. Printer A can print 1/2 page per second. Printer B can print 1/3 page per second. Jacob has 5 pages to print. How long would i
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The two dot plots are missing, so i have attached it.
Answer:
The mean at the beginning of the school year was 9.5 miles and the mean at the end of the school year was 10.2 miles
Step-by-step explanation:
From the attached image, we are told to compare the means for each plot to the nearest tenth.
Mean = Σx/n
Now, from the image, total number of miles run by the 14 students at the beginning of the school year is;
(1 × 7) + (2 × 8) + (4 × 9) + (4 × 10) + (2 × 11) + (1 × 12) = 133
Mean of miles run at the beginning of the school year = 133/14 = 9.5 miles
Again, from the table, total miles run at the end of the school year = (2 × 8) + (2 × 9) + (4 × 10) + (3 × 11) + (3 × 12) = 143
Mean of miles run at the end of the school year = 143/14 = 10.2 miles
Thus;
The mean at the beginning of the school year was 9.5 miles and the mean at the end of the school year was 10.2 miles
In the given diagram, the traingles USW and UTV are similar triangles and thus the following ratio equality applies to them.
..........(Equation 1)
Checking the diagram given, we see that:
VT=y, WS=22, VU=8, ST=x-2
WU=WV+VU=12+8=20
TU=5
SU=ST+TU=(x-2)+5=x+3
Thus, substituting the required values in (Equation 1) we get:
Now, as can be clearly seen, to find y we will use the first two ratios as:
In a similar manner, to find the value of x we can use the last two ratios:
After cross multiplication we get:
Which can be simplified as:
Thus,
Therefore, the required answer is:
x=9.5 and y=8.8
Answer:
a) The 99% CI for the true proportion of voters who prefer the Republican candidate is (0.3658, 0.6001). This means that we are 99% sure that the true population proportion of all voters who prefer the Republican candidate is (0.3658, 0.6001).
b) The upper bound of the confidence interval is above 0.5 = 50%, which meas that the candidate can be confidence of victory.
Step-by-step explanation:
Question a:
In a sample with a number n of people surveyed with a probability of a success of , and a confidence level of
, we have the following confidence interval of proportions.
In which
z is the z-score that has a p-value of .
Sample of 120 registered voters in order to predict the winner of a local election. The Democrat candidate was favored by 62 of the respondents.
So 120 - 62 = 58 favored the Republican candidate, so:
99% confidence level
So , z is the value of Z that has a p-value of
, so
.
The lower limit of this interval is:
The upper limit of this interval is:
The 99% CI for the true proportion of voters who prefer the Republican candidate is (0.3658, 0.6001). This means that we are 99% sure that the true population proportion of all voters who prefer the Republican candidate is (0.3658, 0.6001).
b. If a candidate needs a simple majority of the votes to win the election, can the Republican candidate be confident of victory? Justify your response with an appropriate statistical argument.
The upper bound of the confidence interval is above 0.5 = 50%, which meas that the candidate can be confidence of victory.