Answer:
The gcd(123,76) is equal to 1. The linear combination is [tex]1=-21 \cdot 123+34 \cdot 76[/tex]
The 361x+2109y = 1000 does not have integer solutions because the gcd(2109, 361) is equal to 19 and [tex]19\not| \:1000[/tex]
Step-by-step explanation:
Point a:
The greatest common divisor (GCD) of two numbers is the largest numbers that divide them both. The Euclidean algorithm is an efficient method for computing the GCD without explicitly factoring the two integers.
These are the steps:
Let a=x, b=yGiven x,y, use the division algorithm to write x=yq + r (q = quotient and r = remainder)if r=0, stop and output y; this is the gcd of a,bif r ≠ 0, replace (x,t) by (y,r): Go to step 2The division algorithm is an algorithm in which given 2 integers n and d, it computes their quotient q and remainder r, where [tex]0\leq r<|d|[/tex]. Let's say we have to divide n (dividend) by d (divisor):
Subtract d from n repeatedly.The resulting number is known as the remainder r, and the number of times that d is subtracted is called the quotient q.To compute gcd(123,76), divide the larger number by the smaller number, using the division algorithm we have:
[tex]\frac{123}{76} = 123-76 =47[/tex]
At this point, we cannot subtract 76 again. Hence 1 is the quotient ( we subtract 76 from 123 one time) and 47 is the remainder. We can express this as a linear combination [tex]123=76 \cdot 1+47[/tex].
Using the same reasoning and the steps of the Euclidean algorithm we have:
[tex]gcd(123,76)=\\123=76 \cdot 1 +47\\76=47 \cdot 1 +29\\47=29 \cdot 1 +18\\29=18 \cdot 1 +11\\18=11 \cdot 1 +7\\11=7\cdot 1 +4\\7= 4\cdot 1+3\\4 = 3 \cdot 1 +1\\3 = 1 \cdot 3 +0[/tex]
The gcd(123,76) is equal to 1.
To represent 1 as a linear combination of the integers 123 and 76, we start with the next-to-last of the above equations and successively eliminate the remainders 3, 4, 7, 11, 18, 29, and 47.
[tex]1=4-3 \cdot 1\\1=4-(7-4 \cdot 1) \cdot 1\\1=2\cdot 4 - 7\cdot 1\\1=2\cdot(11 -7 \cdot 1) -7 \cdot 1\\1=2\cdot 11 -7 \cdot 3\\1=2\cdot 11 -(18-11\cdot 1) \cdot 3\\1=5\cdot 11-3\cdot 18\\1=5\cdot (29-18\cdot 1)-3\cdot 18\\1=5\cdot 29 -8\cdot 18\\1=5\cdot 29 -8\cdot (47-29\cdot 1)\\1=13\cdot 29 -8\cdot 47\\1=13 \cdot (76-47 \cdot 1)-8\cdot 47\\1=13 \cdot 76 -21 \cdot 47\\1=13 \cdot 76 -21 \cdot (123-76\cdot 1)\\1=-21 \cdot 123+34 \cdot 76[/tex]
Point b:
We can use this theorem:
When ax + by = c is solvable. Given integers a, b, and c with a and b not both 0, there exist integers x and y such that ax + by = c if and only if (a,b) | c
In this particular expression 361x+2109y = 1000 we need to find the gcd(2109, 361) and check if gcd(2109, 361) | 1000 is true.
[tex]2109=361\cdot 5 +304\\361 = 304 \cdot 1 +57\\304= 57\cdot 5 +19\\57=19\cdot 3 +0[/tex]
The gcd(2109, 361) is equal to 19. We can see that 19 does not divide 1000 ([tex]19\not| \:1000[/tex]), that is the reason 361x+2109y = 1000 does not have integer solutions.
Express the following relations in the set builder notation. Then, determine whether it is reflexive, symmetric, transitive. Please show work.
a.) One number is less than or equal to another.
b.) One integer is a factor of another.
c.) Two integers are unequal.
d.) One set is a subset of another.
Answer:
a)Reflexive, not symmetric, transitive
b)Reflexive, not symmetric, transitive
c)Not reflexive, symmetric, not transitive
d)Reflexive, not symmetric, transitive
Step-by-step explanation:
a)
[tex]R=\left \{ (a,b)\epsilon \mathbb{R} \times \mathbb{R} \mid a \leq b\right \}[/tex]
The relation R is reflexive for
[tex]a\leq a[/tex] for every real number a
it is not symmetric because 0 is less than 1, but 1 is not less than 0
it is transitive
[tex]a\leq[/tex] and [tex] b\leq c\Rightarrow a\leq c[/tex]
So if aRb and bRc, then aRc
b)
[tex]R=\left \{ (m,n)\epsilon \mathbb{Z} \times \mathbb{Z} \mid \exists k\in \mathbb{Z} \ni m=kn \right \}[/tex]
R is reflexive because m=1.m for every integer m
R is not symmetric: 2 is a factor of 4, but 4 is not a factor of 2
R is transitive: if mRn and nRp if m=kn and n=qp, so m=(kq)p and kq is an integer , so mRp
c)
[tex]R=\left \{ (m,n)\epsilon \mathbb{Z} \times \mathbb{Z} \mid m\neq n\right \}[/tex]
R is obviously not reflexive because all numbers equals themselves
R is symmetric: if a different to b, then b different to a
R is not transitive: 1R2 and 2R1 (because 1 different to 2), but 1 = 1
d)
[tex]R=\left \{ A,B\mid A\subseteq B \right \}[/tex]
R is reflexive for every set A is a subset of itself
R is not symmetric {1,2} is a subset of {1,2,3} but {1,2,3} is not a subset of {1,2}
R is transitive: if A is subset of B and B is subset of C, then A is subset of C
A bacteria culture starts with 200 bacteria and grows at a rate proportional to its size. After 6 hours there will be 1200 bacteria (1) Express the population after I hours as a function of t. population: p(tepe (1.066-21) (unction of t) (b) What will be the population after 7 hours? 348125.2 (c) How long will it take for the population to reach 1750 ? Note: You can earn partial credit on this problem.
