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balandron [24]
3 years ago
13

Can you help me with the answer for these questions please

Mathematics
2 answers:
iVinArrow [24]3 years ago
5 0
Hi.
your answers are
22. 90
23. 3
24. 168
hope this helps!!!

marishachu [46]3 years ago
3 0
23. 90
24. Games 1-5 had an average of 8 and games 6-10 had an average of 11, so games 6-10 had an average greater than games 1-5 by 3 points.
25. 168
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2 years ago
The graph represents a distribution of data.
makvit [3.9K]

Answer:

Mean=50

Step-by-step explanation:

The mean of a probability distribution is a <u>measure of central tendency</u>,and gives information about how the possible values of x are distributed.

The vertical axis measures the probability of  finding a specific value of x in the sample.  The probability of finding a value near the mean is high (that is why the value of the function that is depicted in the vertical axis, increases as we get closer to the mean=50): this is because the mean is that value of x around which higher probability of occurrence is associated.

6 0
3 years ago
Read 2 more answers
Multiple Choice
Zigmanuir [339]

Answer:

the slope is POSITIVE (answer A)

Step-by-step explanation:

The line goes through: (-1, -2)  and ( 2, 1)

Then the slope can be calculated as:

slope = (y2 - y1) / (x2 - x1)

in our case:

slope = (1 - -2) / (2 - -1) = (1 + 2) / (2 + 1) = 3 / 3 = 1

Therefore, the slope is POSITIVE (answer A)

8 0
3 years ago
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Write 79/10 as a mixed number
oee [108]

Answer:

7 9/10

Step-by-step explanation:

cuz mah brain big

6 0
3 years ago
in a random sample of 200 people, 154 said that they watched educational television. fine the 90% confidence interval of the tru
melisa1 [442]

Answer:

[0.7210,0.8189] = [72.10%, 81.89%]

Step-by-step explanation:

The sample size is  

n = 200

the proportion is

p = 154/200 = 0.77

<em>Since both np ≥ 10 and n(1-p) ≥ 10 </em>

<em>We can approximate this discrete binomial distribution with the continuous Normal distribution. As the sample size is large enough, not applying the continuity correction factor makes no significant  difference. </em>

The approximation would be to a Normal curve with this parameters:

<em>Mean </em>

p = 0.77

<em>Standard deviation </em>

\bf s=\sqrt{\frac{p(1-p)}{n}}=\sqrt{\frac{0.77*0.23}{200}}=0.0298

The 90% confidence interval for the proportion would be then  

\bf  [0.77-z^*0.0298, 0.77+z^*0.0298]

where \bf z^* is the 10% critical value for the Normal N(0,1) distribution , this is a value such that the area under the N(0,1) curve outside the interval \bf [-z^*,z^*] is 10%=0.1

We can either use a table, a calculator or a spreadsheet to get this value.

In Excel or OpenOffice Calc we use the function

<em>NORMSINV(0.95) and we get a value of 1.645 </em>

The 90% confidence interval for the proportion is then

\bf  [0.77-1.645^*0.0298, 0.77+1.645^*0.0298]=[0.7210,0.8189]

This means there is a 90% probability that the proportion of people who watch educational television is between 72.10% and 81.89%

If the television company wanted to publicize the proportion of viewers, do you think it should use the 90% confidence interval?

Yes, I do.

5 0
3 years ago
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