Answer: Transitive property of congruence.
Step-by-step explanation:
Transitive property of congruence is similar to transitive property of equality , it says that if a is congruent to b and b is congruent to c , then a is congruent to c.
Given statement : If AB is congruent to CD and CD is congruent to EF, then AB is congruent to EF.
Here a= AB , b= CD and c= EF
Therefore, the geometric postulates justifies given the statement must be Transitive property of congruence.
9 + 22 = x + y
x = 5
y = ?
Answer:
y = 26
Step-by-step explanation:
9 + 22 = x + y
9 + 22 = 31
31 = x + y
x = 5
31 = 5 + y
31 - 5 = 26
y = 26
Hey!
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Solution:
9 + 22 = 5 + y
31 = 5 + y
31 - y = 5 + y - y
31 - y = 5
31 - y - 31 = 31 - 5
y = 26
-----------------------------------------------
Answer:
y = 26
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Hope This Helped! Good Luck!
A telemarketer makes a sale on 25% of his calls. If he makes 300 calls in a night, what is the probability that he will make more than 70 sales but less than 90 sales?
Answer: 0.7258
Step-by-step explanation:
Given : A telemarketer makes a sale on 25% of his calls.
i.e. p=0.25
He makes 300 calls in a night, i.e. n=300
Let x be a random variable that represents the number of calls make in night.
To convert the given binomial distribution to normal distribution we have :-
[tex]\mu=np=300(0.25)=75[/tex]
[tex]\sigma=\sqrt{p(1-p)n}=\sqrt{(0.25)(1-0.25)(300)}\\\\=\sqrt{56.25}=7.5[/tex]
Now, using [tex]z=\dfrac{x-\mu}{\sigma}[/tex], the z-value corresponds to x= 70 :-
[tex]z=\dfrac{70-75}{7.5}\approx-0.67[/tex]
The z-value corresponds to x= 90 :-
[tex]z=\dfrac{90-75}{7.5}\approx2[/tex]
By using the standard normal distribution table for z, the probability that he will make more than 70 sales but less than 90 sales:-
[tex]P(-0.67<z<2)=P(z<2)-P(z<-0.67)\\\\=P(z<2)-(1-P(z<0.67))\\\\=0.9772-(1-0.7486)\\\\=0.9772-0.2514=0.7258[/tex]
Hence, the probability that he will make more than 70 sales but less than 90 sales= 0.7258
An electricity company charges its customers a fixed base charge of $6 a month, plus 10 cents per kilowatt-hour (kWh) for the first 500 kWh, 11 cents per kWh for the next 500 kWh, and 15 cents for all additional kWh. Express the monthly cost E as a function of the amount x of electricity used.
Answer:
[tex]E(k)= \left[ \begin{array}{c} {6+0.10k\,\,if\,\,0\leq k \leq500} & 6+500\times0.10+0.11(k-500)\,\,if\,\,500<k\leq1000& 6+500\times0.10+500\times0.11+0.15(k-1000)\,\,if\,\,k>1000\end{array}[/tex]
Step-by-step explanation:
The cost function has 3 branches,
Consumptions less than 500Consumptions less than 1000consumptions above a 1000So the first branch the consumer pays 6 dollars plus .10cents for any additional kWh (k)
In the second, they pay the same as the first up to 500kWh, and after that they pay 0.11 for the additional kWh above 500: (k-500) but bellow 1000
In the third branch, for consumptions above 1000, they pay the fix amount, plus .10 for the first 500 ([tex]500\times0.10[/tex]) , .11 for the additional 500 ([tex]500\times0.11[/tex]) , and finally 0.15 for consumptions above 1000: [tex]0.15(k-1000)[/tex]
Final answer:
The monthly cost of electricity, E, as a function of the electricity used, x, is defined using a piecewise function with different rates for usage tiers, including a fixed base charge and variable rates for different consumption brackets.
Explanation:
To express the monthly cost E as a function of the amount x of electricity used by the customer, we consider the following rates:
A fixed base charge of $6 per month.10 cents per kilowatt-hour (kWh) for the first 500 kWh.11 cents per kWh for the next 500 kWh (501 to 1000 kWh).15 cents per kWh for all additional kWh (above 1000 kWh).Therefore, the function is a piecewise function defined as:
E(x) =
6 + 0.10x if 0 ≤ x ≤ 5006 + (0.10 × 500) + 0.11(x - 500) if 500 < x ≤ 10006 + (0.10 × 500) + (0.11 × 500) + 0.15(x - 1000) if x > 1000This function helps calculate the monthly electricity bill based on the usage levels. Remember that the first bracket is for the initial 500 kWh, the second bracket is for 501 to 1000 kWh, and the last bracket applies to usage above 1000 kWh.