Answer:
We are given that the rate of change is proportional to its size S
So, [tex]\frac{dS}{dt} \propto S[/tex]
[tex]\frac{dS}{dt} = kS[/tex]
[tex]\frac{dS}{S} = kdt[/tex]
Integrating both sides
[tex]\log(S)= kt + log c[/tex]
[tex]\frac{S}{S_0}=e^{kt}[/tex]
[tex]S=S_0 e^{kt}[/tex]
S is the population after t hours
[tex]S_0[/tex] is the initial population
Now we are given that After 6 hours there will be 1200 bacteria
[tex]1200=200 e^{6k}[/tex]
[tex]6=e^{6k}[/tex]
[tex]6^{\frac{1}{6}=e^{k}[/tex]
So, [tex]S=200 \times 6^{\frac{t}{6}[/tex]
a)Now the population after t hours as a function of t; [tex]S=200 \times 6^{\frac{t}{6}[/tex]
b) What will be the population after 7 hours?
Substitute t = 7 hours
A bacteria culture starts with 200 bacteria
[tex]S=200 \times 6^{\frac{7}{6}}[/tex]
[tex]S=1617.607[/tex]
c) How long will it take for the population to reach 1750 ?
[tex]1750=200 \times 6^{\frac{t}{6}[/tex]
[tex]\frac{1750}{200} =6^{\frac{t}{6}[/tex]
[tex]8.75 =6^{\frac{t}{6}[/tex]
[tex]t=7.26[/tex]
So, it will take 7.2 hours for the population to reach 1750
Final answer:
To determine the population growth function in terms of time and find the population after a specific duration, use the exponential growth formula N(t) = N0 x 2^t. Calculate the growth rate using given data points like the initial and final population. Finally, substitute the desired time into the population function to find the population at that specific time.
Explanation:
The population of bacteria after t hours can be represented by the formula N(t) = N0 x 2t.
(a) To express the population after t hours as a function of t, you can use the given data points to find the growth rate. For the provided data, the growth rate is calculated as r = log2(1200/200) / 6 = 0.1333 per hour.
(b) To find the population after 7 hours, substitute t=7 into the function: N(7) = 200 x 27x0.1333 = 3481.49.
c) To find out how long it will take for the population to reach 1750, we get
p(t)=1750 and solving it
t ≈7.846
Please help me with geometry. Theres only 3 questions i need help with
Use the roster method to write each of the given sets. For some exercises you may need to consult a reference, such as the Internet or an encyclopedia. (Enter EMPTY for the empty set.)
The set of natural numbers x that satisfy x + 2 = 1
There are no natural numbers that satisfy the equation x + 2 = 1. Therefore, using the roster method, the set is empty.
Explanation:The question requires us to use the roster method to write the set of natural numbers x that satisfy the equation x + 2 = 1. Natural numbers, by definition, are counting numbers starting from 1. They are non-negative and do not include zero. So, if we try to find a natural number x that satisfies the equation x + 2 = 1, we see that x would need to be -1 (since -1 + 2 equals 1). However, -1 is not a natural number. Therefore, there are no natural numbers that satisfy the equation, so the set is empty.
Learn more about the Roster Method here:https://brainly.com/question/28709089
#SPJ3
The set of natural numbers x that satisfy x + 2 = 1. The correct answer is EMPTY.
To solve the equation x + 2 = 1 for natural numbers x we would first try to isolate x subtracting 2 from both sides of the equation:
x + 2 - 2 = 1 - 2
x = -1
However, natural numbers are defined as the set of positive integers, starting from 1 and increasing indefinitely.
Since the solution to the equation x = -1 is not a positive integer, it does not belong to the set of natural numbers. Therefore, there is no natural number x that satisfies the equation x + 2 = 1
Since there are no elements that satisfy the condition, the set is empty. Hence, the correct representation of the set using the roster method is EMPTY.
what is the value of cos (L)?
A cold tablet contains the following amounts of active ingredients: acetaminophen 325 mg, chlorpheniramine maleate 2 mg, and dextromethorphan hydrobromide 15 mg. How many tablets may be prepared if a manufacturing pharmacist has 1 kg of acetaminophen, 125 g of chlorpheniramine maleate, and unlimited quantities of dextromethorphan hydrobromide?
Answer:
The number of tablets that can be prepared is 3076.
Step-by-step explanation:
The total amount of active ingredients in the tablet is the sum of the amounts provided in the formula:
[tex]325 mg + 2mg+15 mg=342 mg[/tex]
The percentages of each component in the formula are:
Acetaminophen:[tex]\frac{325mg*100}{342mg}=95.03[/tex]%
Chlorpheniramine maleate:[tex]\frac{2mg*100}{342mg} =0.58[/tex]%
Dextromethorphan hydrobromide:[tex]\frac{15mg*100}{342mg}=4.39[/tex]%
If 1 Kg=[tex]10^{6}[/tex] mg of acetaminophen is used, the needed amount of chlorpheniramine maleate would be:
[tex]\frac{10^{6} mg *0.58}{95.03}=6153.85 mg[/tex]
Since there are 125 g = 125000 mg of chlorpheniramine maleate, there is enough of these ingredient to run the available acetaminophen out. Thus, the total amount of active ingredients that can be prepared with 1 kg of acetaminophen is:
[tex]\frac{10^{6}mg*100}{95.03}=1052307.7mg[/tex]
Since each tablet weighs 342 mg, the number of tablets that can be prepared is:
[tex]\frac{1052307.7mg}{342mg}=3076.923[/tex]
Which means that 3076 tablets can be prepared and a there will be a remanent of 0.923*342 mg = 315.69 mg of active ingredients.
