1.3M]

Answer:

The 95% confidence interval would be given (0.0109;0.1651).  

We are confident at 95% that the difference between the two proportions is between 0.0109 \leq p_{elementary} -p_{High school} \leq 0.1651

Step-by-step explanation:

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".  

The margin of error is the range of values below and above the sample statistic in a confidence interval.  

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".  

p_A represent the real population proportion for elementary school  

\hat p_A =\frac{226}{397}=0.569 represent the estimated proportion for elementary school

n_A=397 is the sample size required for Brand A

p_B represent the real population proportion for high school teachers  

\hat p_B =\frac{129}{268}=0.481 represent the estimated proportion for high school teachers

n_B=268 is the sample size required for Brand B

z represent the critical value for the margin of error  

The population proportion have the following distribution  

p \sim N(p,\sqrt{\frac{p(1-p)}{n}})  

The confidence interval for the difference of two proportions would be given by this formula  

(\hat p_A -\hat p_B) \pm z_{\alpha/2} \sqrt{\frac{\hat p_A(1-\hat p_A)}{n_A} +\frac{\hat p_B (1-\hat p_B)}{n_B}}  

For the 95% confidence interval the value of \alpha=1-0.95=0.05 and \alpha/2=0.025, with that value we can find the quantile required for the interval in the normal standard distribution.  

z_{\alpha/2}=1.96  

And replacing into the confidence interval formula we got:  

(0.569-0.481) - 1.96 \sqrt{\frac{0.569(1-0.569)}{397} +\frac{0.481(1-0.481)}{268}}=0.0109  

(0.569-0.481) + 1.96 \sqrt{\frac{0.569(1-0.569)}{397} +\frac{0.481(1-0.481)}{268}}=0.1651  

And the 95% confidence interval would be given (0.0109;0.1651).  

We are confident at 95% that the difference between the two proportions is between 0.0109 \leq p_{elementary} -p_{High school} \leq 0.1651

5 0
3 years ago
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