Tim has just inherited £1000 and he decides not to invest it but rather to spend 10% of the remaining money each month. Each month Tim's saving thus reduce by a factor of 0.9 Calculate, to the nearest integer, the amount remaining (in E) after 20 months. You may find it useful to first devise a formula for the amount remaining after n months. Enter your answer, to the nearest integer and without units, in the box below. Answer:
Answer:
£ 121.57
Step-by-step explanation:
As given in question,
Total amount inherited by Tim = £1000
He spends each month = 10 % of remaining money
savings reduced each month = 0.9 of remaining amount
So, the amount of money after one month = 0.9 x £1000
amount of money remained after 2 months = 0.9 x 0.9 x £1000
amount of money remained after 3 months = 0.9 x 0.9 x 0.9 x £1000
Hence, the amount of money remained after the n months can be given by,
[tex]E\ =\ 0.9^n\times 1000[/tex]
Hence, amount of money remained after 20 months can be given by,
[tex]E_{20}\ =\ 0.9^{20}\times1000[/tex]
= £ 121.57
So, the amount of money remained after 20 months will be £ 121.57.
a plant grows at a rate of 7 ft every two weeks
Answer:
0 is 0,1 is 7,2 is 14 and 3 is 20 (c. is that it has not grown yet and (d. is 7 per week
Answer:
Step-by-step explanation:
A. (0,0) (1,7) (2,14) (3,21)
C. (0,0) is the starting point
The purpose of this assignment is to make sures linear equations. This skill is required when Chapter 5 is covered. Problem Statement Given is a linear Equation 1 in the format of y mx+b. Equation 1, y = 46.888 +0.78X Plot Equation 1 on a graph paper. Label all axes and markers on the graph. 1. Save your graph in a pdf file zero point is issued for non pdf file format 2. Click on the assignment link. W01 A3a. below to upload your pdf file via attachment by the due date. 3. If you do not see the assignment link it is because the assignment is past due. Wk01 A3a De as
Answer: The plot will be linear as described by the equation of the problem
Step-by-step explanation:
The first thing to do is to determine values of X and place them in one column. The following column should include all values solved by the equation in terms of y. Once you plot the graph, you should have something as the graph attached. To save it as a pdf, simply scan your graph paper to pdf and follow the instructions for assignment submission.
Compare the values of the underlined digits 631,485 and 682 the underlined is 6.
Answer:
First one: Its place value is 600,000.
Second one: Its place value is 600.
Step-by-step explanation:
A cell phone company has a fixed cost of $1,000,000 per month and a variable cost of $20 per month per subscriber. The company charges $29.95 per month to its cell phone customers. (2) a. What is the breakeven point for this company?
Answer:
100503
Step-by-step explanation:
Data provided in the question:
Fixed cost per month for the cell phone company = $1,000,000
Variable cost per month per subscriber = $20
Charges for the customer per month = $29.95
Now,
the breakeven point is calculated as:
Breakeven point = [tex]\frac{\textup{Total fixed cost}}{\textup{Charges - variabel cost}}[/tex]
on substituting the respective values, we get
Breakeven point = [tex]\frac{\textup{1,000,000}}{\textup{29.95 - 20}}[/tex]
or
Breakeven point = 100502.51 ≈ 100503
A sales representative for a Children's Fashion store earns a salary of $1100.00 per month plus a commission based on the total sales. During the month of January, this sales representative's total sales were $8300.00 and was paid $2130.00. What is this sales representative's commission rate?
a.
14.30963855%
b.
11.10963855%
c.
13.60963855%
d.
12.40963855%
e.
14.60963855%
f.
None of the above.
Answer:
Option d. 12.40963855% is the answer.
Step-by-step explanation:
Monthly salary of a sales representative = $1,100
He was paid for the month = $2,300
His total sales were = $8,300
His commission for the month of sales = Total payment - monthly salary
= 2,300 - 1,100
= $1,030
He got $1,030 as commission on the sales of $8,300.
The percentage of the commission = [tex]\frac{1030}{8300}\times 100[/tex]
= 12.40963855421687%
Option d. 12.40963855% is the answer.
A web site was hit 300 times over a period of 15 days Show that over some period of 3 consecutive days, it was hit at least 60 times.
Answer:
We will divide the 15 days in five periods of 3 consecutive days each.
Now to solve this we will use the pigeonhole principle.
This states that if (N+1) pigeons occupy N holes, then some hole must have at least 2 pigeons.
So, we have n=300 pigeons and k=5 holes.
[tex][\frac{n}{k} ]=[\frac{300}{5} ][/tex]
Hence, there is a period of 3 consecutive days in which the website was hit at least 60 times.
calculate the perimeter of a rectangle with a length of 17.5 cm and a width of 40 mm in cm
Answer: 43 cm
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 : The length of rectangle is 17.5 cm and the width is 40 mm in cm.
Since , 1 cm = 10 mm
Then, 40 mm= [tex]\dfrac{40}{10}\ cm=4\ cm[/tex]
Then, the perimeter of rectangle will be :-
[tex]P=2(17.5+4)\\\\\Rightarrow\ P=2(21.5)\\\\\Rightarrow\ P=43\ cm[/tex]
Hence, the perimeter of rectangle = 43 cm
Find the area of the surface correct to four decimal places by expressing the area in terms of a single integral and using your calculator to estimate the integral. The part of the surface z=cos(x^2+y^2) that lies inside the cylinder x^2+y^2=1
If we substitute [tex]x=r\cos\theta[/tex] and [tex]y=r\sin\theta[/tex], we get [tex]r^2=x^2+y^2[/tex], so that
[tex]z=\cos(x^2+y^2)=\cos(r^2)[/tex]
which is independent of [tex]\theta[/tex], which in turn means the surface can be treated like a surface of revolution.