Ingredients are used in ratio to prepare a specific product, generally. The number of tablets that can be manufactured for given context is 3076
How to form mathematical expression from the given description?You can represent the unknown amounts by the use of variables. Follow whatever the description is and convert it one by one mathematically. For example if it is asked to increase some item by 4 , then you can add 4 in that item to increase it by 4. If something is for example, doubled, then you can multiply that thing by 2 and so on methods can be used to convert description to mathematical expressions.
We are given that the considered cold tablet consists:
acetaminophen to chlorpheniramine maleate to dextromethorphan hydrobromide as 325 : 2 : 15 in weight.
Since we're given that there is unlimited quantities of dextromethorphan hydrobromid, the first two ingredients will be the one capping(limiting) the number of tablets that can be manufactured.
Suppose that 'n' tablets can be manufactured by the given amount of ingredients, then:
Amount of acetaminophen needed = 325 mg × n ≤ 1 kg (which is available amount)
Amount of chlorpheniramine maleate = 2 mg × n ≤ 125 g
Converting all scales of weight to grams, we get two inequalities:
[tex]0.325 n \leq 1000[/tex][tex]0.002n \leq 125[/tex](since 1 g = 0.001 mg , and 1 kg = 1000 g)
Remember that 'n' is amount of tablets which is going to be a whole number.
Solving the inequalities, we get:
[tex]0.325 n \leq 1000\\\\n \leq \dfrac{1000}{0.325} = 3076.9\\\\n \leq 3076[/tex]
and
[tex]0.002n \leq 125\\\\n \leq \dfrac{125}{0.002} = 62500[/tex]
So, we see that by the given amount of acetaminophen , we can only make 3076 tablets, but we can make 62500 tablets by the second ingredient chlorpheniramine maleate,
since both ingredients are necessary, so after 3076 tablets, first ingredient will exhaust.
Thus, The number of tablets that can be manufactured for given context is 3076
Learn more about inequalities here:
https://brainly.com/question/11901702
Suppose a manufacturer sells a product as $2 per unit. If q units are sold, (a) write the total revenue function, (b) and find the marginal revenue function. What does the constant marginal revenue function mean?
Answer:
We are given that a manufacturer sells a product as $2 per unit.
Quantity = q units
So, Total revenue = [tex]\text{Cost per unit} \times quantity[/tex]
Total revenue = [tex]2q[/tex]
So, the total revenue function is [tex]2q[/tex]
Marginal revenue is the derivative of the revenue functions
So, Marginal revenue = [tex]\frac{dR}{dq} =2[/tex]
The marginal revenue function is 2
The constant marginal revenue function mean that the revenue earned by the addition of the output is constant.
You are looking down at a map. A vector u with |u| = 6 points north and a vector v with |v| = 5 points northeast. The crossproduct u×v points:
(A) south
(B) northwest
(C) up
(D) down
(E) The magnitude |u×v| =
Answer: Hi!
First, UxV = sin(a)*IUI*IVI
where a is the angle between U and V, in this case 45°.
First, the cross product of UxV points:
Here you can use the right hand method,
Put your hand flat, so the point of your fingers point in the same direction that the first vector, in this case U, so your fingers will point to the north.
Now roll your fingers in the direction of the second vector, so here you will roll your fingers in the northeast direction. Now you will see that your thumb is pointing down, so the cross product of UxV points down.
And the magnitude is 6*5*sin(45) = 21.213
The cross-product u×v points: Option D: Down, and its magnitude |u×v| evaluates to [tex]|u \times v| = 15\sqrt{2}[/tex]
How to find the cross product of two vectors?Suppose that two vectors in consideration are u and v, then their cross product is evaluated as:
[tex]u \times v = |u|.|v|.sin(\theta)\hat{n}[/tex]
where [tex]\hat{n}[/tex] is the normal unit vector whose direction is decided by right hand thumb rule, and theta is the angle between u and v vector.
The two bars around a vector represents the magnitude of that vector.
Cross product returns the result as a vector itself.
For this case, we have:
|u| = 6 points, its direction is in north|v| = 5 points, its direction is in northeastThus, as north and northeast have 45 degrees in between them, therefore, we get:
[tex]u \times v = 6\times 5 \times sin(45^\circ) \hat{n} = 15\sqrt{2} \: \hat{n}[/tex]
Directing index finger of right hand to north direction, and middle to northeast makes the thumb go down, therefore, the direction of normal vector (and therefore direction of the resultant cross product vector too) is downside of this whole north south east west plane.
The magnitude of cross product is [tex]|u \times v| = 15\sqrt{2}[/tex]
Thus, the cross-product u×v points: Option D: Down, and its magnitude |u×v| evaluates to [tex]|u \times v| = 15\sqrt{2}[/tex]
Learn more about cross product here:
https://brainly.com/question/1968115
Rosie washes clothes for two families the first family pays her $500 more per month than the second. Her total earning per month is $3200. How much does she earn from each families?
Answer:
Family 1: $1.850Family 2: $1.350Step-by-step explanation:
We know that the total monthly payment is 3200, so if we call Pa (family A`s payment) and Pb (Family B's payment) the payments:
Pa+Pb=3200also, Pa = Pb+500So if we replace Pa in the first ecuation:
Pb+500+Pb=32002Pb= 3200-500Pb=2700/2= $1.350then Pa+Pb=3200 => Pa= 3200-1350= $1.850
Good Luck!
Pollsters are concerned about declining levels of cooperation among persons contacted in surveys. A pollster contacts 8686 people in the 18-21 age bracket and finds that 4343 of them respond and 4343 refuse to respond. When 276276 people in the 22-29 age bracket are contacted, 258258 respond and 1818 refuse to respond. Suppose that one of the 362362 people is randomly selected. Find the probability of getting someone in the 18 dash 2118-21 age bracket or someone who respondedresponded.
