Answer:- 10[tex]m^{2}[/tex] + 3m -9
Step-by-step explanation: Given ;
A= -3 -m
B= 3m -5[tex]m^{2}[/tex]
2B + 3A
solution
2B + 3A
substitute A and B in the formula
2(3m - 5[tex]m^{2}[/tex]) + 3(-3 -m)
6m - 10[tex]m^{2}[/tex] - 9 - 3m group like terms
- 10[tex]m^{2}[/tex] + (6m -3m) -9
- 10[tex]m^{2}[/tex] + 3m -9
A boxcar contains six complex electronic systems. Two of the six are to be randomly selected for thorough testing and then classified as defective or not defective.
a. If two of the six systems are actually defective, find the probability that at least one of the two systems tested will be defective. Find the probability that both are defective.
b. If four of the six systems are actually defective, find the probabilities indicated in part (a).
Answer:
Step-by-step explanation:
Number of electronic systems = 6
(a) Number of defected systems = 2
Probability of getting at least one system is defective
1 defective and 1 non defective + 2 defective
= (2 C 1 ) x (4 C 1) + (2 C 2) / (6 C 2)
= 3 / 5
(b) four defective
Probability of getting at least one system is defective
2 defective and 2 non defective + 3 defective and 1 non defective + 4 defective
= (4 C 2 ) x (2 C 2) + (4 C 3 )(2 C 1) + (4 C 4) / (6 C 4)
= 1
Answer:
(a)P(At least one defective)[tex]=0.6[/tex]
P(Both are defective)[tex]=0.067[/tex]
(b)P(At least one defective)[tex]=14/15[/tex]
P(Both are defective)[tex]=0.4[/tex]
Step-by-step explanation:
We are given that
Total number of complex electronic system, n=6
(a)Defective items=2
Non-defective items=6-2=4
We have to find the probability that at least one of the two systems tested will be defective.
P(At least one defective)=[tex]\frac{2C_1\times 4C_1}{6C_2}+\frac{2C_2\times 4C_0}{6C_2}[/tex]
Using the formula
[tex]P(E)=\frac{favorable\;cases}{total\;number\;of\;cases}[/tex]
P(At least one defective)[tex]=\frac{\frac{2!}{1!1!}\times \frac{4!}{1!3!} }{\frac{6!}{2!4!}}+\frac{\frac{2!}{0!2!}\times \frac{4!}{4!}}{\frac{6!}{2!4!}}[/tex]
Using the formula
[tex]nC_r=\frac{n!}{r!(n-r)!}[/tex]
P(At least one defective)[tex]=\frac{2\times \frac{4\times 3!}{3!}}{\frac{6\times 5\times 4!}{2\times 1\times 4!}}+\frac{1}{\frac{6\times 5\times 4!}{2\times 1\times 4!}}[/tex]
P(At least one defective)[tex]=\frac{2\times 4}{3\times 5}+\frac{1}{3\times 5}[/tex]
P(At least one defective)[tex]=\frac{8}{15}+\frac{1}{15}=\frac{9}{15}[/tex]
P(At least one defective)[tex]=\frac{3}{5}=0.6[/tex]
Now, the probability that both are defective
P(Both are defective)=[tex]\frac{2C_2\times 4C_0}{6C_2}[/tex]
P(Both are defective)=[tex]\frac{\frac{2!}{0!2!}\times \frac{4!}{4!}}{\frac{6!}{2!4!}}[/tex]
P(Both are defective)[tex]=\frac{1}{\frac{6\times 5\times 4!}{2\times 1\times 4!}}[/tex]
P(Both are defective)[tex]=\frac{1}{3\times 5}[/tex]
P(Both are defective)[tex]=0.067[/tex]
(b)
Defective items=4
Non- defective item=6-4=2
P(At least one defective)=[tex]\frac{4C_1\times 2C_1}{6C_2}+\frac{4C_2\times 2C_0}{6C_2}[/tex]
P(At least one defective)[tex]=\frac{\frac{4!}{1!3!}\times \frac{2!}{1!1!} }{\frac{6!}{2!4!}}+\frac{\frac{4!}{2!2!}\times \frac{2!}{2!}}{\frac{6!}{2!4!}}[/tex]
P(At least one defective)[tex]=\frac{2\times \frac{4\times 3!}{3!}}{\frac{6\times 5\times 4!}{2\times 1\times 4!}}+\frac{\frac{4\times 3\times 2!}{2!\times 2\times 1}}{\frac{6\times 5\times 4!}{2\times 1\times 4!}}[/tex]
P(At least one defective)[tex]=\frac{2\times 4}{3\times 5}+\frac{2\times 3}{3\times 5}[/tex]
P(At least one defective)[tex]=\frac{8}{15}+\frac{6}{15}=\frac{8+6}{15}[/tex]
P(At least one defective)[tex]=\frac{14}{15}[/tex]
P(Both are defective)[tex]=\frac{4C_2\times 2C_0}{6C_2}[/tex]
P(Both are defective)[tex]=\frac{\frac{4\times 3\times 2!}{2\times 1\times 2!}\times \frac{2!}{2!}}{\frac{6\times 5\times 4!}{2\times 1\times 4!}}[/tex]
P(Both are defective)[tex]=\frac{\frac{4\times 3\times 2\times 1}{2\times 1\times 2\times 1}}{3\times 5}[/tex]
P(Both are defective)[tex]=\frac{6}{15}=0.4[/tex]
P(Both are defective)[tex]=0.4[/tex]
Write the equation of the line in fully simplified slope-intercept form.