Consider the function [tex]f(t)=\cos(t^2)[/tex] defined over [tex]0\le t\le1[/tex]. Revolve the curve [tex]C[/tex] described by [tex]f(t)[/tex] about the line [tex]t=0[/tex]. The area of the surface obtained in this way is then
[tex]\displaystyle2\pi\int_C\mathrm dS=2\pi\int_0^1\sqrt{1+f'(t)^2}\,\mathrm dt[/tex]
[tex]=\displaystyle2\pi\int_0^1\sqrt{1+(-2t\sin(t^2))^2}\,\mathrm dt[/tex]
[tex]=\displaystyle2\pi\int_0^1\sqrt{1+4t^2\sin^2(t^2)}\,\mathrm dt\approx7.4144[/tex]
The problem is solved by converting the variables in the surface equation into polar coordinates and expressing the surface area as a double integral which is estimated by numerical techniques to give a result of A = 4.5 m².
Explanation:The question is asking to find the surface area of the part of the surface z=cos(x^2+y^2) that lies within the cylinder x^2+y^2=1. To solve this, we must convert the variables into polar coordinates. Using the fact that x^2+y^2=r^2 in polar coordinates, our z function transforms into z = cos(r^2). Our area element in polar coordinates is r * dr * dθ. Plug these values into the surface area formula:
Surface Area = ∫∫ sqrt[1 + ((∂z/∂r)^2 +(∂z/∂θ)^2)] * r dr dθ.
Now it’s a matter of calculating the derivatives, plugging them into the formula and performing the double integral. Unfortunately, in this case, it is not straightforward to calculate the integral analytically so we use numerical techniques (like Riemann sum, Simpson rule, etc.) or a calculator with these built-in functions to estimate this integral, which gives the result A = 4.5 m² accurate up to 2 significant figures.
Learn more about Surface Area here:https://brainly.com/question/36938891
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An eight-sided die, which may or may not be a fair die, has four colors on it; you have been tossing the die for an hour and have recorded the color rolled for each toss. What is the probability you will roll a yellow on your next toss of the die? Express your answer as a simplified fraction or a decimal rounded to four decimal places.
Answer:
Case 1
None of the sides is yellow.
Probability of rolling a yellow on your next toss: zero %
Case 2
All the sides are yellow.
Probability of rolling a yellow on your next toss: 100 %
Case 3
At least one of the sides is yellow.
[tex]\frac{q}{N}\times100\%[/tex]
where
N = total number of tosses in one hour.
q = number of tosses you rolled a yellow.
Step-by-step explanation:
Case 1
None of the sides is yellow.
Probability of rolling a yellow on your next toss: zero %
Case 2
All the sides are yellow.
Probability of rolling a yellow on your next toss: 100 %
Case 3
At least one of the sides is yellow.
As you do not know if the die is fair or not, the only way to approximate a probability of rolling a yellow is by making a table of frequencies and record the times you have rolled yellow.
If the number of tosses made in an hour is big enough as to draw a conclusion, then according to the Law of Large Numbers, the probability of rolling a yellow in one toss of the die should be
[tex]\frac{q}{N}\times100\%[/tex]
where
N = total number of tosses
q = number of tosses you rolled a yellow.
Now, suppose you want to roll the dice once more.
As the event of rolling a die is independent of the previous tosses, this means that the probable outcome of the event does not depend on the previous results. So, the probability remains the same.
Probability of tossing a yellow on your next toss of the die
[tex]\frac{q}{N}\times100\%[/tex]
where
N = total number of previous tosses
q = number of tosses you rolled a yellow.
The propositional variables, p, q, and s have the following truth assignments: p = T, q = T, s = F. Give the truth value for each proposition. (a) p ∧ ¬(q ∨ s) (b) ¬(q ∧ p ∧ ¬s)
Answer: a) FALSE b) FALSE
Step-by-step explanation:
a) For the given proposition p ∧ ¬(q ∨ s) you can solve first (q v s)
q v s is true if either q is true or s is true or both. It is only false if both q and s are false. So, the proposition (q v s) is true because q is true.
Now you can solve the negation: ¬(q ∨ s)
As we know, (q v s) is true then its negation ¬(q ∨ s) is false.
p ∧ ¬(q ∨ s) should be true when both p and ¬(q ∨ s) are true, and false otherwise. So, the proposition is false because p is true and ¬(q ∨ s) is false.
b) For the given proposition ¬(q ∧ p ∧ ¬s)
You can rewrite the expression as: ¬[q ∧ (p ∧ ¬s) ] to solve first each part of the propositions in parenthesis.
The negation of s: ¬s is true because s is false
Now, you can solve (p ∧ ¬s) which is true because both p and ¬s are true.
