Answer:
42 hamburgers
Step-by-step explanation:
Your answer is 42.
6 x 7 = 42
Using a formula estimate the body surface area of a person whose height is 150 cm and who weighs 80 kg.
A.
1.55 m2
B.
1.83 m2
C.
0.47 m2
D.
0.45 m2
Answer:
B. [tex]1.83\text{ m}^2[/tex]
Step-by-step explanation:
We are asked to find the body surface area of a person whose height is 150 cm and who weighs 80 kg.
[tex]\text{Body surface area}( m^2)=\sqrt{\frac{\text{Height (cm)}\times \text{Weight (kg)}}{3600}}[/tex]
Substitute the values:
[tex]\text{Body surface area}( m^2)=\sqrt{\frac{150\text{ cm}\times 80\text{(kg)}}{3600}}[/tex]
[tex]\text{Body surface area}( m^2)=\sqrt{\frac{12,000}{3600}}[/tex]
[tex]\text{Body surface area}( m^2)=\sqrt{3.3333333}[/tex]
[tex]\text{Body surface area}( m^2)=1.825741[/tex]
[tex]\text{Body surface area}( m^2)=1.83[/tex]
Therefore, the body surface area of the person would be 1.83 square meters.
The probability that a lab specimen contains high levels of contamination is 0.15. A group of 3 independent samples are checked. Round your answers to four decimal places (e.g. 98.7654). (a) What is the probability that none contain high levels of contamination? (b) What is the probability that exactly one contains high levels of contamination? (c) What is the probability that at least one contains high levels of contamination?
Answer:
a) There is a 61,41% of none of the samples containing high levels of contamination.
b)There is a 32.52% probability that exactly one sample contains high levels of contamination.
c) There is a 38.59% probability that at least one contains high levels of contamination
Step-by-step explanation:
The probabilities are independent from each other. It means that the probability of selecting a lab specimen being contaminated is always 15%, no matter how many contaminated lab specimen have been chosen.
a) There are 3 independent samples. For each sample, the probability of it not being contaminated is 85%. So, the probability that none of the sample are contaminated is
[tex]P = (0.85)^3 = 0.6141 = 61,41%[/tex]
There is a 61,41% of none of the samples containing high levels of contamination.
b) There are 3 independent samples. For each sample, the probability of it being contaminated is 15% and not contaminated 85%.
So the probability the exactly one sample contains high levels of contamination is:
[tex]P = (0.85)^2(0.15) = 0.1084 = 10,84%[/tex]
There can be 3 orderings of the sample in these conditions.(C-NC-NC, NC-C-NC, NC-NC,C), so the probability that exactly one contains high levels of contamination is
P = 3*0.1084 = 0.3252 = 32.52%.
There is a 32.52% probability that exactly one sample contains high levels of contamination.
c) The sum of the probabilities is always 100%.
In relation to the existence of a contaminated sample, either:
-None of the samples are contaminated.
-At least one of the samples are contaminated.
So, the probability of at least one of the samples being contaminated is 100% - the probability that none of the samples are contaminated, that we have already found in a).
So, it is
100% - 61.41% = 38.59%
There is a 38.59% probability that at least one contains high levels of contamination
Use Gaussian elimination on the augmented matrix, then use back substitution to find the solution of the system of linear equations.
-2x + 3y - 4z = 7
5x - y + 2z = 13
3x + 2y - z = 17
Answer:
x = 4
y = 1
z= -3
Step-by-step explanation:
Given equations are
-2x + 3y - 4z = 7
5x - y + 2z = 13
3x + 2y - z = 17
We can write the above equations in matrix augmented form as
[tex]\left[\begin{array}{ccc}-2&3&-4:7\\5&-1&2:13\\3&2&-1:17\end{array}\right][/tex]
[tex]R_1=>\dfrac{R_1}{-2}[/tex]
[tex]=\ \left[\begin{array}{ccc}1&\dfrac{-3}{2}&2:\dfrac{-7}{2}\\5&-1&2:13\\3&2&-1:17\end{array}\right][/tex]
[tex]R_2=>R_2-5R_1\ and\ R_3=>\ R_3-3R_1[/tex]
[tex]=\ \left[\begin{array}{ccc}1&\dfrac{-3}{2}&2:\dfrac{-7}{2}\\0&-1+\dfrac{15}{2}&-8:13+\dfrac{35}{2}\\0&0&-7:17+\dfrac{21}{2}\end{array}\right][/tex]
[tex]=\ \left[\begin{array}{ccc}1&\dfrac{-3}{2}&2:\dfrac{-7}{2}\\\\0&\dfrac{13}{2}&-8:\dfrac{61}{2}\\\\0&\dfrac{13}{2}&-7:\dfrac{55}{2}+\dfrac{21}{2}\end{array}\right][/tex]
[tex]R_2=>\ \dfrac{2}{13}R_2[/tex]
[tex]=\ \left[\begin{array}{ccc}1&\dfrac{-3}{2}&2:\dfrac{-7}{2}\\\\0&1&\dfrac{-16}{13}:\dfrac{61}{13}\\\\0&\dfrac{13}{2}&-7:\dfrac{55}{2}\end{array}\right][/tex]
[tex]R_3=>R_3-\dfrac{13}{2}R_2[/tex]
[tex]=\ \left[\begin{array}{ccc}1&\dfrac{-3}{2}&2:\dfrac{-7}{2}\\\\0&1&\dfrac{-16}{13}:\dfrac{61}{13}\\\\0&0&1:-3\end{array}\right][/tex]
So, from the above augmented matrix, we can write
[tex]x+\dfrac{-3}{2}y+2z=\dfrac{-7}{2}.......(1)[/tex]
[tex]y+\dfrac{-16}{13}z=\dfrac{61}{13}......(2)[/tex]
z= -3.....(3)
From eq(2) and (3)
[tex]y+\dfrac{-16}{13}(-3)=\dfrac{61}{13}[/tex]
=> y = 1
Now, by putting the value of y and z in equation (1), we will get
[tex]x+\dfrac{-3}{2}(1)+2(-3)=\dfrac{-7}{2}[/tex]
=> x = 4
Hence, the value of
x = 4
y = 1
z= -3
When you don't seem to get any better, you eventually end up in the hospital. The doctor orders 120 mg of medicine to be given twice a day. The nurse comes in with 6 tablets and tells you that there are 30 mg in each tablet. Is this the correct dosage? If not, how many tablets should you get?