Answer:
y = -x+3
Step-by-step explanation:
Slope intercept form => y = mx+b
To find 'm', the slope, pick 2 coordinates.
(0,3)
(2,1)
Use this equation to find the slope using these 2 coordinates: (y1 - y2)/(x1 - x2)
(3 - 1)/(0 - 2) = -1
m = slope = -1
'b' is the y-intersept, or the point when a line passes through the y-axis. That's (0,3).
b = y-intercept = 3
So the equation will be y = -1x + 3, or y = -x + 3
if 3x=y+z, y=6-7, and z+x=8, what is the value of y/z?
Answer:
-4/25
Step-by-step explanation:
3x=y+z
y=6-7=-1
z+x=8
y/z=?
solution
3x=-1 + z
x= -1 + z\3 ...eq(*)
then, z= 8- x
z= 8 - (-1 +z)\3
z= 8 +(1- z )\3
z= 8+1\3 -z \3
= 24+1\3 - z\3
z=25\3-z\3
z+z\3=25\3
4z\3=25\3
4z=25
z=25/4
then,
25/4 +x = 8
x=8- 25/4
x= 32 - 25/4
x=7/4
so that,
y/z=-1 /25/4
=-4/25
find and sketch the domain of the function. f(x,y)=√(4-x^2-y^2) +√(1-x^2)
Answer:
Hello
Step-by-step explanation:
The domain is limited with 2 lines parallel: -1 ≤ x ≤ 1
and the disk ? (inside of a circle) of center (0,0) and radius 2
[tex]dom\ f(x,y)=\{(x,y) \in \mathbb{R} ^2 | \ -1\leq x \leq -1\ and \ ( -\sqrt{4-x^2} \leq \ y \leq \sqrt{4-x^2}\ ) \ \}\\[/tex]
Suppose the method of tree ring dating gave the following dates A.D. for an archaeological excavation site. Assume that the population of x values has an approximately normal distribution.
1241 1210 1267 1314 1211 1299 1246 1280 1291
a. Determine if the data meets the initial conditions to construct a confidence interval.
b. Find the sample mean year x and sample standard deviation σ.
c. What is the maximal margin of error when finding a 90 % confidence interval for the mean of all tree-ring dates from this archaeological site?
Answer:
(1238.845 ;1285.376)
Step-by-step explanation:
Conditions for constructing a confidence interval :
Data must be random
Distribution should be normal and independent ;
Based on the conditions above ; data meets initial conditions ;
C. I = sample mean ± margin of error
Given the data :
1241 1210 1267 1314 1211 1299 1246 1280 1291
Mean, xbar = Σx / n = 11359 / 9 = 1262.11
The standard deviation, s = [√Σ(x - xbar)²/n - 1]
Using a calculator ; s = 37.525
The confidence interval :
C.I = xbar ± [Tcritical * s/√n]
Tcritical(0.10 ; df = n - 1 = 9 - 1 = 8)
Tcritical at 90% = 1.860
C. I = 1262.11 ± [1.860 * 37.525/√9]
C.I = 1262.11 ± 23.266
(1238.845 ;1285.376)
± 23.266
The margin of error :
[Tcritical * s/√n]
[1.860 * 37.525/√9]
C.I = ± 23.266
A website manager has noticed that during the evening hours, about 3.23.2 people per minute check out from their shopping cart and make an online purchase. She believes that each purchase is independent of the others and wants to model the number of purchases per minute.
