1. a. {-2, -1, 0}
2. d. all real numbers (i think)
3. d. all real numbers (i think)
The dollar number that defines where the lowest values falls is
20% percentile value =$20
This is further explained below.
<h3 /><h3>What is the dollar number that defines where the lowest value falls?</h3>
Generally, Now we need to determine the 20th percentile ($20). It indicates that we need to locate a dollar amount that marks the point at which the 20 % data has a value lower than that number.
i=(p/100)*n
Where i is the position of p^th
percentile when the data is presented in ascending order.
i=20/100*50
i=1000/100
i=10
Therefore
n=50
p=20
In conclusion, the 10th position for given data is 20,
Therefore, 20% percentile value =$20
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well, if look at the timetable, She got the 08:30am train from Aberystwyth and arrives on time at Shrewsbury, from the timetable we know she arrived at 10:17am, now she did some rigamarole and got back to the Train station at Shrewsbury 4 hours later. Well, we know she arrived at 10:17am, if we add 4 hours to that that'll make it 1417 or namely 2:17pm.
well, the Train arrives at Shrewsbury a 14 19, or 2:19pm, she is there at 2:17pm, so she's really 2 minutes before the Train arrives at Shrewsbury, she's right on time, possibly with some munchies too.
14 19 > 14 17.
Answer:
c. an observed difference is 0.12 and P < 0.0001
Step-by-step explanation:
Let p1 be the proportion of the students who received musical instruction received a passing grade
Let p2 be the proportion of the students who didn't receive musical instruction received a passing grade
Null and Alternative hypotheses are:
: p1-p2=0
: p1-p2≠0
Test statistic can be found using the equation:
where
- p1 is the sample proportion of the students who received musical instruction received a passing grade (
)
- p2 is the sample proportion of the students who didn't receive musical instruction received a passing grade (
)
- p is the pool proportion of p1 and p2 (
)
- n1 is the sample size of the students who received musical instruction (3239)
- n2 is the sample size of the students who didn't receive musical instruction (2787)
Thus,
≈ 11.84 gives p-value < 0.0001
Observed difference is: 0.87-0.75=0.12