Answer:
Extensive property
Step-by-step explanation:
The intensive properties does not depend on the amount of mass of the system or the size of the system, for example the density [tex]\rho[/tex] is a quantity that is already defined for the system, because the density of water is equal for a drop or for a pool of water.
In the case of extensive properties the value of them is proportional to the size or mass of the system. For example mass is an extensive property because depend on the amount of substance. Other example could be the enegy.
In the case of weigth is a quantity that depends on mass value, then weigth is an extensive property.
Let A and B be sets. Prove that (a) ACB AUB=B; (b) ACB = AnB = A.
For part (a) and (b) suposse [tex]A\subset B[/tex].
To prove part (a) observe that we already have that [tex]B\subset A\cup B[/tex]. So we will prove that [tex]A\cup B \subset B[/tex]. Let [tex]x\in A\cup B[/tex], then [tex]x\in A[/tex] or [tex]x\in B[/tex]. If [tex]x\in B[/tex] we finish the proof, and if [tex]x\in A[/tex] implies [tex]x\in B[/tex] because we assume [tex]A\subset B[/tex], and the proof is complete.
For part (b) we always have [tex]A\cap B\subset A[/tex]. We finish the proof showing [tex]A\subset A\cap B[/tex]. Let [tex]x\in A[/tex], then [tex]x\in B[/tex] by the asumption that [tex]A\subset B[/tex]. So, we have both [tex]x\in A[/tex] and [tex]x\in B[/tex], that implies [tex]x\in A\cap B[/tex]. Therfore [tex]A\subset A\cap B[/tex], which completes the proof.
Suppose the transit system of a large city has a lost and found department that keeps detailed records, including the estimated value in dollars of each lost item. Each year the department publishes a report with a histogram showing the proportion of items lost in various price ranges.
The bar on the histogram with endpoints 100 to 200 has a height of 0.11. What does this mean?
(A) of the lost items, 11% cost $150.
(B) Between 100 and 200 items had an estimated value of 11% of the total cost of all items.
(C) Of the lost items, 11% cost between $100 and $200.
(D) It is impossible to answer the question because it does not say what variable is displayed on the horizontal axis.
(E) Of the lost items, 11% had an average cost of $150.
Answer:
(C) Of the lost items, 11% cost between $100 and $200
Step-by-step explanation:
On the Y-axis is represented the frequency of lost items found.
In this case in the form of a percentage.
On the X-axis is represented the range of prices
See picture attached.
So, this means that 11% of the found items have an estimated price between $100 and $200
T F Any set of vectors containing the zero vector is linearly dependent
Answer:
True
Step-by-step explanation:
Let B={[tex]v_1, v_2,\cdots,v_k,0[/tex]} a set of vectors containing the zero vector.
B is linear dependent if exist scalars [tex]a_0,a_1, a_2,\cdots,a_k[/tex] not all zero such that [tex]a_0*0+a_1*v_1+\cdots+a_k*v_k=0[/tex].
Observe that if [tex]a_1=a_2=\cdots=a_k=0[/tex] and [tex]a_0=\lambda\neq 0[/tex] then
[tex]\lambda*0+0*v_1+\cdots+0*v_k=0[/tex].
Then B is linear dependent.
Final answer:
True, any set of vectors that contains the zero vector is linearly dependent because a nontrivial combination of the vectors (including the zero vector scaled by any scalar) can produce the null vector.
Explanation:
True. Any set of vectors containing the zero vector is linearly dependent. This is because the null vector, which has all components equal to zero (0î + 0ç + 0k) and therefore no length or direction, can always be represented as a linear combination of other vectors in the set by simply scaling it by any scalar. In more technical terms, for a set of vectors that includes the zero vector, you can find a nontrivial combination (not all scalars being zero) such that when these scalars are applied to their respective vectors, their sum equals the zero vector. This is the very definition of linear dependency. For instance, if you consider any scalar 'a' not equal to zero, a∗0 + 0∗0 + ... + 0∗0 = 0, which demonstrates linear dependence because not all the scalars are zero (namely 'a' is not).
Linear independence would require that the only way to represent the zero vector as a linear combination of the set of vectors is to have all scalars that multiply each vector in the combination be zero. Since the set includes the zero vector itself, this condition is violated, and the set is dependent.
4. Suppose a security light is set into the ground and aims its light at the top of the building. The angle of elevation of the light is 60 degrees. The building is 9 feet tall. a. What is the distance of the light's base from the bottom of the building? Use a trig function to calculate the distance. b. How long is the light beam? light beam. Use a trig function to calculate the length of the
Answer:
a) The distance of the light's base from the bottom of the building is approximately: 5.2 ft
b) The length of the beam is approximately: 10.4 ft
Step-by-step explanation:
First, we have to recognize that we may draw a right triangle to picture our problem. Then, in order to find out the distance of the light's base from the bottom of the building, we need to use the tangent trigonometric function:
tan(angle) = opposite side / adjacent side
We know the angle and the opposite side and we want to find the adjacent side:
adjacent side = opposite side / tan(angle) = 9 ft / tan(60°) = 9 ft / = 9 ft / 1.73 = 5.2 ft
In order to find the length of the light beam, we use Pythagoras Theorem:
leg1²+leg2² = hyp²
Since the length of the beam corresponds to the hypotenuse and since we already know the length of the two legs, it is just a matter of substituting the values:
hyp = square_root(leg1²+leg2²) = square_root(9² + 5.2²) ft = square_root(108.4) ft = 10.4 ft
Three pirates must divide 100 gold doubloons. The doubloons cannot be cut into pieces. Pirate A is the strongest, followed by Pirate B, followed by Pirate C. Because of ancient pirate tradition, the coins are divided in the following manner. First Pirate A proposes a division of the coins. The three pirates then vote on whether to accept the proposed division. If the proposal gets a majority vote, it is accepted, and the game is over. If the proposal fails to get a majority vote, Pirate A is executed. It is then Pirate B's turn to propose a division of the coins between the two remaining pirates. The same rules apply, with one exception: if the vote is a tie, Pirate B, being the strongest remaining pirate, gets an additional vote to break the tie.
