The length of a rectangle is 3 in longer than its width.
1 answer:
You might be interested in
Answer:
1) False, to begin converting the equation to standard form, each side must be added by 36. 2) True, to complete the square for the x terms, add 4 to both sides, 3) True, the center of the circle is at (-2, 3), 4) False, the center of the circle is at (-2, 3), 5) False, the radius of the circle is 7 units, 6) False, the radius of the circle is 7 units.
Step-by-step explanation:
Let prove the validity of each choice:
1) To begin converting the equation to standard form, subtract 36 from both sides:
Let consider the following formula and perform the following algebraic operations:
(i) Given
(ii) Commutative Property
(iii) Modulative/Associative Property/Additive Inverse Existence
(iv) Perfect Trinomial Square
(v) Commutative Property/Compatibility with Addition/Additive Inverse Existence/Modulative Property
(vi) Definition of Addition/Result
False, to begin converting the equation to standard form, each side must be added by 36.
2) True, step (iii) on exercise 1) indicates that both side must be added by 4.
3) The general equation for a circle centered at (h, k) is of the form:
Where is the radius of the circle.
By direct comparison, it is evident that circle is centered at (-2,3).
True, step (vi) on exercise 1) indicates that center of the circle is at (-2, 3).
4) False, step (vi) on exercise 1) indicates that the center of the circle is at (-2, 3), not in (4, -6).
5) After comparing both formulas, it is evident that radius of the circle is 7 units.
False, the radius of the circle is 7 units.
6) After comparing both formulas, it is evident that radius of the circle is 7 units.
False, the radius of the circle is 7 units.
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.