Answer:
It isn't the correct dosage, you should get 8 tablets.
Step-by-step explanation:
First, if the doctors orders 120 mg twice a day, it means that you need 240 mg of medicine. That is calculated as:
120 mg * 2 = 240 mg
Then if each tablet has 30 mg, the number of tablets that you should get is calculated as:
[tex]\frac{240mg}{30mg} = 8[/tex]
So, 240 mg of medicine are equivalents to 8 tablets.
8 tablets are different of 6 tablets, so the dosage given by the nurse is incorrect and you should get 8 tablets every day.
Define arcwise(=path) connectedness of a set in a metric space. State a relation between arcwise connectedness and connectedness of a set.
Answer:
See definitions and relation below
Step-by-step explanation:
Given points x and y of a certain set S in a metric space, a path from x to y is a continuous map f:[a,b]-->S of some closed interval [a,b] in the real line into S, such that
f(a)=x and f(b)=y
In this case, we can also say that the points x and y are joined by a path or arc.
A set S in metric space is said to be path connected or arcwise connected if every pair of points x, y of S can be joined by a path.
The relation between arcwise connectedness and connectedness of a set is that every arcwise connected set is also connected, but the converse does not hold; not every connected space is also path connected.
As an example, consider the unit square [0,1]X[0,1] with the dictionary order topology.
It can be proved that this space is connected but not path connected.
Arcwise connectedness is defined as the presence of a continuous path between any two points in a set in a metric space. There is a relation between arcwise connectedness and connectedness, where any path connected set is also connected, but the converse is not necessarily true.
Explanation:A set in a metric space is said to be arcwise connected or path connected if there exists a continuous curve or path that connects any two points in the set.
The relation between arcwise connectedness and connectedness of a set is that any arcwise connected set is also connected, but the converse is not necessarily true. In other words, every path connected set is connected, but not every connected set is path connected.
For example, consider a set consisting of two separate points in a metric space. This set is connected because we cannot find two disjoint open sets that cover the set, but it is not arcwise connected because there is no continuous path connecting the two points.
(Order is anti-symmetric) If a > b and b > a, then a = b. (e) a
Answer:
a=b
Step-by-step explanation:
An antisymmetric relation () satisfies the following property:
If (a, b) is in R and (b, a) is in R, then a = b.
This means that if a|b and b|a then a = b
If a|b then, b can be written as b = an for an integer n
If b|a then a can be written as a= bm for an integer m
Now we have b = (a)n = (bm)n
b = bmn
1 =mn
But since m and n are integers, the only two integers that satisfy this property would be m = n = 1
Therefore, b = an = a (1) = a ⇒b = a
A "child has six blocks, three of which are red and three of which are green". How many patterns can shemake by placing them all in a line?If she is given threewhite blocks, how many total patterns can she make by placing all nine blocks in a line?
Answer: There are 20 ways and 1680 ways respectively.
Step-by-step explanation:
Since we have given that
Total number of blocks = 6
Number of red blocks = 3
Number of green blocks = 3
So, Number of patterns she can make by placing them all in a line is given by
[tex]\dfrac{6!}{3!\times 3!}\\\\=20[/tex]
If there are 3 white blocks
so, total number of white blocks becomes 9
So, Number of total pattern she can make by placing all nine blocks in a line is given by
[tex]\dfrac{9!}{3!\times 3!\times 3!}\\\\=1680\ ways[/tex]
Hence, there are 20 ways and 1680 ways respectively.
The child can create 20 different patterns if she uses just the 6 blocks (3 red, 3 green), and she can create 14,040 different patterns if she uses all 9 blocks (3 red, 3 green, 3 white). This is calculated using a branch of mathematics known as combinatorics.
Explanation:
In this math problem, we are dealing with a concept known as permutations in combination, which is part of combinatorics branch of Mathematics. When placing the blocks in a line, the order in which you arrange them matters, which makes this a permutation problem.
For the first case where she has 6 blocks, 3 red and 3 green, the number of different patterns she can create is calculated by the equation 6! / (3! * 3!). Here, the '!' character means factorial, which is the product of all positive integers up to that number. So, 6! = 6 * 5 * 4 * 3 * 2 * 1 and similarly 3! = 3 * 2 * 1. Plugging these in, the equation becomes 720 / (6*6) = 20 patterns.
For the second case, if she is given 3 white blocks, she then has a total of 9 blocks (3 red, 3 green, 3 white). The number of different patterns she can create is calculated similarly, but this time the equation is 9! / (3! * 3! * 3!). Plugging in the factorials, we have 362,880 / (6 * 6 * 6) = 14,040 patterns.