If we assume that in any proposal there are no doubloons left over, how many different proposals could Pirate A make?
Answer:
[tex]C^{102}_{100}=5151[/tex]
Step-by-step explanation:
Let's imagine the following situation, if we want to distribute 100 coins between three pirates we could represent this situation with a line arrangement. For example if we had7 coins and 3 pirates one possible distribution of coins would be given by CC|CCCC|C, the C's represent coins and the bars the boundaries between two pirates, for the particular line arrangement shown, we have that pirate A has 2 coins, B has 4 coins and C has a single coin. Another possible arrangement is,
|CCC|CCCC, where pirate A has no coin, pirate B has 3 coins and C has 7 coins. If we take notice of the fact that the arrangement representing a distribution is composed of 9 elements, that is 7 C's and 2 | (bars), then a way to make an arrangement would be to fill 9 empty boxes with our available coins and bars in all the possible ways. This means that if we first choose to fill 7 out of 9 boxes with coins then the number of possible combinations is [tex]C^{9}_7=\frac{9!}{7!(9-7)!}36[/tex]. In general if we want to distribute n elements in k boxes, where the boxes can either be filled with any number of elements (including 0 number of elements), we have that the number of possible distributions will be [tex]C^{n+k-1}_{n}=\frac{(n+k-1)!}{n!(k-1)!}[/[tex], where we used the fact that we need k-q bars to represent k boxes. Thus pirate A can choose from [tex]C^{102}_{100}=5151[/tex] possible divisions.
Bonus:
If every pirate wants to have the maximum number of coins possible without being executed, here's how pirate A has to divide the coins in order to keep the largest amount of coins.
We have to think backwards to figure this out. Imagine pirate A was executed and there are only two remaining players. Pirate B should propose to keep all the coins, pirate C could oppose but pirate B's vote would break the vote and keep all the loot. Pirate A, B and C are all aware of this, so pirate A should propose to keep 99 coins and give the remaining gold piece to pirate C, Pirate B will of course oppose the division, but pirate C should accept because if not he would get no coins. Thus the division would be.
A: 99 coins
B: 0 coins
C: 1 coins
What are the values of magnitude ( r) and angle ( θ) for these complex numbers in polar form? a) (5 + 4)(4 +5j) and b) (5 +4j)/(4 +5) Write your answers here a) r b) r=
Answer:
(a) r = 57.628
θ = 51.34°
(b) r = 0.711
θ = 38.65°
Step-by-step explanation:
(a) Let's assume
z = (5 + 4)(4 +5j)
= 9 x (4+5j)
= 36 + 45j
magnitude of z can be given by
[tex]\left | z \right |\ =\ r\ =\sqrt{36^2+45^2}[/tex]
=> r =57.628
angle of z can be given by,
[tex]tan\theta\ =\ \dfrac{45}{36}[/tex]
[tex]=>\ tan\theta=\ \dfrac{5}{4}[/tex]
[tex]=>\ \theta\ =\ tan^{-1}\dfrac{5}{4}[/tex]
θ = 51.34°
(b) Let's assume
[tex]z =\ \dfrac{(5 +4j)}{5+4}[/tex]
[tex]=\ \dfrac{(5+4j)}{9}[/tex]
[tex]=\ \dfrac{5}{9}+\dfrac{4i}{9}[/tex]
magnitude of z can be given by
[tex]\left | z \right |=\ r=\sqrt{(\dfrac{5}{9})^2+(\dfrac{4}{9})^2}[/tex]
[tex]=> r =\dfrac{\sqrt{41}}{9}[/tex]
=> r = 0.711
angle of z can be given by,
[tex]\ tan\theta\ =\ \dfrac{\dfrac{4}{9}}{\dfrac{5}{9}}[/tex]
[tex]=>\ tan\theta=\ \dfrac{4}{5}[/tex]
[tex]=>\ \theta\ =\ tan^{-1}\dfrac{4}{5}[/tex]
=> θ = 38.65°
Ronna has volleyball practice every 4 days. Ronna also has violin lessons every 10 days. She has both activities today after school. When will she have both activities again on the same day?
Answer:
She have both activities again on 20th day from now.
Step-by-step explanation:
Given :Ronna has volleyball practice every 4 days
Ronna also has violin lessons every 10 days.
To Find : When will she have both activities again on the same day?
Solution:
We will find the LCM of 4 and 10
2 | 4 ,10
2 | 2 , 5
5 | 1 , 5
| 1 , 1
So, [tex]LCM = 2\times 2 \times 5[/tex]
[tex]LCM =20[/tex]
Hence she have both activities again on 20th day from now.
Suppose we approximate the value of 1/128 by rounding to 3 decimal places. a) What would be the approximation? b) What is the number of significant digits in the approximation? c) What is the absolute error? d) What is the relative error?
Answer:
solved
Step-by-step explanation:
Calculating i found the value of [tex]\frac{1}{128}[/tex]= 0.0078125
a.therefore the approximate valve of [tex]\frac{1}{128}[/tex] (correct to three decimal places) = 0.008
b. there are three significant digits in my approximation.
c. the absolute error = | exact value- approximate value |
=|0.0078125- 0.008|= |-0.0001875| = 0.0001875
d.relative error= [tex]\frac{ absolute\ error}{exact\ value}[/tex]
= [tex]\frac{0.0001875}{0.0078125}[/tex]= 0.024
a kitchen is 10ft long and 8ft wide.if kitchen floor tiles
are 2.5in by 3in, how many tiles are needed for the
kitchen?