To continue, you have to solve (q ∧ p ∧ ¬s) which is true because both q and (p ∧ ¬s) are true.
To finish, the negation: ¬(q ∧ p ∧ ¬s) is false.
If n is an odd integer, then 5n+ 3 is even.
Construct a know-show table for this statement.
Answer:
If n is an odd integer, then 5n +3 is even.
Step-by-step explanation:
Below is the know-show table for the statement
Determine which of the following sets of three points constitute the vertices of a right triangle: (a) 3 + 5i,2 +2i,5i; (b)2i,3 + 5i,4 + i; (c)6 +4i,7 + 5i, 8 +4i
Answer:
Option (c) is correct
Step-by-step explanation:
Case (a)
A = 3 + 5i = (3, 5)
B = 2 + 2i = (2, 2)
C = 5i = (0, 5)
Use the distance formula to find the distance between two points
[tex]AB = \sqrt{(2-3)^{2}+(2-5)^{2}}=\sqrt{10}[/tex]
[tex]BC = \sqrt{(0-2)^{2}+(5-2)^{2}}=\sqrt{13}[/tex]
[tex]CA = \sqrt{(0-3)^{2}+(5-5)^{2}}=\sqrt{9}[/tex]
For the triangle to be right angles triangle
[tex]BC^{2}=AB^{2}+CA^{2}[/tex]
Here, it is not valid, so these are not the points of a right angled triangle.
Case (b)
A = 2i = (0, 2)
B = 3 + 5i = (3, 5)
C = 4 + i = (4, 1)
Use the distance formula to find the distance between two points
[tex]AB = \sqrt{(3-0)^{2}+(5-2)^{2}}=\sqrt{18}[/tex]
[tex]BC = \sqrt{(4-3)^{2}+(1-5)^{2}}=\sqrt{17}[/tex]
[tex]CA = \sqrt{(4-0)^{2}+(1-2)^{2}}=\sqrt{17}[/tex]
For the triangle to be right angles triangle
[tex]AB^{2}=BC^{2}+CA^{2}[/tex]
Here, it is not valid, so these are not the points of a right angled triangle.
Case (c)
A = 6 + 4i = (6, 4)
B = 7 + 5i = (7, 5)
C = 8 + 4i = (8, 4)
Use the distance formula to find the distance between two points
[tex]AB = \sqrt{(7-6)^{2}+(5-4)^{2}}=\sqrt{2}[/tex]
[tex]BC = \sqrt{(8-7)^{2}+(4-5)^{2}}=\sqrt{2}[/tex]
[tex]CA = \sqrt{(8-6)^{2}+(4-4)^{2}}=\sqrt{4}[/tex]
For the triangle to be right angles triangle
[tex]CA^{2}=BC^{2}+AB^{2}[/tex]
Here, it is valid, so these are the points of a right angled triangle.
Initially tank I contains 100 litres of salt brine with a concentration of 1 kilogram per litre, and tank II contains 100 litres of water. Liquid is pumped from tank I into tank II at a rate of 1 litre per minute, and liquid is pumped from tank II into tank I at a rate of 2 litres per minute. The tanks are kept well stirred. Let A1 be the amount of salt in kilograms in tank I and A2 be the amount of salt in pounds in tank II.
(a) Calculate A1(t) and C1(t). For which range of values of t are the expression for A1(t) and C1(t) valid?
(b) What is the concentration in tank I after 10 minutes?
Answer:
a)[tex]A1(t)=\frac{100000000}{(100-t)(100+t)^{2} } \\C1(t)=\frac{A1(t)}{100+t}[/tex]
b) C1 = 0.8348 [kg/lt]
Step-by-step explanation:
Explanation
First of all, the rate of change of the amount of salt in the tank I is equal to the rate of change of salt incoming less the rate change of the salt leaving, so:
[tex]\frac{dA1(t)}{dt}= R_{in}C_{in}-R_{out}C_{out}[/tex]
We know that the incoming rate is greater than the leaving rate, this means that the fluid in the tank I enters more than It comes out, so the total rate is :
[tex]R_{total}=R_{in}-R_{out}=\frac{2 lt}{min} - \frac{1 lt}{min}= \frac{1 lt}{min}[/tex]
This total rate means that 1 lt of fluid enters each minute to the tank I from the tank II, with the total rate we can calculate the volume in the tank I y tank II as:
[tex]V_{I}=100 lt + Volumen_{in}= 100 lt + (\frac{1lt}{min})(t) =100+t[/tex]
[tex]V_{II}=100 lt - Volumen_{out}= 100 lt - (\frac{1lt}{min})(t) =100-t[/tex]
Now we have the volume of both tanks, the next step is to calculate the incoming and leaving concentration. The concentration is the ratio between the amount of salt and the volume, so:
[tex]C(t)=C_{out} =\frac{A1(t)}{V_{I} }=\frac{A1(t)}{100+t }[/tex]
Since fluid is pumped from tank I into tank II, the concentration of the tank II is a function of the amount of salt of the tank I that enters into the tank II, thus:
[tex]C_{in} =\frac{(A1(t)/V_{I})(t)}{V_{II} }=\frac{A1(t)}{V_{I} V_{II}}(t)[/tex]
[tex]C_{in} =\frac{A1(t)}{(100+t)(100-t)}(t)=\frac{A1(t)}{(10000-t^{2} )}(t)[/tex]
If we substitute the concentrations and the rates into the differential equation we can get:
[tex]\frac{dA1(t)}{dt}= R_{in}C_{in}-R_{out}C_{out}\\\frac{dA1(t)}{dt}= (2)(\frac{(t)A1(t)}{10000-t^{2} })-(1)(\frac{(A1(t)}{100+t })[/tex]
[tex]\frac{dA1(t)}{dt}= A1(t)(\frac{2t}{10000-t^{2} }-\frac{1}{100+t })[/tex]
[tex]\frac{dA1(t)}{dt}- (\frac{2t}{10000-t^{2} }-\frac{1}{100+t })A1(t)=0[/tex]
The obtained equation is a homogeneous differential equation of first order and the solution is:
a) [tex]A1(t)= \frac{100000000}{(100-t)(100+t)^{2} }[/tex]
and the concentration is:
[tex]C1(t)= \frac{100000000}{(100-t)(100+t)^{3}}[/tex]
This equations A1(t) and C1(t) are only valid to 0<=t<100 because to t >=100 minutes the tank II will be empty and mathematically A1(t>=100) tends to the infinite.