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A water tank has a diameter of 15 ft and is 22 ft high. a. What is the volume of the tank in ft?? b. In m?? c. In cm?
Answer:
a) [tex]V = 3887.72 ft^{3}[/tex]
b)[tex]V = 104.97 m^{3}[/tex]
c)[tex]V = 104,968,468.538 cm^{3}[/tex]
Step-by-step explanation:
A tank has the format of a cylinder.
The volume of the cylinder is given by:
[tex]V = \pi r^{2}h[/tex]
In which r is the radius and h is the heigth.
The problem states that the diameter is measured to be 15.00 ft. The radius is half the diameter. So, for this tank
[tex]r = \frac{15}{2} = 7.50[/tex] ft
The height of the tank is 22 ft, so [tex]h = 22[/tex].
a) Volume of the tank in [tex]ft^{3}[/tex]:
[tex]V = \pi r^{2}h[/tex]
[tex]V = pi*(7.5)^2*22[/tex]
[tex]V = 3887.72 ft^{3}[/tex]
b) Volume of the tank in [tex]m^{3}[/tex]:
We must convert both the radius and the height to m.
Each feet has 0.30 m, so:
Radius:
1 feet - 0.30m
7.5 feet - r m
[tex]r = 7.5*0.30[/tex]
[tex]r = 2.25m[/tex]
Height
1 feet - 0.30m
22f - h m
[tex]h = 22*0.30[/tex]
[tex]r = 6.60m[/tex]
The volume is:
[tex]V = \pi r^{2}h[/tex]
[tex]V = pi*(2.25)^2*6.60[/tex]
[tex]V = 104.97 m^{3}[/tex]
c) Volume of the tank in [tex]cm^{3}[/tex]:
Each m has 100 cm.
So [tex]r = 2.25m = 225cm[/tex]
[tex]h = 6.60m = 660cm[/tex]
The volume is:
[tex]V = \pi r^{2}h[/tex]
[tex]V = pi*(225)^2*660[/tex]
[tex]V = 104,968,468.538 cm^{3}[/tex]
A toy manufacturer wants to know how many new toys children buy each year. A sample of 601 children was taken to study their purchasing habits. Construct the 95% confidence interval for the mean number of toys purchased each year if the sample mean was found to be 6.7. Assume that the population standard deviation is 1.5. Round your answers to one decimal place.
Answer:
The confidence interval is 6.6<μ<6.8.
Step-by-step explanation:
We have:
Number of observations = 601
Mean = 6.7
Standard deviation σ = 1.5
The z-score for a 95% confidence interval is 1.96.
The limits of the confidence interval can be calculated as
[tex]X \pm z*\frac{\sigma}{\sqrt{n}}\\\\LL=X-z*\frac{\sigma}{\sqrt{n}}=6.7-1.96*\frac{1.5}{\sqrt{601} } =6.7-0.1199=6.6\\\\UL=X+z*\frac{\sigma}{\sqrt{n}}=6.7+1.96*\frac{1.5}{\sqrt{601} } =6.7+0.1199=6.8[/tex]
The confidence interval is 6.6<μ<6.8.
Where do we use prime numbers every day?
Answer:
1) Prime Factorization
2) Technology
3) Existence of prime number in nature
Step-by-step explanation:
Prime numbers are the numbers whose divisors are 1 and the number itself, For example: 2, 3, 7, 11,...
Prime Numbers are a significant part of our life and are widely used in daily life.
1) Prime Factorization
This method help us to break a number into products of prime Number. This approach help us to find the LCM(Lowest Common Multiple) and GCD(Greatest Common Divisor)
2) Technology
Prime factorization forms the basis oh cryptography. Prime numbers play an important role in password protection and security purposes. They give the basis for many cryptographic algorithms.
3) Existence of prime number in nature
Many scientist have claimed that prime numbers exist in our life in unexpected form. For example, the number of petals in a flower, number of hexes in beehive, the pattern in pineapple are all related to prime number.
If you are asked to provide a set of two or more numeric answers, separate them with commas. For example, to provide the year that Sputnik (the first satellite to be sent into orbit around the Earth) was launched and the year humans first walked on the Moon, you would enter 1957,1969 in the answer box. A rectangle has a length of 5.50 m and a width of 12.0 m . What are the perimeter and area of this rectangle?
Answer: Perimeter = 66 cm and area =[tex]66\ m^2[/tex]
Step-by-step explanation:
The perimeter of rectangle is given by :-
[tex]P=2(l+w)[/tex], where l is length and w is width of the rectangle.
Given : A rectangle has a length of 5.50 m and a width of 12.0 m .
Then, the perimeter of rectangle :
[tex]P=2(12+5.50)\\\\\Rightarrow\ P=2(17.50)=35\ m[/tex]
Also, area of rectangle is given by :-
[tex]A=l\times w=5.50\times12=66 \ m^2[/tex]
Area of rectangle = [tex]66\ m^2[/tex]
Let a = 0.9876 and b = 0.9887 with N = 2, calculate the midpoint.