Answer:
0.9503
Step-by-step explanation:
First of all, there are some wrong figures in the original text. Because there is a total of 362362 people, the figures should be 86086 (people in the 18-21 age bracket), 43043 (people in the 18-21 age bracket who respond) and 43043 people in the 18-21 age bracket who refuse to respond. In the same way, because there are 276276 people in the 22-29 age bracket, it should be 18018 and not 1818 who refuse to respond in this subset of people. Now, let's define the following events:
R: a person respond
A: a person belongs to the 18-21 age bracket. So,
The number of people who respond is 43043 + 258258 = 301301, so
P(R) = 301301/362362 = 0.8315
P(A) = 86086/362362 = 0.2376
P(R | A) = 43043/86086 = 0.5
We are looking for P(A∪R) = P(A) + P(R) - P(A∩R),
P(A∩R) = P(R | A)P(A) = (0.5)(0.2376) = 0.1188, so,
P(A∪B) = 0.2376 + 0.8315 - 0.1188 = 0.9503
In art class Ms smith is working on polygons. She want the students to Create a picture of the polygons. She points out to the class That there is are three sides to a triangle, 4 sides on a quadrilateral , 5 sides on a pentagon, and six sides on a hexagon, How many more side are on a hexagon than on a quadrilateral
Answer:
There are two more sides on a hexagon than on a quadrilateral
Step-by-step explanation:
If the hexagon has 6 sides, and the quadrilateral has 4, then 6-4=2
Janet Woo decided to retire to Florida in 6 years. What amount should Janet invest today so she can withdraw $51,500 at the end of each year for 20 years after she retires? Assume Janet can invest money at 6% compounded annually. (Use the Table 13.2 and Table 12.3.) Present value ??
Answer:
$293,562.707
Step-by-step explanation:
As for the provided details we know,
Janet needs $51,500 from end of 7th year for upcoming 20 years.
The present value of 20 installments of $51,500 shall be @ 6% from year 7 to year 8.0858
Thus total value = $51,500 [tex]\times[/tex] 8.0858 = $416,418.7
Now the compound interest factor for 6 year @ 6 % = 1.4185
Thus, value to be invested today = $416,418.70/1.4185 = $293,562.707
As this when compounded annually will provide the balance as required at the end of 6 years.
which expression has the greatest value |-21|, |14|, |30|, |-45|
Answer:
|-45|
Step-by-step explanation:
In mathematics, the absolute value of a real number is the numeric value of the number, regardless the sign, either this is positive or negative.
The absolute value function can be definied as:
|a|=a si a ≥0|a|=-a si a <0Using this definition, we have:
|-21| = -(-21) = 21
|14| = 14
|30| = 30
|-45| = -(-45) = 45
Therefore, the expression |-45| has the greatest value.
A piece of toast came out of the toaster very overcooked.
What kind of change occurred?
chemical change
change in reaction
phase change
physical change
Answer:
It is a chemical change ⇒ 1st answer
Step-by-step explanation:
* Lets explain the statements to solve the problem
- A chemical change occurs when a new substance is formed through
a chemical reaction
- Ex: cooking an egg
- Change of reaction is the rate of reaction it can be decreases or
increasing
- A phase change is a change from one state to another without a
change in chemical composition
- Ex: Condensation: the substance changes from a gas to a liquid
- A physical change, such as a state change or dissolving, but does
not create a new substance
- Ex: Breaking a glass
* Lets solve the problem
- A piece of toast came out of the toaster very overcooked.
∵ It is like the cooking an egg
∴ It is a chemical change
Answer:
Chemical
Step-by-step explanation:
Formulate the situation as a system of two linear equations in two variables. Be sure to state clearly the meaning of your x- and y-variables. Solve the system by the elimination method. Be sure to state your final answer in terms of the original question.
A lawyer has found 60 investors for a limited partnership to purchase an inner-city apartment building, with each contributing either $4,000 or $8,000. If the partnership raised $348,000, then how many investors contributed $4,000 and how many contributed $8,000?
x = $4,000 investors
y =
$8,000 investors
Solve the system by row-reducing the corresponding augmented matrix. (Enter your answers as a comma-separated list. If the system is inconsistent, enter INCONSISTENT. If the system is dependent, enter DEPENDENT.)
leftbrace2.gif
2x + y = 17
x + y = 13
the situation as a system of two linear equations in two variables. Be sure to state clearly the meaning of your x- and y-variables. Solve the system by the elimination method. Be sure to state your final answer in terms of the original question.
A jar contains 80 nickels and dimes worth $6.80. How many of each kind of coin are in the jar?
x = nickels
y = dimes
Answer:
1) There were 33 $4,000 investors and 27 $8,000 investors.
2) The solution in x = 4, y = 9.
3) There were 24 nickels and 56 dimes.
Step-by-step explanation:
1) A lawyer has found 60 investors for a limited partnership to purchase an inner-city apartment building, with each contributing either $4,000 or $8,000. If the partnership raised $348,000, then how many investors contributed $4,000 and how many contributed $8,000?
I am going to say that:
x is the number of investors that contributed 4,000.
y is the number of investors that contributed 8,000.
Building the system:
There are 60 investors. So:
[tex]x + y = 60[/tex]
In all, the partnership raised $348,000. So:
[tex]4000x + 8000y = 348000[/tex]
I am going to simplify by 4000. So:
[tex]x + 2y = 87[/tex]
Solving the system:
The elimination method is a method in which we can transform the system such that one variable can be canceled by addition. So:
[tex]1)x + y = 60[/tex]
[tex]2)x + 2y = 87[/tex]
I am going to multiply 1) by -1. So we have
[tex]1)-x - y = -60[/tex]
[tex]2)x + 2y = 87[/tex]
By addition, the x are going to cancel each other
[tex]-x + x - y + 2y = -60 + 87[/tex]
[tex]y = 27[/tex]
For x:
[tex]x + y = 60[/tex]
[tex]x = 60-y = 60-27 = 33[/tex]
There were 33 $4,000 investors and 27 $8,000 investors.