1. What model might you suggest to model the number of purchases per minute?
a. Binomial
b. Uniform
c. Poisson
d. Geometric
2. What is the probability that in any one minute at least one purchase is made?
3. What is the probability that no one makes a purchase in the next 2 minutes?
Answer:
1. c. Poisson
2. 0.9592 = 95.92% probability that in any one minute at least one purchase is made.
3. 0.0017 = 0.17% probability that no one makes a purchase in the next 2 minutes.
Step-by-step explanation:
We have only the mean, which means that the Poisson distribution is used to solve this question, and thus the answer to question 1 is given by option c.
Poisson distribution:
In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by the following formula:
[tex]P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}[/tex]
In which
x is the number of sucesses
e = 2.71828 is the Euler number
[tex]\mu[/tex] is the mean in the given interval.
Mean of 3.2 minutes:
This means that [tex]\mu = 3.2n[/tex], in which n is the number of minutes.
2. What is the probability that in any one minute at least one purchase is made?
[tex]n = 1[/tex], so [tex]\mu = 3.2[/tex].
This probability is:
[tex]P(X \geq 1) = 1 - P(X = 0)[/tex]
In which
[tex]P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}[/tex]
[tex]P(X = 0) = \frac{e^{-3.2}*3.2^{0}}{(0)!} = 0.0408[/tex]
So
[tex]P(X \geq 1) = 1 - P(X = 0) = 1 - 0.0408 = 0.9592[/tex]
0.9592 = 95.92% probability that in any one minute at least one purchase is made.
3. What is the probability that no one makes a purchase in the next 2 minutes?
2 minutes, so [tex]n = 2, \mu = 3.2(2) = 6.4[/tex]
This probability is P(X = 0). So
[tex]P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}[/tex]
[tex]P(X = 0) = \frac{e^{-6.4}*6.4^{0}}{(0)!} = 0.0017[/tex]
0.0017 = 0.17% probability that no one makes a purchase in the next 2 minutes.
7b please make the graph look nice and neat and easy to read.
Answer:
Step-by-step explanation:
A trade magazine routinely checks the drive-through service times of fast-food restaurants. 95% confidence interval that results from examining 609 customers in one fast-food chain's drive-through has a lower bound of 160.8 seconds and an upper bound of 160.8 seconds. What does this mean?
Answer:
z= 1.70 Since the test statistic is less than the critical value, we can conclude that the average length of an online video is not more than 8 minutes
Please help with this question
Answer:
im not too sure but try using a cartesuan plane and measure it precisely using a protractor then key in the measurements. Im not entirely sure its the correct method tho
Suppose a rumor is going around a group of 191 people. Initially, only 38 members of the group have heard the rumor, but 3 days later 68 people have heard it. Using a logistic growth model, how many people are expected to have heard the rumor after 6 days total have passed since it was initially spread? (Round your answer to the nearest whole person.)
Answer:
106 people.
Step-by-step explanation:
Logistic equation:
The logistic equation is given by:
[tex]P(t) = \frac{K}{1+Ae^{-kt}}[/tex]
In which
[tex]A = \frac{K - P_0}{P_0}[/tex]
K is the carrying capacity, k is the growth/decay rate, t is the time and P_0 is the initial value.
Suppose a rumor is going around a group of 191 people. Initially, only 38 members of the group have heard the rumor.
This means that [tex]K = 191, P_0 = 38[/tex], so:
[tex]A = \frac{191 - 38}{38} = 4.03[/tex]
Then
[tex]P(t) = \frac{191}{1+4.03e^{-kt}}[/tex]
3 days later 68 people have heard it.
This means that [tex]P(3) = 68[/tex]. We use this to find k.