Answer:
Mid point will be 0.9887
Step-by-step explanation:
We have given a =0.9876 and b = 0.9887
And N = 2
We have to find midpoint
We know that formula for finding mid point that is
Midpoint [tex]=\frac{a+b}{2}[/tex]
So mid point will be
Midpoint [tex]=\frac{0.9876+0.9887}{2}=0.98815[/tex]
So the mid point between a = 0.9876 and b=0.9887 for N =2 will be 0.9887
ack has a collection of 10 pairs of gloves in his wardrobe. Before a business trip, he has to pack his luggage, and he selects 8 gloves, without looking at them. We assume that any set of 8 gloves is equally likely to be chosen. Find the probability that these 8 gloves do not include any matching pair of gloves, that is, that there are no two (left and right) gloves, coming from the same pair.
Answer:
[tex]\frac{{10 \choose 8}2^8}{{20 \choose 8}}\approx 0.091[/tex]
Step-by-step explanation:
We can think of the 10 pairs of gloves as simply being gloves of different colors. Picking no matching pair is the same as picking no 2 gloves of the same color. To compute the probability of doing so, we can compute the number of ways to select 8 gloves from different colors, and divide that by the total number of ways to select 8 random gloves out of the 20 gloves.
To compute the number of ways in which we can select 8 gloves from different colors, we can think of the choosing procedure as follows:
1st step- We choose from which 8 colors are we going to pick gloves from. So we have to pick 8 out of 10 colors. This can be done in [tex]{10 \choose 8}[/tex] ways.
2nd step - We now have to choose which glove are we going to pick from each of the chosen colors. Either the left one or the right one. For the first chosen color we have 2 choices, for the second chosen color we have 2 choices, for the third chosen color we have 2 choices, and so on. Therefore the number of ways in which we could choose gloves from the chosen colors is [tex]2^8[/tex]
And so the total number of ways in which we could choose 8 gloves from different colors is
[tex]{10 \choose 8 }2^8 [/tex]
Now, the total numer of ways in which we could choose 8 gloves out of the 20 gloves is simply [tex] {20 \choose 8}[/tex]
So the probability of picking no mathing pair is
[tex]\frac{{10 \choose 8}2^8}{{20 \choose 8}}\approx 0.091[/tex]
Given a non-linear system: y=x^3 - 3x^2 - 1 a) Find the linear approximation of the system at the point (1, -3) b) Plot the system and its linear approximation on a same plot using Matlab. Make sure your plot is clear and having labeling
[tex]\mbox{First, we compute the derivative of $y$ at $x_0=1$. So, we get}\\$$ y' = 3x^2 - 6x \, , \, y'(1) = -3 $$[/tex].
Therefore, the linear approximation at the point (1,-3) is
[tex]$$ y = -3 - 3(x -1) \ . $$[/tex]
To find the linear approximation of the non-linear system at the point (1, -3), first find the derivative of the function to get the slope of the tangent line at that point. Then, plug the slope and the point into the linearization formula. For the plotting part in Matlab, it should be a separate discussion as this platform does not support programming languages.
Explanation:The subject of this question is a non-linear system given by the equation y=x^3 - 3x^2 - 1. The student is asked to find the linear approximation at the point (1, -3). The linear approximation of a function at a given point is the tangent line to the function at the given point, and it's also the best linear approximation of the function near that point.
Before we begin, let's define some terms. Linear approximation is a process of approximating the values of a nonlinear function using a line near a point. To find the linear approximation, we use the formula for the linearization of a function, L(x) = f(a) + f'(a)(x - a), where 'a' is the x-value of the point of tangency, f(a) is the y-value, and f'(a) is the slope of the tangent line at point 'a'. Tangent line is a straight line that just touches a curve at a given point. The tangent line is the best linear approximation to the curve at that point.
First, we need to find the derivative of the function, f'(x), which is 3x^2 - 6x. Then, evaluate f'(1) to find the slope of the tangent line. Plug these values into the linearization formula to get L(x) = -3 + (3 - 6)(x - 1). Now, you can plot the original function and the linearization on the same graph.
Please note, for the Matlab portion of the question, it should be a separate discussion as this website is designed to walk through problems in a step-by-step manner and doesn't support running such programming languages directly. However, there are many online resources that can provide specific Matlab example codes for plotting functions and their linear approximations.
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Prove the following DeMorgan's laws: if LaTeX: XX, LaTeX: AA and LaTeX: BB are sets and LaTeX: \{A_i: i\in I\} {Ai:i∈I} is a family of sets, then
LaTeX: X-(A\cup B)=(X-A)\cap (X-B)
LaTeX: X-(\cup_{i\in I}A_i)=\cap_{i\in I}(X-A_i)
I'll assume the usual definition of set difference, [tex]X-A=\{x\in X,x\not\in A\}[/tex].
Let [tex]x\in X-(A\cup B)[/tex]. Then [tex]x\in X[/tex] and [tex]x\not\in(A\cup B)[/tex]. If [tex]x\not\in(A\cup B)[/tex], then [tex]x\not\in A[/tex] and [tex]x\not\in B[/tex]. This means [tex]x\in X,x\not\in A[/tex] and [tex]x\in X,x\not\in B[/tex], so it follows that [tex]x\in(X-A)\cap(X-B)[/tex]. Hence [tex]X-(A\cup B)\subset(X-A)\cap(X-B)[/tex].
Now let [tex]x\in(X-A)\cap(X-B)[/tex]. Then [tex]x\in X-A[/tex] and [tex]x\in X-B[/tex]. By definition of set difference, [tex]x\in X,x\not\in A[/tex] and [tex]x\in X,x\not\in B[/tex]. Since [tex]x\not A,x\not\in B[/tex], we have [tex]x\not\in(A\cup B)[/tex], and so [tex]x\in X-(A\cup B)[/tex]. Hence [tex](X-A)\cap(X-B)\subset X-(A\cup B)[/tex].