Answer:
1536 tiles
Step-by-step explanation:
Given,
The length of kitchen = 10 ft,
Width of the kitchen = 8 ft,
∵ The shape of floor of kitchen must be rectangular,
Thus, the area of the kitchen = 10 × 8 = 80 square ft = 11520 in²
( Since, 1 square feet = 144 square inches )
Now, the length of a tile = 2.5 in and its width = 3 in,
So, the area of each tile = 2.5 × 3 = 7.5 in²,
Hence, the number of tiles needed = [tex]\frac{\text{Area of kitchen}}{\text{Area of each tile}}[/tex]
[tex]=\frac{11520}{7.5}[/tex]
= 1536
Approximately 1536 tiles are needed to cover a kitchen floor that is 10 feet by 8 feet using tiles that are 2.5 inches by 3 inches.
To determine how many tiles are needed to cover a kitchen floor that is 10 feet long and 8 feet wide with tiles that are 2.5 inches by 3 inches, we first need to convert the dimensions of the kitchen to inches because the tile dimensions are given in inches. There are 12 inches in a foot, so the kitchen is 120 inches long (10 feet × 12 inches/foot) and 96 inches wide (8 feet × 12 inches/foot).
We now calculate the area of the kitchen floor: Area = Length × Width = 120 inches × 96 inches = 11520 square inches.
Next, we calculate the area of one tile: Tile Area = Length × Width = 2.5 inches × 3 inches = 7.5 square inches.
Now, to find the number of tiles needed, we divide the total area of the kitchen floor by the area of one tile: Number of Tiles = Kitchen Area / Tile Area = 11520 square inches / 7.5 square inches ≈ 1536 tiles.
Therefore, approximately 1536 tiles are needed for the kitchen.
The average weight of an airline passenger’s suitcases is 45 pounds with a standard deviation of 2 pounds. If 16% of the suitcases are overweight, find the maximum weight allowed by the airline. Assume the variable is bell – shaped distributed. Hint: Sketch and label a normal model to help answer the question.
Answer:
we over weighted suitcase will be 46.9889 pound
Step-by-step explanation:
We have given the average weight of passenger's suitcase [tex]\mu =45\ pounds[/tex]
Standard deviation [tex]\sigma =2\ pounds[/tex]
16 % of the suitcase are overweight so 100-16 =84 % of the suitcase will pass without any problem
So probability [tex]p=0.84[/tex]
From z table at p =0.84 z will be z=0.99445
We know that [tex]z=\frac{x-\mu }{\sigma }[/tex]
[tex]0.99445=\frac{x-45}{2}[/tex]
x = 46.9889 pound
So we over weighted suitcase will be 46.9889 pound
Using standard deviation formula to solve the problem. Then the overweighted suitcase is 46.9889 pounds.
What is a standard deviation?It is the measure of the dispersion of statistical data. Dispersion is the extent to which the value is in a variation.
Given
The average [tex](\mu )[/tex] weight of an airline passenger’s suitcases is 45 pounds.
The standard deviation [tex](\sigma)[/tex] of 2 pounds.
16% of the suitcase are overweight. So,
[tex]100 - 16 = 84\%[/tex]
The suitcase will pass without any problem. So the probability will be
[tex]\rm P = 0.84[/tex]
From z table at P = 0.84 will be
z = 0.99445
We know that
[tex]\begin{aligned} z &= \dfrac{x - \mu}{\sigma}\\0.99445 &= \dfrac{x - 45}{2}\\x &= 46.9889\end{aligned}[/tex]
Thus, the overweighted suitcase is 46.9889 pounds.
More about the standard deviation link is given below.
https://brainly.com/question/12402189
Find the length of the vector V = || 2 3 1 ||- Ensure that you use at least 4 decimal place accuracy or exact values when entering your answer.
Answer: [tex]\sqrt{14}[/tex]
Step-by-step explanation:
The given notation is not clear enough, but I will asume that the question is: "Wich is the vector lenght of the vector [tex](2,3,1)[/tex] in a 3 dimensional euclidean space"
Vector lenght of a vector [tex](a,b,c)[/tex] is given by the next formula:
║[tex](a,b,c)[/tex]║=[tex]\sqrt{a^2+b^2+c^2}[/tex]
In this case we have a=2, b=3 and c=1. We should substitute these values in the formula given before to obtain:
║[tex](2,3,1)[/tex]║=[tex]\sqrt{2^2+3^2+1^2}=\sqrt{14}=3.74165[/tex]
Wich is the answer to the problem.
Find x if a line passes through the points ( 3 , -14 ) and ( x , 6 ) with a slope of -4
Answer: -8
Step-by-step explanation:
The slope of a line passing through two points (a,b) and (c,d) is given by:-
[tex]m=\dfrac{d-b}{c-a}[/tex]
Given : A line passes through the points ( 3 , -14 ) and ( x , 6 ) with a slope of -4, then we have
[tex]-4=\dfrac{6-(-14)}{x-3}\\\\\Rightarrow\ (x-3)(-4)=6+14\\\\\Rightarrow\ -4(x-3)=20\\\\\Rightarrow\ (x-3)=\dfrac{20}{-4}=-5\\\\\Rightarrow\ x-3=-5\\\\\Rightarrow\ x=-5-3=-8[/tex]
Hence, the value of x= -8
Consider the vectors x=(1,2,3,4) and y=(4,2,4,5) in R4 . Find the angle between them.