2) Solve the system by row-reducing the corresponding augmented matrix.
[tex]2x + y = 17[/tex]
[tex]x + y = 13[/tex]
This system has the following augmented matrix:
[tex]\left[\begin{array}{ccc}2&1&17\\1&1&13\end{array}\right][/tex]
To help the row reducing, i am going to swap the first with the second line:
[tex]L1 <-> L2[/tex]
So we have:
[tex]\left[\begin{array}{ccc}1&1&13\\2&1&17\end{array}\right][/tex]
Now, reducing the first column.
[tex]L2 = L2 - 2L1[/tex]
So we have:
[tex]\left[\begin{array}{ccc}1&1&13\\0&-1&-9\end{array}\right][/tex]
Now we do:
[tex]L2 = -L2[/tex]
And the matrix is:
[tex]\left[\begin{array}{ccc}1&1&13\\0&1&9\end{array}\right][/tex]
Now to reduce the second column, we do:
[tex]L1 = L1 - L2[/tex]
[tex]\left[\begin{array}{ccc}1&0&4\\0&1&9\end{array}\right][/tex].
So the solution is:
x = 4, y = 9.
3) A jar contains 80 nickels and dimes worth $6.80. How many of each kind of coin are in the jar?
I am going to say that x is the number of nickels and y is the number of dimes.
Each nickel is worth 5 cents and each dime is worth 10 cents.
Building the system:
There are 80 coins in all:
[tex]x + y = 80[/tex]
They are worth $6.80. So:
[tex]0.05x + 0.10y = 6.80[/tex]
Solving the system:
[tex]1)x + y = 80[/tex]
[tex]2)0.05x + 0.10y = 6.80[/tex]
I am going to divide 1) by -10, so we can cancel y. So:
[tex]1)-0.10x - 0.10y = -8[/tex]
[tex]2)0.05x + 0.10y = 6.80[/tex]
Adding:
[tex]-0.10x + 0.05x - 0.10y + 0.10y = -8 + 6.80[/tex]
[tex]-0.05x = -1.2[/tex] *(-100)
[tex]5x = 120[/tex]
[tex]x = \frac{120}{5}[/tex]
[tex]x = 24[/tex]
Also
[tex]x + y = 80[/tex]
[tex]y = 80-x = 80-24 = 56[/tex]
There were 24 nickels and 56 dimes.
Solve the initial value problem: dydx+5y=7 y(0)=0
Answer:
Given differential equation,
[tex]\frac{dy}{dx}+5y=7[/tex]
[tex]\frac{dy}{dx}=7-5y[/tex]
[tex]\implies \frac{dy}{7-5y}=dx[/tex]
Taking integration both sides,
[tex]\int \frac{dy}{7-5y}=\int dx[/tex]
Put 7 - 5y = u ⇒ -5 dy = du ⇒ dy = -du/5,
[tex]-\frac{1}{5} \int \frac{du}{u} = \log x + C[/tex]
[tex]-\frac{1}{5} \log u = \log x + C[/tex]
[tex]-\frac{1}{5}\log(7-5y) = \log x + C---(1)[/tex]
Here, x = 0, y = 0
[tex]\implies -\frac{1}{5} \log 7= C[/tex]
Hence, from equation (1),
[tex]-\frac{1}{5}\log(7-5y)=\log x -\frac{1}{5}log 7[/tex]
[tex]\log(7-5y)=\log (\frac{x}{7^\frac{1}{5}})[/tex]
[tex]7-5y=\frac{x}{7^\frac{1}{5}}[/tex]
[tex]7-\frac{x}{7^\frac{1}{5}}=5y[/tex]
[tex]\implies y=\frac{1}{5}(7-\frac{x}{7^\frac{1}{5}})[/tex]
The 1992 world speed record for a bicycle (human-powered vehicle) was set by Chris Huber. His time through the measured 200 m stretch was a sizzling 6.509 s, at which he commented,"Cogito ergo zoom!" (I think, therefore I go fast!). a.) What was Chris Huber’s speed in meters per second(m/s)? b) In 2001, Sam Whittingham beat Huber’s record by 19.0 km/h. What was Whittingham’s time through the 200 m? (answer hours)
Answer:
a) 30.726m/s and b) 5.5549s
Step-by-step explanation:
a.) What was Chris Huber’s speed in meters per second(m/s)?
Given the distance and time, the formula to obtain the speed is
[tex]v=\frac{d}{t}[/tex].
Applying this to our problem we have that
[tex]v=\frac{200m}{6.509s}= 30.726m/s[/tex].
So, Chris Huber’s speed in meters per second(m/s) was 30.726m/s.
b) What was Whittingham’s time through the 200 m?
In a) we stated that [tex]v=\frac{d}{t}[/tex]. This formula implies that
[tex]t=\frac{d}{v}[/tex].First, observer that [tex]19\frac{km}{h} =19,000\frac{m}{h}=\frac{19,000}{3,600}m/s= 5.2777m/s[/tex].
Then, Sam Whittingham speed was equal to Chris Huber’s speed plus 5.2777 m/s. So, [tex]v=30.726\frac{m}{s} +5.2777\frac{m}{s}= 36.003 m/s.[/tex]
Then, applying 1) we have that
[tex]t=\frac{200m}{36.003m/s}=5.5549s.[/tex]
So, Sam Whittingham’s time through the 200 m was 5.5549s.
suppose you deposit $1000 in an account paying 4.6% annual interest compounded continuously. How long will it take for the money to double?
Answer: About 16 years
Step-by-step explanation:
The formula to find the compound amount if compounded continuously is given by :-
[tex]A=Pe^{rt}[/tex], where P is Principal amount, r is the rate of interest ( in decimal) and t is time ( in years).