The two sets are subsets of one another, so they must be equal.
[tex]X-\left(\bigcup\limits_{i\in I}A_i\right)=\bigcap\limits_{i\in I}(X-A_i)[/tex]The proof of this is the same as above, you just have to indicate that membership, of lack thereof, holds for all indices [tex]i\in I[/tex].
Proof of one direction for example:
Let [tex]x\in X-\left(\bigcup\limits_{i\in I}A_i\right)[/tex]. Then [tex]x\in X[/tex] and [tex]x\not\in\bigcup\limits_{i\in I}A_i[/tex], which in turn means [tex]x\not\in A_i[/tex] for all [tex]i\in I[/tex]. This means [tex]x\in X,x\not\in A_{i_1}[/tex], and [tex]x\in X,x\not\in A_{i_2}[/tex], and so on, where [tex]\{i_1,i_2,\ldots\}\subset I[/tex], for all [tex]i\in I[/tex]. This means [tex]x\in X-A_{i_1}[/tex], and [tex]x\in X-A_{i_2}[/tex], and so on, so [tex]x\in\bigcap\limits_{i\in I}(X-A_i)[/tex]. Hence [tex]X-\left(\bigcup\limits_{i\in I}A_i\right)\subset\bigcap\limits_{i\in I}(X-A_i)[/tex].
A home improvement company expects the daily supply equation of their most popular cordless drill kit to be: S(q) = 10q2 + 100q +20 where S(q) is the price in dollars at which q units are supplied. Find the price at which the company should set the drill kit if they plan to supply 16 a day?
Answer:
Price of the drill kit should be set as $4180.
Step-by-step explanation:
Daily supply of the most popular cordless drill kit is represented by the equation
S(q) = 100q² + 100q + 20
where S(q) = price of the kits at which q units are supplied
q = number of drill kits supplied
Now we have to calculate the price of the drill kits if company plans to supply 16 kits a day.
S(16) = 10(16)² + 100(16) + 20
= 10×256 + 1600 + 20
= 2560 + 1600 + 20
= $4180
Therefore, cost of the drill set should be set as $4180.
To determine the price for 16 drill kits, substitute q = 16 into the supply equation S(q) = 10q^2 + 100q + 20, resulting in a price of $4180.
To find the price at which the home improvement company should set the drill kit if they plan to supply 16 a day, we need to plug the quantity (q) into the given supply equation S(q) = 10q2 + 100q + 20.
Substituting q = 16, we get:
S(16) = 10(16)2 + 100(16) + 20
= 10(256) + 1600 + 20
= 2560 + 1600 + 20
= 4180.
So, the company should set the price of the cordless drill kit at $4180 if they plan to supply 16 units a day.
Chef needs to prepare 32 Servings of cooked beans, each portion 3 oz. How 29. any 12 oz. cans should Chef order, if the Yield factor per can is 0.8 a. 8 cans b. 10 cans c. 6.4 cans d. 120 cans e. NONE of the above answers are corredt
Answer:
10 cans
Step-by-step explanation:
Number of servings to be prepared = 32
Weight of each pound = 3 oz
Yield factor per can = 0.8
Now,
Total weight of the cans = 32 × 3 = 96 oz
Actual weight required with yield factor 0.8 = [tex]\frac{96\ oz}{0.8}[/tex] = 120 oz
Therefore,
The number of 12 oz cans required = [tex]\frac{120}{12}[/tex] = 10 cans
The Chef should order 10 cans.
Explanation:To determine how many 12 oz. cans the Chef should order, we need to calculate the total amount of cooked beans required.
Since each serving is 3 oz. and there are 32 servings, the total amount needed is
3 oz. * 32 = 96 oz..
The yield factor per can is 0.8, so each can provides 0.8 * 12 oz. = 9.6 oz.
Therefore, the Chef should order 96 oz. / 9.6 oz. per can = 10 cans.
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Show that the given curve c(t) is a flow line of the given velocity vector field F(x, y, z).
c(t) = (t^2, 2t − 6, 3sqrt(t)), t > 0; F(x,y,z) = ( y+6, 2, 9/2z)
c'(t) = ?
F(c(t)) = ?
Answer:
The given curve c(t) is a is a flow line of given velocity vector field F(x, y, z).
Step-by-step explanation:
We are given the following information in the question:
[tex]c(t) = (t^2, 2t-6, 3\sqrt{t}), t > 0\\\\ F(x, y, z) =(y+6, 2, \frac{9}{2z} )[/tex]
Now, we evaluate the following:
[tex]c'(t) = \frac{d(c(t))}{dt} = (2t, 2, \frac{3}{2\sqrt{t}} )[/tex]
Now, we have to evaluate:
[tex]F(c(t)) = (2t-6+6, 2, \frac{9}{6\sqrt{t}} ) = (2t, 2, \frac{3}{2\sqrt{t}} )[/tex]
When F(c(t)) = c'(t), then c(t) is a flow line of given velocity vector field F(x, y, z).
Since, [tex]F(c(t)) = c'(t)[/tex], we can say that c(t) is a flow line of given velocity vector field F(x, y, z).
The derivative of c(t) is calculated by differentiating each component with respect to t, resulting in c'(t)=(2t, 2, 3/2*t^(-1/2)). The velocity field F(c(t)) is found by substituting the equation of c(t) into F(x, y, z), resulting in F(c(t))=(2t+6, 2, 3/2*t^(-1/2)). As the results are equivalent, it's confirmed that c(t) is a flow line of the velocity vector field F(x, y, z).