Answer:
Angle between the two vectors is 20.7718°.
Step-by-step explanation:
The given two vectors are x = (1,2,3,4) and y = (4,2,4,5)
Now the dot product between two vector is defined as
x.y = ║x║║y║cosθ
Now, x.y( dot product of x and y) = 1×4 + 2×2 + 3×4 + 4×5 = 40
║x║= [tex]\sqrt{1+4+9+16}[/tex] = [tex]\sqrt{30}[/tex]
║y║= [tex]\SQRT{16+4+16+25} = \sqrt{61}[/tex] '
Thus, we get
40 = [tex]\sqrt{30}[/tex]×[tex]\sqrt{61}[/tex]×Cos θ
Thus, Cos θ = [tex]\frac{40}{\sqrt{30}\sqrt{61}}[/tex]
Cos θ = 0.935
θ = [tex]Cos^{-1} (0.935)[/tex]
θ = 20.7718°
You track your commute times for three weeks (15 days), recording the following times in minutes: 17, 16, 20, 24, 22, 15, 21, 25, 17, 90, 16, 17, 22, 20, 20. Enter the data into a 3-by-5 matrix, where rows are weeks and columns are weekdays. Please answer the following questions and provide the necessary code. (a) Name each column with the weekday abbreviations, i.e., Mon, Tue, Wed, Thu, and Fri (b) Find the shortest, the longest, the average, and the median commute time within those three weeks. (c) The 90 was a mistake. It should have been 19. How can you fix this? Do so, and then find the new average and median. (d) How many times was your commute 20 minutes or more? (e) What percent of your commutes are less than 18 minutes long? (f) In average, which weekday you spend the most commute time?
Answer:
b) 15,90,24.133,20
c) New average = 19.4, New Median = 20
d) 8
e) 40%
f) Wednesday
Step-by-step explanation:
a) Mon Tue Wed Thu Fri
Week 1 17 16 20 24 22
Week 2 15 21 25 17 90
Week 3 16 17 22 20 20
This is a 3×5 matrix where the rows gives us the weeks(Week 1, Week 2, Week 3) and the columns are the weekdays(Mon, Tue, Wed, Thu, Fri).
b)The shortest duration out of the data given for fifteen days is 15 minutes.
The longest duration is 90 minutes.
To calculate the average of given data, we use the following formula:
Average = [tex]\frac{\text{Sum of all observations}}{\text{Total no of observations}}[/tex]
=[tex]\frac{17+16+20+24+22+15+21+25+17+90+16+17+22+20+20}{15}[/tex]
=[tex]\frac{362}{15}[/tex]
=24.133
Median = the value that divides the data into two equal parts]
Ascending order: 15,16,16,17,17,17,20,20,20,21,22,22,24,25,90
Median = [tex](\frac{n+1}{2})\text{th term}[/tex] = 8th term = 20
c) Mon Tue Wed Thu Fri
Week 1 17 16 20 24 22
Week 2 15 21 25 17 19
Week 3 16 17 22 20 20
New average = [tex]\frac{\text{Sum of new observations}}{\text{Total no of observations}}[/tex]
=[tex]\frac{17+16+20+24+22+15+21+25+17+19+16+17+22+20+20}{15}[/tex]
=[tex]\frac{362}{15}[/tex]
=19.4
New Median
Ascending order: 15,16,16,17,17,17,19,20,20,20,21,22,22,24,25
Median = [tex](\frac{n+1}{2})\text{th term}[/tex] = 8th term = 20
d) In order to know how many times we commute for 20 minutes or more, we check the entries that are greater than or equal to 20 minutes. We have 8 such entries. Thus, 8 times the commute was 20 minutes or more longer.
e) Percent of commutes less than 18 minutes = [tex]\frac{\text{Commutes that are less than 18 minutes}}{\text{Total number of commutes} }[/tex]× 100%
=[tex]\frac{6}{15}[/tex]×100% = 40%
f) Average = [tex]\frac{\text{Sum of all observations}}{\text{Total no of observations}}[/tex]
Average time spent in commuting on Monday = 16
Average time spent in commuting on Tuesday = 18
Average time spent in commuting on Wednesday = 22.3
Average time spent in commuting on Thursday = 20.3
Average time spent in commuting on Friday= 20.3
On Wednesday, we spent the most on commute time.
A manufacturer of pickup trucks is required to recall all the trucks manufactured in a given year for the repair of possible defects in the steering column and defects in the brake linings. Dealers have been notified that 3% of the trucks have defective steering only, and that 6% of the trucks have defective brake linings only. If 87% of the trucks have neither defect, what percentage of the trucks have both defects?
Answer: Probability of trucks having both defects is 4%.
Step-by-step explanation:
Since we have given that
Probability of trucks having defective brake linings only = 6%
Probability of trucks having defective steering only = 3%
Probability of trucks having neither defect = 87%
Probability of trucks having either defect - 100 - 87 = 13%
Let the probability of trucks having both defects be 'x'.
P(defective brake lining) = P(B) = 0.06+x
P(defective steering) = P(S) = 0.03+x
So, Probability of the trucks having both defects is given by
[tex]P(S\cup B)=P(S)+P(B)-P(S\cap B)\\\\0.13=0.03+x+0.06+x-x\\\\0.13-0.03-0.06=2x-x\\\\0.04=x[/tex]
So, Probability of trucks having both defects is 4%.
The percentage of trucks with both defects is 0%.
Explanation:To solve this problem, we will use the principle of inclusion-exclusion. Let's assume that the percentage of trucks with both defects is x. The percentage of trucks with only defective steering is 3%, the percentage of trucks with only defective brake linings is 6%, and the percentage of trucks with neither defect is 87%. We want to find the percentage of trucks with both defects.
