Which of the following are written in point-slope
2 answers:
You might be interested in
Answer:
dy/dx at x=6 is 0.334
Step-by-step explanation:
The center difference method requires that the values of the function are given in equal intervals which is the case, and allows one to find the value for x = 6 using those of the function for x = 5.5 and for x = 6.5 as follows:
TASK 1
a) Data and frequency table are attached in picture #1.
b) The histogram is attached in picture #2
c) As the intervals increase the number of pieces of mail received, the frequency decreases, therefore we can say that are received few pieces of mail <span>are received more often.
TASK 2
a) The list of breeds and weights are reported in picture #3.
</span>
In order to find the mean, sum up all the weights and divide by 9:
m = (<span>8 + 10 + 15 + 29 + 32 + 75 + 77 + 80 + <span>85) / 9 = 45.67 lbs
In order to find the median, you need to select the central value of your distribution: since you have 9 values, the central one will be in the fifth position:
M = 32 lbs
In order to find the mode, you need to select the value which shows more frequently: in this case, every value shows only once, therefore the mode cannot be determined.
The data are best described by the mean because they are almost symmetrically distributed between the least and the biggest values, while the median is more towards the small-weight side of the distribution.
b) In order to have an average of 250 lbs, the tenth dog should weight 2089 lbs.
Indeed, if the average is 250 lbs for 10 dogs, it means that the total sum of their weights is 250 × 10 = 2500 lbs. We know that the sum of the first 9 dogs is 411 lbs, therefore, the tenth dog should weight 2500 - 411 = 2089 lbs.
TASK 3
The city picked is Baltimore, the year 2017.
The table is attached in picture #4.
</span></span>
a) The box-and-whisker plot is attached in picture #5.
b) The median is the central value of the distribution, which is: <span>
42.4, 45.3, 46.8 | 54.7, 57.6, 66.0 | 69.1, 76.5, 80.6 | 85.8, 87.8, 90.0 </span>
where we marked with a tally the quartiles.
Since we have 12 values, the median will be the average between the two central values:
M = (66.0 + 69.1) / 2 = 67.1 °F
c) 7<span>5% of the temperatures are below 83.2 °F.
Indeed, this value represents the third quartile. The position of the third quartile can be found by the formula:
3/4 </span>· (n + 1) = 3/4 · (12 + 1) = 3/4 <span>· 13 = 9.75
Therefore, since we did not get in integer position, the third quartile will be the average between the numbers in position 9 and 10:
q</span>₀.₇₅ = (80.6 + 85.8) / 2 = 83.2 °F
d) 7<span>5% of the temperatures are above 50.8 °F.
Indeed, this value represents the first quartile. Similarly to point c), the position can be found by the formula:
1</span>/4 · (n + 1) = 1/4 <span>· 13 = 3.25
And therefore:
</span>q₀.₂₅ = (<span><span><span>46.8 + </span><span>54.7) / 2 = 50.8 °F.</span></span></span>
e) Baltimore in 2017 had a median high temperature of 67.1 °F.
25% of the year the high temperatures were warmer than 83.2 °F, with no outlier above 90.0 °F which was the hottest temperature.
25% of the high temperatures were colder than 50.8 °F, with no outlier below 42.4 °F, which was the coldest high temperature.
a) Data and frequency table are attached in picture #1.
b) The histogram is attached in picture #2
c) As the intervals increase the number of pieces of mail received, the frequency decreases, therefore we can say that are received few pieces of mail <span>are received more often.
TASK 2
a) The list of breeds and weights are reported in picture #3.
</span>
In order to find the mean, sum up all the weights and divide by 9:
m = (<span>8 + 10 + 15 + 29 + 32 + 75 + 77 + 80 + <span>85) / 9 = 45.67 lbs
In order to find the median, you need to select the central value of your distribution: since you have 9 values, the central one will be in the fifth position:
M = 32 lbs
In order to find the mode, you need to select the value which shows more frequently: in this case, every value shows only once, therefore the mode cannot be determined.
The data are best described by the mean because they are almost symmetrically distributed between the least and the biggest values, while the median is more towards the small-weight side of the distribution.
b) In order to have an average of 250 lbs, the tenth dog should weight 2089 lbs.
Indeed, if the average is 250 lbs for 10 dogs, it means that the total sum of their weights is 250 × 10 = 2500 lbs. We know that the sum of the first 9 dogs is 411 lbs, therefore, the tenth dog should weight 2500 - 411 = 2089 lbs.
TASK 3
The city picked is Baltimore, the year 2017.
The table is attached in picture #4.
</span></span>
a) The box-and-whisker plot is attached in picture #5.
b) The median is the central value of the distribution, which is: <span>
42.4, 45.3, 46.8 | 54.7, 57.6, 66.0 | 69.1, 76.5, 80.6 | 85.8, 87.8, 90.0 </span>
where we marked with a tally the quartiles.
Since we have 12 values, the median will be the average between the two central values:
M = (66.0 + 69.1) / 2 = 67.1 °F
c) 7<span>5% of the temperatures are below 83.2 °F.
Indeed, this value represents the third quartile. The position of the third quartile can be found by the formula:
3/4 </span>· (n + 1) = 3/4 · (12 + 1) = 3/4 <span>· 13 = 9.75
Therefore, since we did not get in integer position, the third quartile will be the average between the numbers in position 9 and 10:
q</span>₀.₇₅ = (80.6 + 85.8) / 2 = 83.2 °F
d) 7<span>5% of the temperatures are above 50.8 °F.
Indeed, this value represents the first quartile. Similarly to point c), the position can be found by the formula:
1</span>/4 · (n + 1) = 1/4 <span>· 13 = 3.25
And therefore:
</span>q₀.₂₅ = (<span><span><span>46.8 + </span><span>54.7) / 2 = 50.8 °F.</span></span></span>
e) Baltimore in 2017 had a median high temperature of 67.1 °F.
25% of the year the high temperatures were warmer than 83.2 °F, with no outlier above 90.0 °F which was the hottest temperature.
25% of the high temperatures were colder than 50.8 °F, with no outlier below 42.4 °F, which was the coldest high temperature.