Explanation:In order to show that the curve c(t) is a flow line of the velocity vector field F(x, y, z), we first need to find c'(t), the derivative of c(t), and F(c(t)), the velocity field evaluated at the points along the curve c(t).
First, let's find c'(t). c(t) is given by (t^2, 2t − 6, 3sqrt(t)), so its derivative c'(t) is given by differentiating each component with respect to t: c'(t)=(2t, 2, 3/2*t^(-1/2)).
Next, let's find F(c(t)). F(x,y,z) is given by (y+6, 2, 9/2*z) so F(c(t)) is evaluated by substituting the equation of c(t) into F(x, y, z). Thus F(c(t))=(2t+6,2,3/2*t^(-1/2)).
Since F(c(t)) and c'(t) are equivalent, c(t) is indeed a flow line of the velocity vector field F(x, y, z).
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A city is holding a referendum on increasing property taxes to pay for a new high school. In a survey of 458 likely voters, 254 said that they would vote "yes" on the referendum. Create a 95% confidence interval for the proportion of likely voters who will vote "yes" on the referendum. Use Excel to create the confidence interval, rounding to four decimal places.
Answer:
The confidence interval is -5.3444 to 6.453 .
Step-by-step explanation:
We are given that In a survey of 458 likely voters, 254 said that they would vote "yes" on the referendum.
So, n = 458
x = 254
We will use sample proportion over here
[tex]\widehat{p}=\frac{x}{n}[/tex]
[tex]\widehat{p}=\frac{254}{458}[/tex]
[tex]\widehat{p}=0.5545[/tex]
Confidence level = 95% = 0.95
Level of significance = 1-0.95 = 0.05
z value at 0.05 significance level = 1.96
Formula of confidence interval : [tex]\widehat{p}-x\times \sqrt{\frac{\widehat{p} \times (1-\widehat{p})}{n}[/tex] to [tex]\widehat{p}+x\times \sqrt{\frac{\widehat{p} \times (1-\widehat{p})}{n}[/tex]
Confidence interval : [tex]0.5545-254\times \sqrt{\frac{0.5545\times (1-0.5545)}{458}}[/tex] to [tex]0.5545+254\times \sqrt{\frac{0.5545\times (1-0.5545)}{458}}[/tex]
Confidence interval : [tex]-5.3444[/tex] to [tex]6.453[/tex]
Hence The confidence interval is -5.3444 to 6.453 .
Answer:
Step-by-step explanation:
We have given,
x=254
n=458
Estimate for sample proportion
Level of significance is =1-0.95=0.05
Z critical value(using Z table)=1.96
Confidence interval formula is
=(0.5091,0.6001)
Lower limit for confidence interval=0.5091
Upper limit for confidence interval=0.6001
An investment project involves an immediate outlay of $8 million. The net cash flows received at the end of years 1, 2, and 3 will be $3 million, $4 million, and $2 million. A 10% discount rate is applicable so that the present value factors for years 1, 2, and 3 are 0.9091, 0.8264, and 0.7513. The NPV of the investment will be: (a) (b) (c) (d) $1 million $9 million $0.46 million $7.54 million
Answer:
(c) 0.46 million
Step-by-step explanation:
As provided immediate cash outlay = $8 million.
This will represent cash outflow at period 0, as it is made immediately, no time period has lapsed.
Cash inflows as provided and the respective present value factor are:
Year Cash Inflow Factor Discounted Value
1 $3 million 0.9091 $2,727,300
2 $4 million 0.8264 $3,305,600
3 $2 million 0.7513 $1,502,600
Total present value of cash inflow = $7,535,500
Therefore, net present value = $7,535,500 - $8,000,000 = - $464,500
That is - 0.46 million
Correct option is
(c) 0.46 million
The USA Today reports that the average expenditure on Valentine's Day is $100.89. Do male and female consumers differ in the amounts they spend? The average expenditure in a sample survey of 50 male consumers was $135.67, and the average expenditure in a sample survey of 38 female consumers was $68.64. Based on past surveys, the standard deviation for male consumers is assumed to be $40, and the standard deviation for female consumers is assumed to be $23. What is the point estimate of the difference between the population mean expenditure for males and the population mean expenditure for females (to 2 decimals)?
Answer: $ 67.03
Step-by-step explanation:
Given : The average expenditure in a sample survey of 50 male consumers was $135.67, and the average expenditure in a sample survey of 38 female consumers was $68.64.
i.e. [tex]\overline{x}_1=\$135.67\ \ \&\ \ \overline{x}_2=\$68.64[/tex]
The best point estimate of the difference between the two population means is given by :-
[tex]\overline{x}_1-\overline{x}_2\\\\=135.67-68.64=67.03[/tex]
Hence, the point estimate of the difference between the population mean expenditure for males and the population mean expenditure for females : $ 67.03
The point estimate of the difference between the average expenditure of male and female consumers for Valentine's Day is $67.03.
Explanation:The subject of your question is related to comparative statistical analysis between two groups, in this case, male and female consumers on Valentine's Day expenditures. Your question focuses on finding the point estimate for the difference between the population mean expenditure of males and females.
The point estimate is calculated by simply subtracting one mean from the other. According to your data, the average expenditure of the male consumers is $135.67 and of female consumers is $68.64. So, the calculation looks like this: $135.67 - $68.64 = $67.03. Therefore, the point estimate of the difference between the population mean expenditure for males and the population mean expenditure for females is $67.03.