Given : P= $1000 ; r= 4.6%=0.046
let t be the time it will take to double the amount, the we have
[tex]2(1000)=(1000)e^{0.046\times t}[/tex]
Dividing 1000 both sides, we get
[tex]2=e^{0.046 t}[/tex]
Taking natural log on each side, we get
[tex]\ln2=\ln(0.046\times t)\\\\\Rightarrow\ 0.6931=0.046t\\\\\Rightarrow\ t=\dfrac{0.6931}{0.046}=15.0673913043\approx16\text{ years}[/tex]
Hence, it will take about 16 years to double the amount.
A snorkeler dives for a shell on a reef. After entering the water, the diver decends 11/3 ft in one second. Write an equation that models the divers position with respect to time.
Answer:
[tex]h(t)=-\dfrac{11}{3}t[/tex]
Step-by-step explanation:
A snorkeler dives for a shell on a reef. After entering the water, the diver decends [tex]\frac{11}{3}[/tex] ft in one second.
Let t be the time passed after entering the water, in seconds, and h(t) be the position of the snorkeler under the water, in feet.
The initial position of the snorkeler was 0 feet under the water.
An equation that models the divers position with respect to time is
[tex]h(t)=0-\dfrac{11}{3}t\\ \\h(t)=-\dfrac{11}{3}t[/tex]
Here the position is negative, because the diver decends (he deepens under the water)
Clearly write what it means for a set to be closed under an operation
Answer:
A set S is closed under an operation * (we're denoting the operation as asterisk) IF for any two elements a,b in S, the result a*b is also in S.
Step-by-step explanation:
A set being closed under an operation means that whenever you operate elements from the set, the result you get out of it is ALWAYS inside the set. For example, think of the set Z of integer numbers and the operation + (usual addition). If we add ANY two integers, we're going to get another integer. Or said in terms of sets, for any two numbers a,b in Z, a+b is also in Z.
On the other side, not being closed under an operation means you do NOT ALWAYS get results inside the same set. Think of the set of natural numbers N, and the operation - (usual difference). If we do the operation 5-12, we get -7 which is NOT in the set of natural numbers. So N is not closed under subtraction.
Plot another sin function of 20% higher frequency over the same range.
Step-by-step explanation:
The frequency of sine function is given by the number of periods in a given range. For example:
Frequency for [tex]sin(x)[/tex] is 1 in the interval [tex][0,2\pi][/tex].This means that, if we want another sine function with frequency 20% higher, we need that function to have a frequency of 1.2 in the interval [tex][0,2\pi][/tex].
To be easier to see we will consider interval [tex][0,10\pi][/tex] instead of [tex][0,2\pi][/tex]. In this interval [tex]sin(x)[/tex] has 5 periods, therefore our new sine function should have 6 periods.
Finally, as we can see in the graph, the function [tex]sin(\frac{6}{5}x )[/tex] (in blue) has a frequency 20% higher than [tex]sin(x)[/tex] (in red). This can be easily seen counting the number of periods between 0 and [tex]10\pi[/tex] for both functions. 5 for [tex]sin(x)[/tex] and 6 for [tex]sin(\frac{6}{5} x)[/tex].
At lunchtime, Ciaran buys a sandwich.
He can choose white bread or brown bread.
What is the probability that he chooses brown bread?
Answer:
1/2.
Step-by-step explanation:
There are 2 choices and he has to choose 1 , so the answer is 1/2.
Assuming Ciaran has no preference for white or brown bread, and each choice is equally likely, the probability of choosing brown bread is 1/2 or 50%.
The probability that Ciaran chooses brown bread depends on the assumption that he has no preference and that the choices are equally likely. If the only options available to Ciaran are white bread or brown bread, and each choice is equally likely, then the probability of choosing one over the other is 1 out of the total number of options.
In this case, there are 2 options (white or brown), so the probability that Ciaran will choose brown bread is 1/2 or 0.5, which can also be expressed as a 50% chance.
You're driving into Canada and trying to decide whether to fill your gas tank before or after crossing the border. Gas in the United States costs $2.58/gallon, in Canada it's $1.29/L, and the Canadian dollar is worth 79¢ in U.S. currency. Where you should fill up?
Answer:
It is more convenient to fill up in the United States.
Step-by-step explanation:
We convert a US gallon to liters:
1 Gallon equals 3.78541 liters.
Therefore, 1 US Gallon costs (3.78541) x (1.29) = 4.8831789 Canadian dollars.
Now we convert the price of a US gallon in Canadian territory to US dollars:
4.8831789 * 0.79 = 3.85771133 US dollars.
Conclusion: A gallon purchased in the United States costs 2.58 US dollars, while a gallon in Canada is equivalent to 3.85771133 US dollars. This way it is more convenient to fill up in the United States.
It would be more economical to fill up the gas tank in the United States before crossing into Canada.
To determine whether it is more cost-effective to fill up the gas tank in the United States or Canada, we need to convert the Canadian gas price into U.S. dollars per gallon. The price of gas in Canada is $1.29 per liter. Since 1 gallon is equal to 3.78541 liters, the price per gallon in Canada would be $1.29 × 3.78541 = $4.88238 CAD. Now, we need to convert this price into USD, knowing the Canadian dollar is worth 79¢ in U.S. currency. Therefore, $4.88238 CAD × 0.79 = $3.85728 USD per gallon.
The price of gas in the United States is $2.58 per gallon. When comparing the two prices, it is clear that $2.58 per gallon in the United States is cheaper than the converted price of gas in Canada ($3.85728 per gallon in USD). Hence, it would be more economical to fill up the gas tank in the United States before crossing into Canada.
A long-distance telephone company claims that the mean duration of long-distance telephone calls originating in one town was greater than 9.4 minutes, which is the average for the state. Determine the conclusion of the hypothesis test assuming that the results of the sampling don’t lead to rejection of the null hypothesis.
(A) Conclusion: Support the claim that the mean is less than 9.4 minutes.
(B) Conclusion: Support the claim that the mean is greater than 9.4 minutes.