b) To calculate the concentration in the tank I after 10 minutes we have to substitute t=10 in C1(t), thus:
[tex]C1(10)= \frac{100000000}{(100-10)(100+10)^{3}}=0.8348 kg/lt[/tex]
Show that any integer n > 12 can be written as a sum 4r + 5s for some nonnegative integers r, s. (This problem is sometimes called a postage stamp problem. It says that any postage greater than 11 cents can be formed using 4 cent and 5 cent stamps.)
Answer:
Use induction for the prove
Step-by-step explanation:
Mathemathical induction is an useful method to prove things over natural numbers, you check for the first case, supose for the n and prove using your hypothesis for n+1
there says any integer bigger than 12 can be written as 4r+5s
so first number n can be is 13.
we can check n=13 = 4*2+5*1 r=2 and s=1 give 13.
Now we suppose n can be written as 4r+5s
and we can check if n+1=4r'+5s' with r' and s' integers.
we replace n as 4r+5s because that is our hypotesis
n+1=4r+5s+1
if we write that 1 as 5-4
4r+5s+1
4r+5s+5-4
then we can write
4(r-1)+5(s+1) , we got n+1= 4 (r-1) +5(s+1) where r-1 and s+1 are non negative integers. because r and s were no negative integers ( if r is not 0)
what if r=0?
if r is 0 , n is a multiple of 5 and n+1 can be written as 5s+1
first multiple of 5 we can write is 15 since n is bigger than 12 , then smaller s is 3.
for any n+1 we can write
n+1=5s+1=5 (s-3) +3*5+1=5(s-3)+4*4, s-3 is 0 or bigger.
(check 3*5+1 is 16, the same as 4*4)
Final answer:
Any integer greater than 12 can be expressed as the sum 4r + 5s with nonnegative integers r and s by providing concrete examples for n values from 13 to 17 and then proving for any n > 17 using number theory.
Explanation:
To show that any integer n greater than 12 can be written as the sum 4r + 5s for some nonnegative integers r and s, we can provide examples and create a general proof. We know that for n = 13 through n = 17, specific values of r and s can be found that satisfy the equation:
n = 13 = 4(1) + 5(1)
n = 14 = 4(3) + 5(0)
n = 15 = 4(0) + 5(3)
n = 16 = 4(4) + 5(0)
n = 17 = 4(1) + 5(3)
For n > 17, we can write n = 17 + k, where k is a nonnegative integer. Since every nonnegative integer can be expressed as a multiple of 4 plus an additional 0, 1, 2, or 3 (as k = 4q + r, with 0 ≤ r < 4), we can substitute into our equation for n to n = 17 + 4q + r. The key is to notice that for any additional 4 we add to the sum, we can simply increase r (our existing count of 4-cent stamps), ensuring that the equation 4r + 5s will always hold.
This proof shows that we can always add enough 4-cent stamps (or subtract them and add a 5-cent stamp) to reach any postage amount above 12 cents. This is a classic example of a problem that uses the concept of number theory and diophantine equations.
A company estimates that the marginal cost (in dollars per item) of producing x items is 1.65 − 0.002x. If the cost of producing one item is $570, find the cost of producing 100 items. (Round your answer to two decimal places.)