[tex]P(t) = \frac{191}{1+4.03e^{-kt}}[/tex]
[tex]68 = \frac{191}{1+4.03e^{-3k}}[/tex]
[tex]68 + 274.04e^{-3k} = 191[/tex]
[tex]e^{-3k} = \frac{191-68}{274.04}[/tex]
[tex]e^{-3k} = 0.4484[/tex]
[tex]\ln{e^{-3k}} = \ln{0.4484}[/tex]
[tex]-3k = \ln{0.4484}[/tex]
[tex]k = -\frac{\ln{0.4484}}{3}[/tex]
[tex]k = 0.2674[/tex]
Then
[tex]P(t) = \frac{191}{1+4.03e^{-0.2674t}}[/tex]
How many people are expected to have heard the rumor after 6 days total have passed since it was initially spread?
This is P(6). So
[tex]P(6) = \frac{191}{1+4.03e^{-0.2674*6}} = 105.52[/tex]
Rounding to the nearest whole number, 106 people.
The cost of producing a custom-made clock includes an initial set-up fee of $1,200 plus an additional $20 per unit made. Each clock sells for $60. Find the number of clocks that must be produced and sold for the costs to equal the revenue generated. (Enter a numerical value.)
Answer:
30 clocks
Step-by-step explanation:
Set up an equation:
Variable x = number of clocks
1200 + 20x = 60x
Isolate variable x:
1200 = 60x - 20x
1200 = 40x
Divide both sides by 40:
30 = x
Check your work:
1200 + 20(30) = 60(30)
1200 + 600 = 1800
1800 = 1800
Correct!
Anthony read 46 pages of a book in 23 minutes.
To find the unit rate, use
.
Anthony read
pages per minute.
Answer:
2 pages per minute
Step-by-step explanation:
Take the number of pages and divide by the number of minutes
46 pages / 23 minutes
2 pages per minute
2 Pages per Minute
Solutions:46 ÷ 23 = 2
Final Answer:Anthony can read 2 pages per minute.
plzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzz help i will give
brainliest
Answer:
55
Step-by-step explanation:
55 appears 3 times, which is the most repetition in the data set
Answer:
55
Step-by-step explanation:
Mode = number that appears most often
The number 55 appears 3 times which is the most out of the other numbers
Hence mode = 55
The Wowo Novelty Company makes three basic types of noisemakers: Toot, Wheet and honk. A toot can be made in 30 minutes and has a feather attached to it. A wheet requires 15 minutes, has two feathers and is sprinkled with 0.5 oz of sequin powder. A honk requires 30 minutes, has 3 feathers and 1 oz of sequin powder. The net profit is P0.40 per toot, P0.5 per wheet and P0.80 per honk. The following resources are available: 80 hours of labor, 360 feathers and 90 oz of sequin powder. Determine the quantity of each type of noisemakers that maximizes profit.
Answer:
P104
Step-by-step explanation:
Let x represent the number of toot, y represent the number of wheet and z represent the number of honk.
Since a toot is made in 30 minutes (0.5 hours), wheet in 15 minutes (0.25 hour), honk in 30 min (0.5 hr). There is 80 hours of labor, hence:
0.5x + 0.25y + 0.5z ≤ 80 (1)
There are 360 feathers, hence:
x + 2y + 3z ≤ 360 (2)
There is 90 oz of powder, hence:
0.5y + z ≤ 90 (3)
solving equation 1, 2 and 3, gives:
x = 70, y= 40, z = 70
The profit is given by:
Profit = 0.4x + 0.5y + 0.8z
substituting x, y and z gives:
Profit = 0.4(70) + 0.5(40) + 0.8(70) = P104
Two angles of a triangle have the same measure and the third one is 36 degrees greater than the measure of each of the other two. Find the measure of the LARGEST angle in the triangle.
Answer:
Largest angle is 84
Step-by-step explanation:
Let the smallest angle be x, ATQ, x+x+x+36=180, 3x+36=180, x=48
Please help with this question
9514 1404 393
Answer:
(d) -1/32
Step-by-step explanation:
It may be easier to rearrange the expression so it has positive exponents.
[tex]\dfrac{1}{2^{-2}x^{-3}y^5}=\dfrac{2^2x^3}{y^5}=\dfrac{4(2)^3}{(-4)^5}=-\dfrac{4\cdot8}{1024}=\boxed{-\dfrac{1}{32}}[/tex]
A chemist has three different acid solutions.
The first solution contains 25% acid, the second contains 35%acid, and the third contains 55% acid.
She created 120 liters of a 40% acid mixture, using all three solutions. The number of liters of 55% solution used is 3 times the number of liters of 35% solution used.