(C) Conclusion: Support the claim that the mean is equal to 9.4 minutes.
(D) Conclusion: Do not support the claim that the mean is greater than 9.4 minutes.
Answer:
(D) Conclusion: Do not support the claim that the mean is greater than 9.4 minutes
Step-by-step explanation:
Let X be the mean duration of long distance telephone class originating in one town.
[tex]H_0: x bar = 9.4\\H_a: x bar >9.4[/tex]
(one tailed test)
The conclusion of the hypothesis test assuming that the results of the sampling don’t lead to rejection of the null hypothesis.
This means that there is no statistical evidence to support the alternate claim.
Hence option D is right.
(D) Conclusion: Do not support the claim that the mean is greater than 9.4 minutes
Write x'" = x + t as a first order system
Answer:
y = x'
z = y'
z' = x + t
Step-by-step explanation:
Hi!
You need to define two new variables y and z:
y = x'
z = y'
Then:
z = y' = x''
z' = x''' = x + t
Now you have a system of 3 equations with only first derivatives
Problem 4.28: People with type O-negative blood are
universaldonors. That is, any patient can receive a transfusion
ofO-negative blood. Only 7% of the American population
haveO-negative blood. If 10 people appear at random to give blood,
whatis the probability that at least 1 of them is a
universaldonor?
Answer: 0.516
Step-by-step explanation:
Binomial probability distribution formula to find the probability of getting success in x trial:-
[tex]P(x)=^nC_xp^x(1-p)^{n-x}[/tex], where n is the number of trials , p is the probability of getting success in each trial.
Given : People with type O-negative blood are universal donors.
The proportion of the American population have O-negative blood =0.07
For n=10, the probability that at least 1 of them is a O-negative blood :-
[tex]P(x\geq1)=1-P(x=0)\\\\=1-[^{10}C_0(0.07)^0(1-0.07)^{10}]\\\\=1-[(1)(1)(0.93)^{10}]\ \ \ [\text{ Since}^nC_0=1]\\\\=1-0.483982307179\approx1-0.4840=0.516[/tex]
Hence, the probability that at least 1 of them is a universal donor = 0.516
A local fraternity is conducting a raffle where 50 tickets areto be sold--one per customer. There are three prizes to beawarded. If the four organizers of the raffle each buy oneticket, what is the probability that the four organizers
a) win all of the prizes?
b) win exactly two of the prizes?
c) win exactly one of the prizes?
d) win none of the prizes?
The answers:
a) (4) / (19600)
b) (276) / (19600)
c) (4140) / (19600)
d) (15180) / 19600)
Answer:
The answers are the same you stated.
The calculations are in the step-by-step explanation
Step-by-step explanation:
There are 3 prizes to be distributed among 50 tickets. The order these prizes are distributed does not matter. So the total number of prizes is a combination of 3 from 50.
The formula for a combination of n from m is:
[tex]C(m,n) = \frac{m!}{n!(m-n)!}[/tex]
So, the total number of prizes is:
[tex]T = C(50,3) = \frac{50!}{3!(50-3)!} = \frac{50*49*48*47!}{3!*47!} = 19600[/tex]
what is the probability that the four organizers
a) win all of the prizes?
The number of ways that the four organizers can will all of the prizes is a combination of 3 from 4.
[tex]C(4,3) = \frac{4!}{3!1!} = 4[/tex]
The probability that the win all of the prizes is the number of ways that they can win all the prizes divided by the total numbers of ways that the prizes can be distributed.
[tex]P = \frac{4}{19600}[/tex]
b) win exactly two of the prizes?
The total outcomes(total number of ways that the prizes can be distributed) is 19600.
For them to win exactly two of the prizes, we have a combination of 2 from 4(two organizers win prizes) multiplied by a combination of one from 46(one non-organizers wins a prize), so:
[tex]C(4,2)*C(46,1) = \frac{4!}{2!2!}*\frac{46!}{1! 45!} = 6*46 = 276[/tex]
The probability that they win exactly two of the prizes is
[tex]P = \frac{276}{19600}[/tex]
c) win exactly one of the prizes?
The total outcomes(total number of ways that the prizes can be distributed) is 19600.
For them to win exactly one of the prizes, we have a combination of 1 from 4(one organizer wins a prize) multiplied by a combination of two from 46(two non-organizers win prizes), so:
[tex]C(4,1)*C(46,2) = \frac{4!}{1!3!}*\frac{46!}{2! 44!} = 4*1035 = 4150[/tex]
The probability that they win exactly one prize is
[tex]P = \frac{4150}{19600}[/tex]
d) win none of the prizes?
The total outcomes(total number of ways that the prizes can be distributed) is 19600.
For them to win none of the prizes, we have a combination of 3 from 46(3 prizes distributed among 46 non-organizers). So:
[tex]C(46,3) = \frac{46!}{43!3!} = 15180[/tex]
The probability that they don't win any prize is:
[tex]P = \frac{15180}{19600}[/tex]
The probabilities of winning all of the prizes, winning exactly two of the prizes, winning exactly one of the prizes, and winning none of the prizes are [tex]\rm \dfrac{4}{19600},\dfrac{276}{19600},\dfrac{4140}{19600}, \ and \ \dfrac{15180}{19600}[/tex] respectively.
What is probability?Probability means possibility. It deals with the occurrence of a random event. Its basic meaning is something is likely to happen. It is the ratio of the favorable event to the total number of events.
Given
A local fraternity is conducting a raffle where 50 tickets are to be sold one per customer.
There are three prizes to be awarded.
If the four organizers of the raffle each buy one ticket.
The total outcomes will be
[tex]\rm ^{50}C_3 = \dfrac{50!}{47! *3!}\\\\\rm ^{50}C_3 = 19600[/tex]
a) win all of the prizes.