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List all subsets of ta, b, c, d, e) containing a but not containing b
Answer:
(a), (a,c), (a,d), (a,e), (a,c,d), (a,c,e), (a,d,e), (a,c,d,e)
Step-by-step explanation:
We are given the set (a,b,c,d,e).
Total number of subsets of the above set are [tex]2^5[/tex] = 32
Subsets:
φ
(a,b,c,d,e)
(a), (b), (c), (d), (e)
(a,b), (a,c), (a,d), (a,e), (b,c), (b,d), (b,e), (c,d), (c,e), (d,e)
(a,b,c), (a,b,d), (a,b,e), (a,c,d), (a,c,e), (a,d,e), ( b,c,d), (b,c,e), (b,d,e), (c,d,e)
(a,b,c,d), (a,b,c,e), (a,b,d,e), (a,c,d,e), (b,c,d,e)
Subset having a but not b :
(a), (a,c), (a,d), (a,e), (a,c,d), (a,c,e), (a,d,e), (a,c,d,e)
Convert 26inches into meters
Answer:
0.6604 m
Step-by-step explanation:
The convertion from inches to meters is 1 inch= 0.024 meters, so:
26 inches = 26 inch* 0.024 meters/inch = 0.6604 meters
Good luck!
Draw Conclusions The decimal 0.3 represents What type of number best describes 0.9, which is 3.0.3? Explain
Answer:
The product represent the number [tex]3\times 0.3=0.9[/tex]
Step-by-step explanation:
To find : The decimal 0.3 represents What type of number best describes 0.9, which is 3.0.3?
Solution :
0.3 represents [tex]0.3=\frac{3}{10}[/tex]
0.9 represents [tex]0.9=\frac{9}{10}[/tex]
If we multiply 0.3 by 3 we get 0.9
As, [tex]3\times 0.3=3\times \frac{3}{10}[/tex]
[tex]3\times 0.3=\frac{9}{10}[/tex]
[tex]3\times 0.3=0.9[/tex]
Therefore, The product represent the number [tex]3\times 0.3=0.9[/tex]
The first difference of a sequence is the arithmetic sequence 1, 3, 5, 7, 9, .... Find the first six terms of the original sequence in each of the following cases. a. The first term of the original sequence is 1. b. The sum of the first two terms in the original sequence is 5. c. The fifth term in the original sequence is 28.
Answer:
Step-by-step explanation:
Given that the first difference of a sequence is the arithmetic sequence 1, 3, 5, 7, 9, ....
a) When I term a =1
[tex]a_2 =1+1 =2\\a_3 = 4+5 =9\\a_4 = 9+7 =16\\a_5 =16+9 =25\\a_6=25+11 =36[/tex]
Thus first 6 terms are
1,2,5,12,21,32.....
b) Here [tex]a_1+a_2=5\\a_2-a_1 =3\\-------------\\2a_2=8\\a_2 =4\\a_1 =1[/tex]
[tex]a_2 =1+3 =4\\a_3 = 4+5 =9\\a_4 = 9+7 =16\\a_5 =16+9 =25\\a_6=25+11 =36[/tex]
So sequence would be
3,4,9,16,25, 36,...
c) When 5th term is 28
we have the sequences as
a1, a1+1,a1+1+3, ...a1+1+3+5+7
When 5th term is 28 we have
[tex]a_1 +16 =28\\a_1 =12\\[/tex]
Hence first 6 terms would be
12, 13, 16, 21, 28, 37,...
Suppose a four-period weighted average is being used to forecast demand. Weights for the periods are as follows: wt-4 = 0.1, wt-3 = 0.2, wt-2 = 0.3 and wt-1 = 0.4. Demand observed in the previous four periods was as follows: At-4 = 380, At-3 = 410, At-2 = 390, At-1 = 400. What will be the demand forecast for period t?
To find the demand forecast for period t using the four-period weighted average, multiply each demand observation by its corresponding weight and sum the results.
Explanation:To find the demand forecast for period t using the four-period weighted average, we multiply each demand observation by its corresponding weight and sum the results. In this case, we have:
Adding these weighted demands together gives us the demand forecast for period t:
Demand forecast for period t = 38 + 82 + 117 + 160 = 397.
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Find the 100th AND the nth term for the following sequence. Please show work.
a. 197+7 x 3^27, 197+8 x 3^27, 197+9 x 3^27
Answer:
nth term of this sequence is [tex](197+(n+6)\times 3^{27})[/tex]
and 100th term is [tex](197+106\times 3^{27})[/tex].
Step-by-step explanation:
The given sequence is [tex](197+7\times 3^{27}),(197+8\times 3^{27}),(197+9\times 3^{27})[/tex]
Now we will find the difference between each successive term.
Second term - First term = [tex](197+8\times 3^{27})-(197+7\times 3^{27})[/tex]
= [tex](8\times 3^{27}-7\times 3^{27})[/tex]
= [tex]3^{27}(8-7)[/tex]
= [tex]3^{27}[/tex]
Similarly third term - second term = [tex]3^{27}[/tex]
So there is a common difference of [tex]3^{27}[/tex].
It is an arithmetic sequence for which the explicit formula will be
[tex]T_{n}[/tex]=a + (n - 1)d
where [tex]T_{n}[/tex] = nth term of the arithmetic sequence
where a = first term of the arithmetic sequence
n = number of term
d = common difference in each successive term
Now we plug in the values to get the 100th term of the sequence.