How many liters of each solution was used?
Let x, y, and z be the amounts (in liters, L) of the 25%, 35%, and 55% solutions that the chemist used.
She ended up with 120 L of solution, so
x + y + z = 120 … … … [1]
x L of 25% acid solution contains 0.25x L of acid. Similarly, y L of 35% solution contains 0.35y L of acid, and z L of 55% solution contains 0.55z L of acid. The concentration of the new solution is 40%, so that it contains 0.40 (120 L) = 48 L of acid, which means
0.25x + 0.35y + 0.55z = 48 … … … [2]
Lastly,
z = 3y … … … [3]
since the chemist used 3 times as much of the 55% solution as she did the 35% solution.
Substitute equation [3] into equations [1] and [2] to eliminate z :
x + y + 3y = 120
x + 4y = 120 … … … [4]
0.25x + 0.35y + 0.55 (3y) = 48
0.25x + 2y = 48 … … … [5]
Multiply through equation [5] by -2 and add that to [4] to eliminate y and solve for x :
(x + 4y) - 2 (0.25x + 2y) = 120 - 2 (48)
0.5x = 24
x = 48
Solve for y :
x + 4y = 120
4y = 72
y = 18
Solve for z :
z = 3y
z = 54
HELP ASAP PLEASE! I tried inputting the numbers into the standard deviation equation but I did not get the right answer to find z. Can someone please help me? Thank you for your time!
Answer:
Z = -1.60
it is low ... it appears that for this problem 2 standard deviations below must be reached to be considered "unusual"
Step-by-step explanation:
[(2021-Y)-5]*X-X=XX cho biết X,Y,XX là gì?
(3) If a tire rotates at 400 revolutions per minute when the car is traveling 72km/h, what is the circumference of the tire?
Show all your steps.
Answer:
3 meters.
Step-by-step explanation:
400 rev / minute = 400 × 60 rev / 60 minutes
= 24,000 rev / hour
24,000 × C = 72,000 m : C is the circumference
C = 3 meters
Answer:
3 meters
Step-by-step explanation:
72 km / hour * 1 hour/ 60 min * 1000m/ 1 km
72000 meters /60 minute
1200 meters / minute
velocity = radius * w
Where w is 2*pi * the revolutions per minute
1200 = r * 2 * pi *400
1200 / 800 pi = r
1.5 /pi = r meters
We want to find the circumference
C = 2 * pi *r
C = 2* pi ( 1.5 / pi)
C = 3 meters
URGENT HELP!!!!
Picture included
Answer:
Length (L) = 72 feet
Step-by-step explanation:
From the question given above, the following data were obtained:
Period (T) = 9.42 s
Pi (π) = 3.14
Length (L) =?
The length of the pendulum can be obtained as follow:
T = 2π √(L/32)
9.42 = (2 × 3.14) √(L/32)
9.42 = 6.28 √(L/32)
Divide both side by 6.28
√(L/32) = 9.42 / 6.28
Take the square of both side
L/32 = (9.42 / 6.28)²
Cross multiply
L = 32 × (9.42 / 6.28)²
L = 72 feet
Thus, the Lenght is 72 feet
3.52 A coin is tossed twice. Let Z denote the number of heads on the first toss and W the total number of heads on the 2 tosses. If the coin is unbalanced and a head has a 40% chance of occurring, find (a) the joint probability distribution of W and Z; (b) the marginal distribution of W; (c) the marginal distribution of Z
Answer:
a) The joint probability distribution
P(0,0) = 0.36, P(1,0) = 0.24, P(2,0) = 0, P(0,1) = 0, P(1,1) = 0.24, P(2,1)= 0.16
b) P( W = 0 ) = 0.36, P(W = 1 ) = 0.48, P(W = 2 ) = 0.16
c) P ( z = 0 ) = 0.6
P ( z = 1 ) = 0.4
Step-by-step explanation:
Number of head on first toss = Z
Total Number of heads on 2 tosses = W
% of head occurring = 40%
% of tail occurring = 60%
P ( head ) = 2/5 , P( tail ) = 3/5
a) Determine the joint probability distribution of W and Z
P( W =0 |Z = 0 ) = 0.6 P( W = 0 | Z = 1 ) = 0
P( W = 1 | Z = 0 ) = 0.4 P( W = 1 | Z = 1 ) = 0.6
P( W = 1 | Z = 0 ) = 0 P( W = 2 | Z = 1 ) = 0.4
The joint probability distribution