The number of ways the four organizers can win all the prizes is given by
[tex]\rm ^4C_3 = \dfrac{4!}{3! *1!}\\\\\rm ^4C_3 = 4[/tex]
Then the probability of winning all the prizes will be
[tex]\rm Probability = \dfrac{4}{19600}[/tex]
b) win exactly two of the prizes.
The number of ways is 2 from organizers and 1 from non-organizers given by
[tex]\rm ^4C_2 * ^{46}C_1 = \dfrac{4!}{2! *2!}*\dfrac{46!}{45! *1!}\\\\\rm ^4C_2 * ^{46}C_1 = 276[/tex]
Then the probability of 2 from organizers and 1 from non-organizers will be
[tex]\rm Probability = \dfrac{276}{19600}[/tex]
c) win exactly one of the prizes.
The number of ways is 1 from organizers and 2 from non-organizers given by
[tex]\rm ^4C_1 * ^{46}C_2 = \dfrac{4!}{1! *3!}*\dfrac{46!}{44! *2!}\\\\\rm ^4C_1 * ^{46}C_2 = 4140[/tex]
Then the probability of 1 from organizers and 2 from non-organizers will be
[tex]\rm Probability = \dfrac{4140}{19600}[/tex]
d) win none of the prizes.
The number of ways is 3 from non-organizers given by
[tex]\rm ^{46}C_3 = \dfrac{46!}{43! *3!}\\\\\rm ^{46}C_3 = 15180[/tex]
Then the probability of 3 from non-organizers will be
[tex]\rm Probability = \dfrac{15180}{19600}[/tex]
Thus, The probabilities of winning all of the prizes, winning exactly two of the prizes, winning exactly one of the prizes, and winning none of the prizes are [tex]\rm \dfrac{4}{19600},\dfrac{276}{19600},\dfrac{4140}{19600}, \ and \ \dfrac{15180}{19600}[/tex] respectively.
More about the probability link is given below.
https://brainly.com/question/795909
Determine the sum of the first k odd positive integers for a number of values of k. What generalizations occur to you? Are your inferences correct for all positive integers k?
Answer:
[tex]S_{n} = \sum_{k=1}^{n} (2k-1) = n^2[/tex]
Step-by-step explanation:
Let's take a look at the first few odd numbers and their sum.
Lets define [tex]O_{k}[/tex] as the [tex]kth[/tex] odd number as:
[tex]O_{k} = 2k-1[/tex]
So we have:
[tex]O_{1} = 1\\O_{2} = 3\\O_{3} = 5\\O_{4} = 7\\[/tex]
And lets define the sum of all the odd numbers from [tex]k=1[/tex] to [tex]k=n[/tex] as:
[tex]S_{n} = \sum_{k=1}^n O_{k} = \sum_{k=1}^n (2k-1)[/tex]
Lets now check some values of said sum:
[tex]S_{1} = 1\\S_{2} = 1 + 3 = 4\\S_{3} = 1 + 3 + 5 = 9\\S_{4} = 1+3+5+7 = 16[/tex]
We can then observe than the sum up to [tex]n[/tex] equals [tex]n^2[/tex]
Let us then prove that this is the case by Induction.
First of all, we can prove this by an Induction Proof because we are taking all positive Integers. This is, we are working with the set of natural numbers [tex]\mathbb{N}[/tex].
We want to prove that
[tex]P(n) = S_{n} = \sum_{k=1}^n = n^2 \forall n\in \mathbb{N}[/tex]
This is, we want to prove that the sum of all odd numbers from [tex]1[/tex] to [tex]n[/tex] equals [tex]n^2[/tex] for all natural numbers.
Now, in order to prove something by Induction we need to check 2 things:
[tex]1) The\ base\ case . \ The\ statement\ holds\ for\ n=1\\2) The\ inductive\ step.\ Prove\ that\ if\ the\ statement\ holds\ for\ n\ then\ it\ must\ hold\ for\ n+1\\[/tex]
[tex]P(1)[/tex] is immediate:
[tex]P(1) = \sum_{k=1}^1 2k-1 = 1 = 1^2[/tex]
Now let's assume the statement holds for [tex]P(n)[/tex] and let's take a look at [tex]P(n+1)[/tex]
[tex]P(n+1) = \sum_{k=1}^{n+1} 2k-1[/tex]
And we can rewrite it by taking the last term out as:
[tex]P(n+1) = \sum_{k=1}^n 2k-1 \ + 2.(n+1) - 1[/tex]
And by inductive hypothesis we know that [tex]\sum_{k=1}^n 2k-1 = n^2[/tex]
and then:
[tex]P(n+1) = \sum_{k=1}^n 2k-1 \ + 2.(n+1) -1 = n^2 + 2n +2 -1 = n^2 +2n +1 = (n+1)^2[/tex]
And we have the proof we were looking for!
Which of the following angles have equal measure when a pair of parallel lines are crossed by a transversal?
supplementary angles
complementary angles
corresponding angles
adjacent angles
Answer:
corresponding angles
Step-by-step explanation:
Corresponding angles are congruent where a transversal crosses parallel lines. Such a geometry has 4 pairs of corresponding angles. The corresponding angles of each pair are congruent.
Answer:
corresponding angles
Step-by-step explanation:
I need to find the standard for Hamilton's method to figure out how many teachers should be at each school
Answer:
The standard divisor is 22.48.
Step-by-step explanation:
There are a total of 3259 students at the 5 schools. Then dividing that number by the number of teachers (145) we get the "standard divisor" of ...
3259/145 ≈ 22.48
__
By Hamilton's method, that divisor is used to divide the number of students at each school, and the result is rounded down. This is the initial allocation of teachers to schools. The remainders from the division are examined. Starting with the largest and working down, one additional teacher is assigned until all the unassigned teachers have been assigned.
For this problem, the initial assignment results in 142 teachers being assigned, so there are 3 more that can be allocated. In order, the highest three remainders are associated with the number of students at East, Central, and South. Each of those schools gets one more teacher than the number initially assigned. The final allocation of teachers is highlighted in the attachment.