In a library, the ratio of maths books to science books is 3 : 2 and the ratio of science books to history books is 1 : 3. If th
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Hi,
To find the number of small cylinders that fit inside the big cylinder. You need to divide volume of small cylinder with the volume of big cylinder. We do this by using formula to calculate volume of cylinder:
To find radius just divide diameter with 2.
Now just division:
Answer: You could fit 374.8 smaller cylinders inside a big cylinder.
Hope this helps.
r3t40
To find the number of small cylinders that fit inside the big cylinder. You need to divide volume of small cylinder with the volume of big cylinder. We do this by using formula to calculate volume of cylinder:
To find radius just divide diameter with 2.
Now just division:
Answer: You could fit 374.8 smaller cylinders inside a big cylinder.
Hope this helps.
r3t40
Using the uniform distribution, it is found that:
A,B) 0.3 of the time does it take longer than 27 minutes for Josh’s bag to arrive in baggage claim.
C) 20% of the time does Josh’s bag arrive in less than 22 minutes.
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An uniform distribution has two bounds, a and b.
The probability of finding a value of at lower than x is:
The probability of finding a value between c and d is:
The probability of finding a value above x is:
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- In the graph, we have that the distribution is uniform between 20 and 30 minutes, thus
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Itens a and b:
- Above 27 minutes, thus:
0.3 of the time does it take longer than 27 minutes for Josh’s bag to arrive in baggage claim.
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Item c:
- Less than 22 minutes, thus:
0.2*100 = 20%
20% of the time does Josh’s bag arrive in less than 22 minutes.
A similar problem is given at brainly.com/question/15855314
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>
The 90% confidence interval for the proportion would be then
where 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
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
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.