According to the principle of inclusion-exclusion, the total percentage of trucks with at least one defect is given by:
Total = Percentage with defective steering + Percentage with defective brake linings - Percentage with both defects
Using the given percentages, we can set up the equation:
100% = 3% + 6% - x
Simplifying the equation, we get:
x = 9% - 100% = -91%
Since a negative percentage does not make sense in this context, there are no trucks with both defects.
"I fear not the man who has practiced 10,000 kicks once, but I fear the man who has practiced one kick 10,000 times."
what can you infer from this quote related to mathematics practice.
when considering the quote, indicate how you think this relates to how you study and how you practice math.
Answer:
Step-by-step explanation:
Given is a phrase
"I fear not the man who has practiced 10,000 kicks once, but I fear the man who has practiced one kick 10,000 times."
This implies instead of attempting so many things without depth one thing indepth training is very much good and feared by others.
This applies to knowledge and also the subject Mathematics
Instead of practising many sums without getting answers, learning one concept and trying to get the difficult problem will help go a long way in learning.
In depth learning helps to master other allied concepts also very easily and this helps extending to other topics
Determine the greatest mass among the following measurements.
a) 2.0 X106 milli gram b) 5.0X1012 pico gram c) 9.0X109 micro gram and d) 2.0X102 kilo gram
b) 5.0X1012 pico gram
d) 2.0X102 kilo gram
c) 9.0X109 micro gram
a) 2.0 X106 milli gram
Answer: [tex]d) 2.0*10^2[/tex]kilo gram
Step-by-step explanation:
To know the greatest mass we have to convert all these measurements to the same unit, the gram. For that we have to state the next units:
[tex]mili = 10^-^3\\micro = 10^-^6\\pico = 10^-^1^2\\kilo= 10^3[/tex]
Now we can convert the measurements to the same unit:
[tex]b)5.0*10^1^2*10^-^1^2gram=5.0gram\\d)2.0*10^2*10^3gram=2.0*10^5gram\\c)9.0*10^9*10^-^6gram=9.0*10^3gram\\a)2.0*10^6*10^-^3gram=2.0*10^3gram[/tex]
Comparating the values, we can see that the option d) is the greatest mass.
Answer:
The greatest mass among the following measurements is:
[tex]2*10^{2}kg = 200kg[/tex]
Step-by-step explanation:
I am going to convert each one of these measurements to kilogram, then the biggest value in kg is the greatest mass.
a) [tex]2*10^{6}mg[/tex]
Each kg has [tex]10^{6} kg[/tex].
So
1 kg - [tex]10^{6} kg[/tex]
x kg - [tex]2*10^{6}mg[/tex]
x = 2.
So
[tex]2*10^{6}mg = 2kg[/tex]
b)[tex]5*10^{12}pg[/tex]
Each kg has [tex]10^{15}pg[/tex].
So
1 kg - [tex]10^{15}pg[/tex]
x kg - [tex]5*10^{12}pg[/tex]
[tex]x = 5*10^{-3}kg[/tex]
This is a smaller mass than a)
c) [tex]9.0*10^{9} mcg[/tex]
Each kg has [tex]10^{9}mcg[/tex].
So
1 kg - [tex]10^{9}mcg[/tex]
x kg - [tex]9*10^{9} mcg[/tex]
x = 9kg.
[tex]9.0*10^{9} mcg = 9kg[/tex]
For now, this is the greatest mass.
d) [tex]2*10^{2}kg = 200kg[/tex]
This is the greatest mass.
A pharmacist attempts to weigh 120 mg of codeine sulfate on a balance with a sensitivity requirement of 6 mg. Calculate the maximum potential error in terms of percentage.
Answer:
5%.
Step-by-step explanation:
We have been given that a pharmacist attempts to weigh 120 mg of codeine sulfate on a balance with a sensitivity requirement of 6 mg.
To find the maximum potential error, we need to figure out 6 mg is what percentage of 120 mg.
[tex]\text{The maximum potential error}=\frac{6}{120}\times 100\%[/tex]
[tex]\text{The maximum potential error}=0.05\times 100\%[/tex]
[tex]\text{The maximum potential error}=5\%[/tex]
Therefore, the maximum potential error is 5%.
Find m∠ABC. 20 POINTS AND BRAINLIEST FOR CORRECT ANSWER
A. 10°
B. 21°
C. 37°
D. 53°
The measure of angle ABC in the figure is 53 degrees.
Option D) 53° is the correct answer.
From the figure in the image;
Angle ABC and angle CBD are complementary angles, hence, their sum equals 180 degrees.
From the figure:
Angle ABC = ( 5x + 3 )
Angle CBD = ( 3x + 7 )
First, we solve for x:
Since the two angles are complementary:
Angle ABC + Angle CBD = 90
( 5x + 3 ) + ( 3x + 7 ) = 90
5x + 3x + 3 + 7 = 90
8x + 10 = 90
8x = 90 - 10
8x = 80
x = 80/8
x = 10
Now, we find the measure of angle ABC:
Angle ABC = ( 5x + 3 )
Plug in x = 10:
Angle ABC = 5(10) + 3
Angle ABC = 50 + 3
Angle ABC = 53°
Therefore, angle ABC measures 53 degrees.
The correct option is D) 53°
Suppose that you are organizing your texts for the semester at your desk consist-ing of 3 Computer Science, 2 Mathematics, and 2 Sociology textbooks. Assume alltextbooks are distinct. In how many ways can this be done ifa) the books can be arranged in any order?b) the Computer Science books must be together, but all other books can bearranged in any order?c) the Mathematics books and the Computer Science books must be together?