Answer:
The cost of producing 100 items is $723.35
Step-by-step explanation:
The marginal cost is the derivative of the total cost function, so we have
[tex]C^{'}(x)=1.65-0.002x[/tex]
To find the total cost function we need to do integration
[tex]C(x)= \int\, C^{'}(x)dx \\C(x)=\int\,(1.65-0.002x) dx[/tex]
Apply the sum rule to find the integral
[tex]\int f\left(x\right)\pm g\left(x\right)dx=\int f\left(x\right)dx\pm \int g\left(x\right)[/tex]
[tex]\int \:1.65dx=1.65x\\\int \:0.002xdx=0.001x^2[/tex]
[tex]C(x)=\int\,(1.65-0.002x) dx = 1.65x-0.001x^2+D [/tex]
D is the constant of integration
We are given that C(1) = $570, we can use this to find the value of the constant in the total cost function
[tex]C(1)=570=1.65*(1)+0.001*(1)^2+D\\D=570-1.649=568.351[/tex]
So the total cost function is [tex]C(x)=1.65x-0.001x^2+568.351 [/tex] and the cost of producing 100 items is
x=100
[tex]C(100)=1.65*(100)-0.001*(100)^2+568.351 = 723.35[/tex]
Find the area of the region described.
The region between the line y=x and the curve y=2x√(25 - x^2) in the first quadrant. The total area of the shaded region is ___ (from 7087 to 100).
Answer:
The area is [tex]\frac{567}{8}u^2[/tex]
Step-by-step explanation:
The area of a flat region bounded by the graphs of two functions f (x) and g (x), with f (x)> g (x) can be found through the integral:
[tex]\int\limits^b_a {[f(x) - g(x)]} \, dx[/tex]
The integration limits are given by the intersection points of the graphs of the functions in the first quadrant. Then, the cut points are:
[tex]g(x) = x\\f(x) = 2x\sqrt{25-x^2}[/tex]
[tex]x=2x\sqrt{25-x^2}\\x^2=4x^2(25-x^2)\\x^2(1-100+4x^2)=0\\x_1=0\\x_2=\frac{3\sqrt{11}}{2}[/tex]
The area of the region is:
[tex]\int\limits^b_a {[f(x) - g(x)]} \, dx = \int\limits^{\frac{3\sqrt{11}}{2}}_0 {x(2\sqrt{25-x^2}-1)} \, dx = \frac{567}{8}u^2[/tex]
The point P(1, 1/6) lies on the curve y = x/(5 + x). If Q is the point (x, x/(5 + x)), use a scientific calculator to find the slope of the secant line PQ (correct to six decimal places) for the following values of x.
Answer:
[tex]m(x)=\frac{5}{6(5+x)}[/tex]
Step-by-step explanation:
Slope of the secant line PQ:
P : (1, 1/6)
Q : (x, x/(5 + x))
[tex]m(x)=\frac{y_{Q}-y_{P}}{x_{Q}-x_{P}}=\frac{x/(5 + x)-1/6}{x-1}=\frac{5(x-1)}{6(5+x)(x-1)}[/tex]
The slope of the secant line PQ is 0.
Explanation:To find the slope of the secant line PQ, we need to determine the coordinates of Q and P. Given that Q has the coordinates (x, x/(5+x)), we can substitute x=1 into the equation to find Q. So, Q is (1, 1/6).
The slope of a line passing through two points can be found using the formula:
slope = (y2 - y1) / (x2 - x1)
Substituting the coordinates of P and Q into the formula:
m = (1/6 - 1/(5+1)) / (1 - 1) = (1/6 - 1/6)/(0) = 0
Therefore, the slope of the secant line PQ is 0.
The Big Wave Surf Shop sells 8 ounce containers of sunscreen for $8 each. The total daily cost associated with the sale of this item yesterday was $200. What is the shop’s profit if it sold 150 units of sunscreen yesterday?
Answer:
The shop’s profit if it sold 150 units of sunscreen yesterday is $1000
Step-by-step explanation:
The total daily cost associated with the sale of this item yesterday was $200.
Cost of 1 unit = $8
So, Cost of 150 units = [tex]8 \times 150[/tex]
= [tex]1200[/tex]
So, revenue = $1200
Cost = $200
So Profit = Revenue - Cost
Profit = $1200-$200
Profit = $1000
Hence the shop’s profit if it sold 150 units of sunscreen yesterday is $1000
(x-5)^2=81, solve for x
Answer:
x ∈ {-4, 14}
Step-by-step explanation:
Take the square root and add 5.
(x -5) = ±√81 = ±9
x = 5 ± 9
x ∈ {-4, 14} . . . . . . x may be either of -4 or 14
There are 129 schools in the NCAA's Division 1 Football Bowl Subdivision. Of these, 115 have nicknames that end in "s" (like the UCLA Bruins) 19 have nicknames that involve a color (the Stanford Cardinal), and · 13 nicknames involve both a color and end in "s" (the California Golden Bears). How many teams have nicknames without a color and don't end in "s?
Answer:
There are 8 teams that have nicknames without a color and don't end in "s.
Step-by-step explanation:
This can be solved by treating each value as a set, and building the Venn Diagram of this.
-I am going to say that set A are the teams that have nicknames that end in S.
-Set B are those whose nicknames involve a color.
-Set C are those who have nicknames without a color and don't end in "s.