P(0,0) = 0.36, P(1,0) = 0.24, P(2,0) = 0, P(0,1) = 0, P(1,1) = 0.24, P(2,1)= 0.16
B) Marginal distribution of W
P( W = 0 ) = 0.36, P(W = 1 ) = 0.48, P(W = 2 ) = 0.16
C) Marginal distribution of Z ( pmf of Z )
P ( z = 0 ) = 0.6
P ( z = 1 ) = 0.4
Part(a): The required joint probability of W and Z is ,
[tex]P(0,0)=0.36,P(1,0)=0.24,P(2,0)=0,P(0,1)=0,P(1,1)=0.24,\\\\P(2,1)=0.16[/tex]
Part(b): The pmf (marginal distribution) of W is,
[tex]P(w=0)=0.36,P(w=1)=0.48,P(w=2)=0.16[/tex]
Part(c): The pmf (marginal distribution) of Z is,
[tex]P(z=0)=0.6,P(z=1)=0.4[/tex]
Part(a):
The joint distribution is,
[tex]P(w=0\z=0)=0.6,P(w=1|z=0)=0.4,P(w=2|z=0)=0[/tex]
Also,
[tex]P(w=0\z=1)=0,P(w=1|z=1)=0.6,P(w=2|z=1)=0.4[/tex]
Therefore,
[tex]P(0,0)=0.36,P(1,0)=0.24,P(2,0)=0,P(0,1)=0,P(1,1)=0.24,\\\\P(2,1)=0.16[/tex]
Learn More: https://brainly.com/question/13127182
A wire 9 meters long is cut into two pieces. One piece is bent into a equilateral triangle for a frame for a stained glass ornament, while the other piece is bent into a circle for a TV antenna. To reduce storage space, where should the wire be cut to minimize the total area of both figures? Give the length of wire used for each: For the equilateral triangle:
The length of wire used for the equilateral triangle is approximately 5.61 meters.
The remaining length of wire used for the circle will be 9 - 5.61 ≈ 3.39 meters.
Here,
To minimize the total area of both figures, we need to find the optimal cut point for the wire.
Let's assume the length of the wire used for the equilateral triangle is x meters, and the remaining length of the wire used for the circle is (9 - x) meters.
For the equilateral triangle:
An equilateral triangle has all three sides equal in length.
Let's call each side of the triangle s meters. Since the total length of the wire is x meters, each side will be x/3 meters.
The formula to find the area of an equilateral triangle with side length s is:
Area = (√(3)/4) * s²
Substitute s = x/3 into the area formula:
Area = (√(3)/4) * (x/3)²
Area = (√(3)/4) * (x²/9)
Now, for the circle:
The circumference (perimeter) of a circle is given by the formula:
Circumference = 2 * π * r
Since the remaining length of wire is (9 - x) meters, the circumference of the circle will be 2π(9 - x) meters.
The formula to find the area of a circle with radius r is:
Area = π * r²
To find the area of the circle, we need to find the radius.
Since the circumference is equal to 2πr, we can set up the equation:
2πr = 2π(9 - x)
Now, solve for r:
r = (9 - x)
Now, substitute r = (9 - x) into the area formula for the circle:
Area = π * (9 - x)²
Now, we want to minimize the total area, which is the sum of the areas of the triangle and the circle:
Total Area = (√(3)/4) * (x²/9) + π * (9 - x)²
To find the optimal value of x that minimizes the total area, we can take the derivative of the total area with respect to x, set it to zero, and solve for x.
d(Total Area)/dx = 0
Now, find the critical points and determine which one yields the minimum area.
Taking the derivative and setting it to zero:
d(Total Area)/dx = (√(3)/4) * (2x/9) - 2π * (9 - x)
Setting it to zero:
(√(3)/4) * (2x/9) - 2π * (9 - x) = 0
Now, solve for x:
(√(3)/4) * (2x/9) = 2π * (9 - x)
x/9 = (8π - 2πx) / (√(3))
Now, isolate x:
x = 9 * (8π - 2πx) / (√(3))
x(√(3)) = 9 * (8π - 2πx)
x(√(3) + 2π) = 9 * 8π
x = (9 * 8π) / (√(3) + 2π)
Now, we can calculate the value of x:
x ≈ 5.61 meters
So, the length of wire used for the equilateral triangle is approximately 5.61 meters.