Answer : (a) 5040 b) 720 c) 288
Step-to step explanation:-
Given : There are 3 Computer Science, 2 Mathematics, and 2 Sociology textbooks.
Total books = 7
a) The number of ways the books can be arranged in any order is given by :-
[tex]7!=7\times6\times5\times4\times3\times2=5040[/tex]
b) We consider all computer books as one thing, then, the total number of things to arrange = 1+2+2=5
Now, the number of ways to arrange books such that Computer Science books must be together, but all other books can be arranged in any order:-
[tex]3!\times5!\\\\=6\times120=720[/tex]
c) We consider all computer books as one thing and math book as one thing, then, the total number of things to arrange = 1+1+2=4
Now, the number of ways to arrange books such that Computer Science books must be together, but all other books can be arranged in any order:-
[tex]4!\times 3!\times2!=24\times6\times2=288[/tex]
Communicate Mathematical Ideas Irrational numbers can never be precisely represented in decimal form. Why is this? 20 Unit1
Answer:
An irrational number can never be represented precisely in decimal form.
Step-by-step explanation:
A number can be precisely represented in decimal form if you can give a rule for the construction of its decimal part,
for example:
2.246973973973973... (an infinite tail of 973's repeated over and over)
7.35 (a tail of zeroes)
If this the case, then the number is a RATIONAL NUMBER, i.e, the QUOTIENT OF TWO INTEGERS.
Let's show this for the first example and then a way to show the general situation will arise naturally.
Suppose N = 2.246973973973...
You can always multiply by a suitable power of 10 until you get a number with only the repeated chain in the tail
for example:
1) [tex]N.10^3=2246.973973973...[/tex]
but also
2) [tex]N.10^6=2246973.973973...[/tex]
Subtracting 1) from 2) we get
[tex]N.10^6-N.10^3=2246973.973973973... - 2246.973973973...[/tex]
Now, the infinite tail disappears
[tex]N.10^6-N.10^3=2246973 - 2246=226747[/tex]
But
[tex]N.10^6-N.10^3=N.(10^6-10^3)=N.(1000000-1000)=999000.N[/tex]
We have then
999000.N=226747
and
[tex]N=\frac{226747}{999000}[/tex]
We do not need to simplify this fraction, because we only wanted to show that N is a quotient of two integers.
We arrive then to the following conclusion:
If an irrational number could be represented precisely in decimal form, then it would have to be the quotient of two integers, which is a contradiction.
So, an irrational number can never be represented precisely in decimal form.
If you had 454g of shrimp/sq ft in a tank with an 11 sq meter bottom area, how many total kilograms of shrimp do you have?
Answer:
53.69 kilograms.
Step-by-step explanation:
You had 454 grams of shrimp per square feet in a tank.
Converting square feet in square meter.
1 square foot = 0.092903 square meter
Means you had 454 grams of shrimp per 0.093 square meter in a tank.
The bottom area of the tank is 11 square meter.
So, grams of shrimp in 11 square meter = [tex]\frac{454\times11}{0.093}[/tex]
= 53698.92 grams
In kg it will be [tex]\frac{53698.92}{1000}[/tex] = 53.69 kg
240 ml of orange juice (1 cup) provides about 110 kcal. How many kcalories are in 500 ml of orange juice?
Answer:
229.17 kilo-calories
Step-by-step explanation:
We have been given that 240 ml of orange juice (1 cup) provides about 110 kcal.
First of all, we will divide 500 by 240 to find amount of cups in 500 ml.
[tex]\text{Number of cups}=\frac{500}{240}[/tex]
[tex]\text{Number of cups}=2.08333[/tex]
Now, we will multiply 2.08333 by 110 to find amount of k-cal in orange juice.
[tex]\text{Number of k-cal}=2.08333\times 110\text{ k-cal}[/tex]
[tex]\text{Number of k-cal}=229.16666\text{ k-cal}[/tex]
[tex]\text{Number of k-cal}\approx 229.17\text{ k-cal}[/tex]
Therefore, there are approximately 229.17 kilo-calories in 500 ml of orange juice.
Final answer:
To find the number of kcalories in 500 ml of orange juice, set up a proportion using the known ratio of 240 ml to 110 kcal. Cross multiply and solve for the unknown value to find that there are approximately 229 kcalories in 500 ml of orange juice.
Explanation:
To calculate the number of kcalories in 500 ml of orange juice, we can use the ratio method. First, we know that 240 ml (1 cup) of orange juice provides 110 kcal. We can set up a proportion to find the number of kcalories in 500 ml of orange juice.
240 ml / 110 kcal = 500 ml / x kcal
Cross multiplying, we get 240x = 55000. Dividing both sides by 240, we find that x is approximately equal to 229.17. Therefore, there are about 229 kcalories in 500 ml of orange juice.
Find a parametric equation for the line through the point ( - 5, - 6, - 6) and perpendicular to the plane 2x + 3y + 9z = 17 which is written using the coordinates of the given point and the coefficients of x, y, and z in the given equation of the plane. Type expressions using t as the variable.)
Answer:
x=-5+2t, y=-6+3t, z=-6+9t
Step-by-step explanation:
We need to find parametric equations for the line through the point ( - 5, - 6, - 6) and perpendicular to the plane 2x + 3y + 9z = 17 .