We have that:
[tex]A = a + (A \cap B)[/tex]
In which a are those that have nickname ending in "s", but no color, and [tex]A \cap B[/tex] are those whose nickname involves a color and and in "s".
By the same logic, we have
[tex]B = b + (A \cap B)[/tex]
In which b are those that nicknames involves a color but does not end in s.
We have the following subsets:
[tex]a,b, (A \cap B), C[/tex]
There are 129 schools, so:
[tex]a + b + (A \cap B) + C = 129[/tex]
Lets find the values, starting from the intersection.
The problem states that:
13 nicknames involve both a color and end in "s". So:
[tex]A \cap B = 13[/tex]
19 have nicknames that involve a color. So:
[tex]B = 19[/tex]
[tex]B = b + (A \cap B)[/tex]
[tex]b + 13 = 19[/tex]
[tex]b = 6[/tex]
115 have nicknames that end in "s". So:
[tex]A = 115[/tex]
[tex]A = a + (A \cap B)[/tex]
[tex]a + 13 = 115[/tex]
[tex]a = 102[/tex]
Now, we just have to find the value of C, in the following equation:
[tex]a + b + (A \cap B) + C = 129[/tex]
[tex]102 + 6 + 13 + C = 129[/tex]
[tex]C = 129 - 121[/tex]
[tex]C = 8[/tex]
There are 8 teams that have nicknames without a color and don't end in "s.
Suppose you invest $750.00 in a fund earning 6% simple discount. A certain time later you withdraw the investment (principal and interest) and invest it in another fund earning 3.5% compound interest for two years. How much total time (INCLUDING THE TWO YEARS earning compound interest) will be required for the original $750.00 to accumulate to $1,000.00? (two decimal places)
Answer:
4.08 + 2 = 6.08 years
Step-by-step explanation:
we know that
Simple Interest(S.I.) = (P × R × T) ÷ 100
where, P = Principal = 750
R = Rate = 6%
T = unknown
⇒ S.I. = (750 × 6 × t)÷ 100
⇒ S.I. = 45t
Also, Amount = S.I + Principal
⇒ Amount = 750 + 45t
Now Formula for Compound Interest is:
[tex]A = P(1+\frac{r}{100})^{t}[/tex]
where A = Amount
=1000
P = Principle
r = rate
t = total number of year
Here, P = 750 + 45t, r = 3.5% , and t = 2.
Putting all these values in above formula:
[tex]1000 = (750 + 45t)(1+\frac{3.5}{100})^{2}[/tex]
⇒ [tex]1000 = (750 + 45t)(1.071)[/tex]
⇒ t = 4.08
Hence, total time required will be 2 + 4.08 = 6.08 years.
Which correlation coefficient corresponds to the best-fit line that most closely models it’s set of data. A. -0.87, B. -0.15. C. 0.13. D. 0.84
Answer:
A. -0.87
Step-by-step explanation:
The correlation coefficient is a measure of the strength and nature of the linear relationship between two variables, one dependent and one independent. This coefficient takes values between -1 and 1, indicating with its sign whether the relationship is direct or inverse between the variables involved and with its absolute value indicates the strength of the linear relationship between them. A coefficient with absolute value close to 1 indicates great strength and better fit.
Conclusion: The best coefficient, of the propuetso in the problem, is that of -0.87, which indicates a strong relationship between the variables, a good fit and an inverse relationship between them.
A ferry leaves Nanaimo to make the 22 km trip to Vancouver at the same time as a forry leaves Vancouver for Nanaimo. The ferry leaving Nanaimo travels 2 km/h faster than the other ferry How far are they from Vancouver when they meet 45 minutes later?
Answer:
They are 11.7475km away from Vancouver when they meet.
Step-by-step explanation:
The first step to solve this problem is modeling the position of each ferry. The position can be modeled by a first order equation in the following format:
[tex]S(t) = S_{0} + vt[/tex], in which [tex]S_{0}[/tex] is the initial position of the ferry, t is the time in hours and v is the speed in km/h.
I am going to say that the positive direction is from Nanaimo to Vancouver, and that Nanaimo is the position 0 and Vancouver the position 22.
First ferry:
Leaves Nanaimo, so [tex]S_{0} = 0[/tex]. It is 2km/h faster than the second ferry, so i am going to say that [tex]v = v + 2[/tex]. It moves in the positive direction, so v is positive. The equation of the position of this train is modeled as:
[tex]S_{1}(t) = 0 + (v+2)t[/tex],
Second ferry:
Leaves Vancouver, so [tex]S_{0} = 22[/tex]. It has a speed of v, that is negative, since it moves in the negative direction. So
[tex]S_{2}(t) = 22 - vt[/tex]
The problem states that they meet in 45 minutes. Here we have to pay attention. Since the speed is in km/h, the time needs to be in h. So 45 minutes = 0.75h.
They meet in 0.75h. It means that
[tex]S_{1}(0.75) = S_{2}(0.75)[/tex]
With this we find the value o v, and replace in the equation of [tex]S_{2}[/tex] to see how far they are from Vancouver when they meet.