The remaining length of wire used for the circle will be 9 - 5.61 ≈ 3.39 meters.
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A bus driver makes roughly $3280 every month. How much does he make in one week at this rate.
Answer:
I think around $36
Hope it helps!
Answer:
It depends...
Step-by-step explanation:
It depends how much weeks are in the month if there are three weeks and no extra days then you would have an answer of about 1093 (exact: 1093.33333333). just divide the number of weeks by the number of money.
There were 2,300 applicants for enrollment to the freshman class at a small college in the year 2010. The number of applicants has risen linearly by roughly 170 per year. The number of applications f(x) is given by f(x) = 2,300 + 170x, where x is the number of years since 2010. a. Determine if the function g(x) = * = 2,300 is the inverse of f. 170 b. Interpret the meaning of function g in the context of the problem.
a. No
b. The value g(x) represents the number of years since the year 2010 based on the number of applicants to the freshman class, x.
a. Yes
b. The value 8(x) represents the number of applicants to the freshman class based on the number of years since 2010,
a. No
b. The value slx) represents the number of applicants to the freshman class based on the number of years since 2010,
a. Yes
b. The value six) represents the number of years since the year 2010 based on the number of applicants to the freshman class x
Answer:
The inverse function is [tex]g(x) = \frac{x - 2300}{170}[/tex]
The value of g(x) represents the number of applicants to the freshman class based on the number of years since 2010.
Step-by-step explanation:
Number of applicants in x years after 2010:
Is given by the following function:
[tex]f(x) = 2300 + 170x[/tex]
Inverse function:
We exchange the values of y = f(x) and x in the original function, and then find y. So
[tex]x = 2300 + 170y[/tex]
[tex]170y = x - 2300[/tex]
[tex]y = \frac{x - 2300}{170}[/tex]
[tex]g(x) = \frac{x - 2300}{170}[/tex]
The inverse function is [tex]g(x) = \frac{x - 2300}{170}[/tex]
Meaning of g:
f(x): Number of students in x years:
g(x): Inverse of f(x), is the number of years it takes for there to be x applicants, so the answer is:
The value of g(x) represents the number of applicants to the freshman class based on the number of years since 2010.
A life insurance policy cost $8.52 for every $1000 of insurance at this rate what is the cost for 20,000 worth of life insurance
Answer:
$170.40
Step-by-step explanation:
cost of policy=$8.52
Groups of $1000=20,00÷1000
=20
20,000 worth=$8.52×20
$170.40
(3.5 x 10 ^ -4) ÷ (5 x 10 ^ 5) in standard form PLZZ ANSWER QUICK
Answer:
7x10 ^-10
Step-by-step explanation:
Simplify this expression 3^-3
ASAPPPP PLSSSS
Step-by-step explanation:
-27 okay 3^-3 its same as 3^3
Answer: A)
[tex]3^{-3}[/tex]
[tex]3^{-3}=\frac{1}{3^3}[/tex]
[tex]=\frac{1}{3^3}[/tex]
[tex]3^3=27[/tex]
[tex]=\frac{1}{27}[/tex]
OAmalOHopeO
What is the area of the circle in terms of [tex]\pi[/tex]?
a. 3.4225[tex]\pi[/tex] m²
b. 6.845[tex]\pi[/tex] m²
c. 7.4[tex]\pi[/tex] m²
d. 13.69[tex]\pi[/tex] m²
[tex] \sf \: d \: = 3.7m \\ \sf \: r \: = \frac{3.7}{2} = 1.85 \: m\\ \\ \sf \: c \: = \pi {r}^{2} \\ \\ \sf \: c \: = \pi ({1.85})^{2} \\ \sf c = 1.85 \times 1.85 \times \pi \\ \sf \: c = \boxed {\underline{ \bf a. \: 3.4225\pi \: m ^{2} }}[/tex]
How many subsets of at least one element does a set of seven elements have?
[tex]\boxed{\large{\bold{\blue{ANSWER~:) }}}}[/tex]
For each subset it can either contain or not contain an element. For each element, there are 2 possibilities. Multiplying these together we get 27 or 128 subsets. For generalisation the total number of subsets of a set containing n elements is 2 to the power n.
total subsets
2^n2⁷128