[tex]T_{n}=(197+7\times 3^{27})+(n-1)\times 3^{27}[/tex]
= [tex](197+(n+6)\times 3^{27})[/tex]
[tex]T_{100}=(197+7\times 3^{27})+(100-1)\times 3^{27}[/tex]
= [tex]197+7\times 3^{27}+99\times 3^{27}[/tex]
= [tex]197+106\times 3^{27}[/tex]
Therefore, nth term of this sequence is [tex](197+(n+6)\times 3^{27})[/tex]
and 100th term is [tex](197+106\times 3^{27})[/tex].
Which terms are rational in the expansion of (\sqrt{3} + \frac{1}{\sqrt[4]{6}})^{15} . List the rational terms and justify why the others are not rational.
Answer:
[tex](\sqrt{3} + \frac{1}{\sqrt[4]{6}})^{15}[/tex]
Binomial expansion formula,
[tex](a+b)^n=\sum_{r=0}^{n} ^nC_r (a)^{n-r} (b)^r[/tex]
Where,
[tex]^nC_r=\frac{n!}{r!(n-r)!}[/tex]
[tex]\implies (\sqrt{3} + \frac{1}{2})^{15}=\sum_{r=0}^{15} ^{15}C_r (\sqrt{3})^{15-r} (\frac{1}{\sqrt[4]{6}})^r[/tex]
[tex]=(\sqrt{3})^{15}+15(\sqrt{3})^{14}(\frac{1}{\sqrt[4]{6}})^1+105(\sqrt{3})^{13}(\frac{1}{\sqrt[4]{6}})^2+455(\sqrt{3})^{12}(\frac{1}{\sqrt[4]{6}})^3+1365(\sqrt{3})^{11}(\frac{1}{\sqrt[4]{6}})^4+3003(\sqrt{3})^{10}(\frac{1}{\sqrt[4]{6}})^5+5005(\sqrt{3})^{9}(\frac{1}{\sqrt[4]{6}})^6+6435(\sqrt{3})^{8}(\frac{1}{\sqrt[4]{6}})^7+6435(\sqrt{3})^{7}(\frac{1}{\sqrt[4]{6}})^8+5005(\sqrt{3})^{6}(\frac{1}{\sqrt[4]{6}})^9+3003(\sqrt{3})^{5}(\frac{1}{\sqrt[4]{6}})^{10}+1365(\sqrt{3})^{4}(\frac{1}{\sqrt[4]{6}})^{11}+455(\sqrt{3})^{3}(\frac{1}{\sqrt[4]{6}})^{12}+105(\sqrt{3})^{2}(\frac{1}{\sqrt[4]{6}})^{13}+15(\sqrt{3})^{1}(\frac{1}{\sqrt[4]{6}})^{14}+(\frac{1}{\sqrt[4]{6}})^{15}[/tex]
∵ both [tex]\sqrt{3}[/tex] and [tex]\frac{1}{\sqrt[4]{6}}[/tex] are irrational numbers,
And, if the power of √3 is even, it converted to a rational number,
If its power is odd it remained as irrational number,
But, the product of a rational number and irrational number is irrational,
Thus, all terms in the above expansion are irrational. ( which can not expressed in the form of p/q, where, p and q are integers s.t. q ≠ 0 )
DOES ANYONE KNOW HOW TO DO THESE?????????
Answer:
a) Degree of E = 2
b) Even vertices: B, C, E
Odd vertices : A, D
c) Vertices A, C, and E are adjacent to D
Step-by-step explanation:
a) The degree of a vertex is given by the number of segments that end there, so in the case of vertex E, there are only two segments that connect it, therefore its degree is 2
b) Following the same idea of degree of a vertex, we can find the number of segments that end on each one of the 5 vertices shown and assign to them their degree:
A (3), B (2), C (4), D (3), E (2)
Therefore the odd vertices are: A and D (both of degree 3)
The even vertices are: B, E (both of degree 2, and C (degree 4)
c) the vertices adjacent to vertex D are those connected directly to it via a segment: that is, vertices A, C, and E
Let the universal set be the set of integers and let A = {x | x^2 ≤ 5}. Write A using the roster method.
A = { } --use commas to separate elements in the set
*Finite Math question
Answer:
Step-by-step explanation:
Given that Z the set of integers is the universal set and
A is given in set builder form.
[tex]A = {x | x^2 ≤ 5}[/tex]
To convert this into roster form, we can find solutions for x
When [tex]x^2\leq 5\\|x|\leq \sqrt{5} =2.236[/tex]
i.e. all integers lying between -2.236 and 2.236
The only integers satisfying this conditions are
-2,-1,0,1,2
Hence A in roster form is
A=[tex]{-2,-1,0,1,2}[/tex]
Final answer:
The set A = {x | x^2 ≤ 5}, which includes all integers whose squares are less than or equal to 5, is expressed using the roster method as A = { -2, -1, 0, 1, 2 }.
Explanation:
The set A includes all integers x such that x squared is less than or equal to 5. To list the set using the roster method, we identify all integers which, when squared, give a result that does not exceed 5.
The integers satisfying x2 ≤ 5 are -2, -1, 0, 1, and 2 because:
(-2)2 = 4, which is less than or equal to 5,(-1)2 = 1, which is less than or equal to 5,02 = 0, which is less than or equal to 5,12 = 1, which is less than or equal to 5,(2)2 = 4, which is less than or equal to 5.Therefore, using the roster method, the set A is written as A = { -2, -1, 0, 1, 2 }.