If a line passes through a point [tex](x_1,y_1,z_1)[/tex] and perpendicular to the plane [tex]ax+by+cz=d[/tex], then the Cartesian form of line is
[tex]\frac{x-x_1}{a}=\frac{y-y_1}{b}=\frac{z-z_1}{c}[/tex]
and parametric equations are
[tex]x=x_1+at,y=y_1+bt,z=z_1+ct[/tex]
For the given information, [tex]x_1=-5,y_1=-6,z_1=-6,a=2,b=3,c=9[/tex]. So, the Cartesian form of line is
[tex]\frac{x+5}{2}=\frac{y+6}{3}=\frac{z+6}{9}[/tex]
Parametric equations of the line are
[tex]x=-5+2t[/tex]
[tex]y=-6+3t[/tex]
[tex]z=-6+9t[/tex]
Therefore the parametric equation for the line are x=-5+2t, y=-6+3t, z=-6+9t.
Final answer:
The parametric equations for the line are found using the point (-5, -6, -6) and the normal vector of the plane (2, 3, 9) as the direction vector, resulting in the equations x = -5 + 2t, y = -6 + 3t, and z = -6 + 9t.
Explanation:
To find a parametric equation for the line through the point (-5, -6, -6) and perpendicular to the plane 2x + 3y + 9z = 17, we use the fact that the normal vector of the plane (2, 3, 9) gives the direction of the line that is perpendicular to the plane. The general form of a parametric equation of a line is x = [tex]x_{0}[/tex] + at, y = [tex]y_{0}[/tex] + bt, and z = [tex]z_{0}[/tex] + ct, where [tex](x_{0},y_{0},z_{0})[/tex] is a point on the line and (a, b, c) is the direction vector of the line. In this case, the point given is (-5, -6, -6) and the direction vector is (2, 3, 9), so the parametric equations of the line are x = -5 + 2t, y = -6 + 3t, and z = -6 + 9t.
The students in Lucy's grade voted to select a guest speaker. 25 students voted for a famous athlete. The other 75 students in Lucy's grade voted for a famous actor. What percentage of the students voted for the athlete?
Write your answer using a percent sign (%).
Final answer:
To find the percentage of students who voted for the athlete, divide the number who voted for the athlete (25) by the total number of students (100), and multiply by 100%. Therefore, 25% of the students voted for the athlete.
Explanation:
To calculate the percentage of students who voted for the athlete, we divide the number of students who voted for the athlete by the total number of students who voted, and then multiply the result by 100 to convert it to a percentage.
Total number of students = 25 (voted for the athlete) + 75 (voted for the actor) = 100 students
Now, to find the percentage who voted for the athlete:
Percentage voting for the athlete = (Number of students voting for the athlete ÷ Total number of students) × 100%
Percentage voting for the athlete = (25 ÷ 100) × 100%
Percentage voting for the athlete = 0.25 × 100%
Percentage voting for the athlete = 25%
25% of the students voted for the athlete.
A space is totally disconnected if its connected spaces are one-point-sets.Show that a finite Hausdorff space is totally disconnected.
Step-by-step explanation:
If X is a finite Hausdorff space then every two points of X can be separated by open neighborhoods. Say the points of X are [tex]x_1, x_2, ..., x_n[/tex]. So there are disjoint open neighborhoods [tex]U_{12}[/tex] and [tex]U_2[/tex], of [tex]x_1[/tex] and [tex]x_2[/tex] respectively (that's the definition of Hausdorff space). There are also open disjoint neighborhoods [tex]U_{13}[/tex] and [tex]U_3[/tex] of [tex]x_1[/tex] and [tex]x_3[/tex] respectively, and disjoint open neighborhoods [tex]U_{14}[/tex] and [tex]U_4[/tex] of [tex]x_1[/tex] and [tex]x_4[/tex], and so on, all the way to disjoint open neighborhoods [tex]U_{1n}[/tex], and [tex]U_n[/tex] of [tex]x_1[/tex] and [tex]x_n[/tex] respectively. So [tex]U=U_2 \cup U_3 \cup ... \cup U_n[/tex] has every element of [tex]X[/tex] in it, except for [tex]x_1[/tex]. Since [tex]U[/tex] is union of open sets, it is open, and so [tex]U^c[/tex], which is the singleton [tex]\{ x_1\}[/tex], is closed. Therefore every singleton is closed.
Now, remember finite union of closed sets is closed, so [tex]\{ x_2\} \cup \{ x_3\} \cup ... \cup \{ x_n\}[/tex] is closed, and so its complemented, which is [tex]\{ x_1\}[/tex] is open. Therefore every singleton is also open.
That means any two points of [tex]X[/tex] belong to different connected components (since we can express X as the union of the open sets [tex] \{ x_1\} \cup \{ x_2,...,x_n\}[/tex], so that [tex]x_1[/tex] is in a different connected component than [tex]x_2,...,x_n [/tex], and same could be done with any [tex]x_i[/tex]), and so each point is in its own connected component. And so the space is totally disconnected.
Lemma 2.3.2 (Multiplication is commutative). Let n,m be natural питbers. hen n x m = mхп.
Answer:
Multiplication is commutative.
Step-by-step explanation:
Let n,m be two natural numbers.
We have to show that [tex]n\times m=m\times n[/tex]
We will prove this by induction
For n = 1, we have [tex]n\times 1 = 1\times n = n[/tex], which is true.
Let the given statement be true for n-1 natural numbers, that is,
[tex](n-1)\times m = m\times (n-1)[/tex]
Now, we have to show that it is true for n natural numbers.