[tex]S_{1}(0.75) = S_{2}(0.75)[/tex]
[tex]0.75(v+2) = 22 - 0.75v[/tex]
[tex]0.75v + 1.50 = 22 - 0.75v[/tex]
[tex]1.50v = 20.50[/tex]
[tex]v = \frac{20.50}{1.50}[/tex]
[tex]v = 13.67[/tex]km/h.
[tex]S_{2}(t) = 22 - vt[/tex]
[tex]S_{2}(0.75) = 22 - 13.67*0.75 = 11.7475[/tex]
They are 11.7475km away from Vancouver when they meet.
The ferries are 10.25 km away from Vancouver when they meet after traveling for 45 minutes, with the ferry from Nanaimo moving at 15.67 km/h and the one from Vancouver at 13.67 km/h.
Distance Between Ferries
Let's denote the speed of the slower ferry as s km/h. Therefore, the speed of the faster ferry leaving Nanaimo will be s + 2 km/h. As they both start at the same time, we can express the distance they travel in terms of their speeds multiplied by the time, which is 45 minutes (or 0.75 hours when converted to hours).
Now, let's set up the equation for the distance each ferry has traveled when they meet: the distance traveled by the slower ferry plus the distance traveled by the faster ferry equals the total distance between Nanaimo and Vancouver, which is 22 km.
Distance of slower ferry: s (0.75) km
Distance of faster ferry: (s + 2)(0.75) km
The sum of both distances is 22 km, so we have:
s (0.75) + (s + 2)(0.75) = 22
Rearranging terms and solving for s gives us:
1.5s + 1.5 = 22
s = (22 - 1.5)/1.5
Calculating the value of s and consequently the distance each ferry traveled before meeting will give us the answer.
After solving, we can determine that the speed of the slower ferry is 13.67 km/h and the faster ferry is 15.67 km/h. To find the distance from Vancouver to the meeting point, we only need to calculate the distance traveled by the slower ferry:
Distance from Vancouver = (13.67 km/h) (0.75 h) = 10.25 km
Therefore, the ferries are 10.25 km away from Vancouver when they meet.
A town's population is currently 30,000. If the population doubles every 45 years, what will the population be 90 years from now?
A. 240,000 people
B. 150,000 people
C. 120,000 people
D. 60,000 people
Answer:
C. 120,000 people
Step-by-step explanation:
First thing I did was divide 90 by 45 which gave me 2 as my answer. So now I know that the population 90 years from now will be doubled twice (x4). So I did...
[tex]30,000 \times 4 = 120,000[/tex]
Answer:
80,000 people
Step-by-step explanation:
First, find out how many times the population will double. Divide the number of years by how long it takes for the population to double.
68÷34=2
The population will double 2 times.
Now figure out what the population will be after it doubles 2 times. Multiply the population by 2 a total of 2 times.
20,00022=80,000
That calculation could also be written with exponents:
20,00022=80,000
After 68 years, the population will be 80,000 people.
One group of scientists (Group A) has measured a time for a particular chemical reaction to be completed to be 7.34 ± 0.05 s. A second group of scientists (Group B) complete a similar experiment and measure the time to be 7.38 ± 0.03s. A third group (Group C) measured the time to be 7.46 ±0.06s. Do results of Group A and B agree with each other within the experimental uncertainty? Do results of Group A and C agree with each other within the experimental uncertainty?
Answer:
1. Group A and B agree with each other.
2. Group A and C do not agree with each other.
Step-by-step explanation:
When we are analizing this problem, we will see what are the ranges of this measured times. Since we are taking into account the error we can see that :
Group A varies from 7.34-0.05 to 7.34+0.05. So the limits are (7.29 ;7.39)Group B varies from 7.38-0.03 to 7.38+0.03. So the limits are (7.35; 7.41)Group C varies from 7.46-0.06 to 7.46+0.06. So the limits are (7.40; 7.52)Question 1 is about the overlapping response in Group A and Group B. And yes, we have an overlap between 7.35 to 7.39. Among this times both group A and B are in agree with each other within the experimental uncertainty.
Question 2 is now referring to Group A and Group C. And no, there isn't any common time where both groups agree with each other.
15.) A market
research worker interviewed a random sample of 18people about their
use of a certain product. The results, in termsof Yes (Y) or No (N)
are as follows:Y-N-N-Y-Y-Y-N-Y-N-Y-Y-Y-N-Y-N-Y-Y-N. Estimate the
populationproportion of users of the product.
Answer: [tex]p=0.610[/tex]
Step-by-step explanation:
Given : A market research worker interviewed a random sample of 18 people about their use of a certain product.
The results, in terms of Yes (Y) or No (N) are as follows:
Y-N-N-Y-Y-Y-N-Y-N-Y-Y-Y-N-Y-N-Y-Y-N.
The number of people said "Yes" for the product= 11
Then, the sample proportion for the users of the product =[tex]\hat{p}=\dfrac{11}{18}0.611111111111\approx0.61[/tex]
We know that the sample proportion is the best estimate for the population proportion.
Thus the point estimate for population proportion : [tex]p=\hat{p}=0.610[/tex]