Adult tickets to the aquarium costs $22 each while child tickets cost 15 each. If a large group buys 32 tickets and paid 543 how
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We are given the following data:
Average = m = 123.6
Population standard deviation = σ= psd = 36.8
Sample Size = n = 22
We are to find the confidence intervals for 90%, 95% and 98% confidence level.
Since the population standard deviation is known, and sample size is not too small, we can use standard normal distribution to find the confidence intervals.
Part 1) 90% Confidence Interval
z value for 90% confidence interval = 1.645
Lower end of confidence interval =
Using the values, we get:
Lower end of confidence interval=
Upper end of confidence interval =
Using the values, we get:
Upper end of confidence interval=
Thus the 90% confidence interval will be (110.69, 136.51)
Part 2) 95% Confidence Interval
z value for 95% confidence interval = 1.96
Lower end of confidence interval =
Using the values, we get:
Lower end of confidence interval=
Upper end of confidence interval =
Using the values, we get:
Upper end of confidence interval=
Thus the 95% confidence interval will be (108.22, 138.98)
Part 3) 98% Confidence Interval
z value for 98% confidence interval = 2.327
Lower end of confidence interval =
Using the values, we get:
Lower end of confidence interval=
Upper end of confidence interval =
Using the values, we get:
Upper end of confidence interval=
Thus the 98% confidence interval will be (105.34, 141.86)
Part 4) Comparison of Confidence Intervals
The 90% confidence interval is: (110.69, 136.51)
The 95% confidence interval is: (108.22, 138.98)
The 98% confidence interval is: (105.34, 141.86)
As the level of confidence is increasing, the width of confidence interval is also increasing. So we can conclude that increasing the confidence level increases the width of confidence intervals.
Average = m = 123.6
Population standard deviation = σ= psd = 36.8
Sample Size = n = 22
We are to find the confidence intervals for 90%, 95% and 98% confidence level.
Since the population standard deviation is known, and sample size is not too small, we can use standard normal distribution to find the confidence intervals.
Part 1) 90% Confidence Interval
z value for 90% confidence interval = 1.645
Lower end of confidence interval =
Using the values, we get:
Lower end of confidence interval=
Upper end of confidence interval =
Using the values, we get:
Upper end of confidence interval=
Thus the 90% confidence interval will be (110.69, 136.51)
Part 2) 95% Confidence Interval
z value for 95% confidence interval = 1.96
Lower end of confidence interval =
Using the values, we get:
Lower end of confidence interval=
Upper end of confidence interval =
Using the values, we get:
Upper end of confidence interval=
Thus the 95% confidence interval will be (108.22, 138.98)
Part 3) 98% Confidence Interval
z value for 98% confidence interval = 2.327
Lower end of confidence interval =
Using the values, we get:
Lower end of confidence interval=
Upper end of confidence interval =
Using the values, we get:
Upper end of confidence interval=
Thus the 98% confidence interval will be (105.34, 141.86)
Part 4) Comparison of Confidence Intervals
The 90% confidence interval is: (110.69, 136.51)
The 95% confidence interval is: (108.22, 138.98)
The 98% confidence interval is: (105.34, 141.86)
As the level of confidence is increasing, the width of confidence interval is also increasing. So we can conclude that increasing the confidence level increases the width of confidence intervals.
Answer:
The length of segment QM' = 6
Step-by-step explanation:
Given:
Q is the center of dilation
Pre-image (original image) = segment LM
New image = segment L'M'
The length of LQ = 4
The length of QM = 3
The length of LL' = 4
The original image was dilated with scale factor = 2
QM' = ?
To determine segment QM', first we would draw the diagram obtained from the given information.
Find attached the diagram
When a figure is dilated, we would have similar shape in thus cars similar triangles.
Segment L'M' = scale factor × length of LM
Let LM = x
L'M' = 2x
Using similar triangles theorem, ratio of their corresponding sides are equal.
QM/LM = QM'/L'M'
3/x = QM'/2x
6x = QM' × x
Q'M' = 6
The length of segment QM' = 6