[tex]nm = (n-1)m + m = m(n-1) + m[/tex]
which is equal to
[tex](m-1)(n-1) + (n-1) + m[/tex]
[tex]= (m-1)(n-1) + (m-1) + n[/tex]
[tex]= n(m-1) + n[/tex]
[tex]= (m-1)n + n[/tex]
[tex]= mn[/tex]
Hence, by principal of mathematical induction, the given statement is true for all natural numbers.
f the equation of a circle is (x + 4)2 + (y - 6)2 = 25, its radius is 5 10 25
Answer:
5 = radius
Step-by-step explanation:
According to one of the Circle Equations, (X - H)² + (Y - K)² = R², you take the square root of your squared radius:
√25 = √R²
5 = R
* When we are talking about radii and diameters, we want the NON-NEGATIVE root.
I am joyous to assist you anytime.
Find the roots of the equation f(x) = x3 - 0.2589x2 + 0.02262x -0.001122 = 0
Answer:
The root of the equation [tex]x^3-0.2589x^{2}+0.02262x-0.001122=0[/tex] is x ≈ 0.162035
Step-by-step explanation:
To find the roots of the equation [tex]x^3-0.2589x^{2}+0.02262x-0.001122=0[/tex] you can use the Newton-Raphson method.
It is a way to find a good approximation for the root of a real-valued function f(x) = 0. The method starts with a function f(x) defined over the real numbers, the function derivative f', and an initial guess [tex]x_{0}[/tex] for a root of the function. It uses the idea that a continuous and differentiable function can be approximated by a straight line tangent to it.
This is the expression that we need to use
[tex]x_{n+1}=x_{n} -\frac{f(x_{n})}{f(x_{n})'}[/tex]
For the information given:
[tex]f(x) = x^3-0.2589x^{2}+0.02262x-0.001122=0\\f(x)'=3x^2-0.5178x+0.02262[/tex]
For the initial value [tex]x_{0}[/tex] you can choose [tex]x_{0}=0[/tex] although you can choose any value that you want.
So for approximation [tex]x_{1}[/tex]
[tex]x_{1}=x_{0}-\frac{f(x_{0})}{f(x_{0})'} \\x_{1}=0-\frac{0^3-0.2589\cdot0^2+0.02262\cdot 0-0.001122}{3\cdot 0^2-0.5178\cdot 0+0.02262} \\x_{1}=0.0496021[/tex]
Next, with [tex]x_{1}=0.0496021[/tex] you put it into the equation
[tex]f(0.0496021)=(0.0496021)^3-0.2589\cdot (0.0496021)^2+0.02262\cdot 0.0496021-0.001122 = -0.0005150[/tex], you can see that this value is close to 0 but we need to refine our solution.
For approximation [tex]x_{2}[/tex]
[tex]x_{2}=x_{1}-\frac{f(x_{1})}{f(x_{1})'} \\x_{1}=0-\frac{0.0496021^3-0.2589\cdot 0.0496021^2+0.02262\cdot 0.0496021-0.001122}{3\cdot 0.0496021^2-0.5178\cdot 0.0496021+0.02262} \\x_{1}=0.168883[/tex]
Again we put [tex]x_{2}=0.168883[/tex] into the equation
[tex]f(0.168883)=(0.168883)^3-0.2589\cdot (0.168883)^2+0.02262\cdot 0.168883-0.001122=0.0001307[/tex] this value is close to 0 but again we need to refine our solution.
We can summarize this process in the following table.
The approximation [tex]x_{5}[/tex] gives you the root of the equation.
When you plot the equation you find that only have one real root and is approximate to the value found.
If the sum of three vectors in R is zero, must they lie in the same plane? Explain.
Answer with Step-by-step explanation:
We are given that sum of three vectors in [tex]R^3[/tex] is zero.
We have to tell and explain that if sum of three vectors in [tex]R^3[/tex] is zero then they must lie in the same plane or not.
We know that if three or more vectors lie in the same then they are coplanar.
If three vectors are co-planar then their scalar product is zero.
According to given condition
Let (x,0,0), (-x,0,0) and (0,0,0)
[tex]\vec{u}=x\hat{i}[/tex]
[tex]\vec{v}=-x\hat{i}[/tex]
[tex]\vec{w}=0[/tex]
Sum of three vectors=[tex](x-x+0)\hat{i}+(0+0+)\hat{j}+(0+0+0)\hat{k}=0[/tex]
[tex]u\cdot (v\times w)=\begin{vmatrix}x_1&y_1&z_1\\x_2&y_2&z_2\\x_3&y_3&z_3\end{vmatrix}[/tex]
[tex]u\cdot (v\times w)=\begin{vmatrix}x&0&0\\-x&0&0\\0&0&0\end{vmatrix}[/tex]
When all elements of one row or column are zero of square matrix A then det(A)=0
[tex]u\cdot (v\times w)=0[/tex]
Therefore, vectors u,v and w are co-planar.
Hence, if the sum of three vectors in [tex]R^3[/tex] is zero then they must lie in the same plane.
If the sum of three vectors in three-dimensional space is zero, they typically lie in the same plane, forming a triangle with their magnitudes and directions. They must also be linearly dependent since their linear combination can produce the zero vector.
If the sum of three vectors is zero in R^3 (three-dimensional space), it is a common situation for these vectors to lie in the same plane. This condition implies that they can be represented as three sides of a triangle taken in sequence in terms of magnitude and direction. If we consider two vectors in sequence, such as AB and BC, the third vector must be the closing side CA, taken in the opposite direction, in order to satisfy the triangle law of vector addition. In this case, the sum of vectors AB + BC + CA equals zero.
Furthermore, vectors are homogeneous when added, which means that they must have the same nature and dimension. However, three vectors can still form a linearly dependent set in three-dimensional space even if they lie in the same plane. For instance, if we take vectors a, b, and c, and if a linear combination of these vectors yields the zero vector (i.e., if we can find scalar coefficients x, y, and z such that xa + yb + zc = 0), then these vectors are not linearly independent and must lie in the same plane.