Answer:
2100
Step-by-step explanation:
50*18=900
900+1,200=2100
To find the population in 2018, we calculate the total increase from 2000 to 2018 by multiplying the yearly increase of 50 people by 18 years, resulting in an additional 900 people. Adding this to the initial population of 1,200 people gives us a total population of 2,100 people in 2018.
If a population is recorded at 1,200 in the year 2000 and increases at a steady rate of 50 people each year, we can calculate the population in 2018 using a linear growth model. First, we need to determine the number of years between 2000 and 2018, which is 18 years. Next, we multiply the annual increase (50 people) by the number of years (18) to find the total increase over this period.
The calculation would be as follows:
Total Increase = Annual Increase times Number of YearsTotal Increase = 50 people/year times 18 yearsTotal Increase = 900 peopleWe then add this total increase to the initial population to get the population in 2018:
Population in 2018 = Initial Population + Total IncreasePopulation in 2018 = 1,200 people + 900 peoplePopulation in 2018 = 2,100 peopleThe population in 2018 would be 2,100 people.
In a certain country, the true probability of a baby being a boy is 0.524. Among the next five randomly selected births in the country, what is the probability that at least one of them is a girl?
Final answer:
The probability of at least one of the next five births being a girl is approximately 92.85%.
Explanation:
To find the probability that at least one of the next five births is a girl, we will find the probability of none of the next five births being a girl and then subtract it from 1.
The probability of a baby being a girl is 1 - 0.524 = 0.476. Therefore, the probability of a baby being a boy is 0.524.
The probability of none of the next five births being a girl is (0.524)^5 = 0.07150816.
Therefore, the probability of at least one of the next five births being a girl is 1 - 0.07150816 = 0.92849184, or approximately 92.85%.
In a learning curve application, 658.5 work hours are required for the third production unit and 615.7 work hours are required for the fourth production unit. Determine the value of n (and therefore s) in the equation Z U=K(u^n ), where u=the output unit number; Z_=the number of input resource units to produce output unit u; K=the number of input resource units to produce the first output unit; s=the learning curve slope parameter expressed as a decimal (s=0.9 for a 90% learning curve); n=logs/log2 =the learning curve exponent.
Answer:
n ≈ -0.2336s ≈ 0.8505Step-by-step explanation:
We can put the given numbers into the given formula and solve for n.
658.5 = k·3^n
615.7 = k·4^n
Dividing the first equation by the second, we get ...
658.5/615.7 = (3/4)^n
The log of this is ...
log(658.5/615.7) = n·log(3/4)
n = log(658.5/615.7)/log(3/4) ≈ 0.0291866/-0.124939
n ≈ -0.233607
Then we can find s from ...
log(s) = n·log(2)
s = 2^n
s ≈ 0.850506
The learning curve helps in estimating the reduction in the time required to produce units as experience is gained in manufacturing. In this problem, you need to solve a system of two equations based on the given work hours for the third and fourth production units, to find the value of exponent n and hence the slope parameter s.
Explanation:This problem involves understanding of the concept of learning curve used in production and operations management. The learning curve predicts the time required to produce subsequent units given the time consumed by previous units. The initial units take more time to produce due to learning, but as the team gains experience, the time required to produce each subsequent unit decreases.
Above, you've given two data points - the third production unit requires 658.5 work hours and the fourth production unit requires 615.7 work hours. You want to find n in the equation Z_u=K(u^n), where this 'n' is the learning curve exponent which strongly influences how rapidly production time decreases as experience is gained.
We cannot directly calculate n because we do not have a value for K. However, we can set up a system of equations using both data points and solve for n. If you equate the two expressions for Z3 and Z4 and solve for n, you can determine its value and hence find the slope parameter s as well using the relationship s=logs/log2.
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74% of freshmen entering public high schools in 2006 graduated with their class in 2010. A random sample of 81 freshmen is selected. Find the probability that the proportion of students who graduated is greater than 0.743 .
Answer: Probability that the proportion of students who graduated is greater than 0.743 is P = 0.4755
Step-by-step explanation:
Given that,
Probability of freshmen entering public high schools in 2006 graduated with their class in 2010, p = 0.74
Random sample of freshman, n = 81
Utilizing central limit theorem,
[tex]P(\hat{p}<p) = P(Z<\hat{p} - \frac{p}{\sqrt{\frac{p(1-p)}{n} } } )[/tex]
So,
[tex](P(\hat{p}>0.743) = P(Z>0.743 - \frac{0.74}{\sqrt{\frac{0.74(1-0.74)}{81} } } )[/tex]
= P( Z > 0.0616)
= 0.4755 ⇒ probability that the proportion of students who graduated is greater than 0.743.
In a recent study, the following responses were obtained to the question, “Do you favor recycling in your neighborhood?”. Given that a randomly selected response was "No Opinion", what is the probability that it came from a female? Round your result to 2 significant places after the decimal (For example, 0.86732 should be entered as 0.87).
Yes No No Opinion
Male 92 48 47
Female 78 63 49
Answer with Step-by-step explanation:
Yes No No Opinion
Male 92 48 47
Female 78 63 49
Given that a randomly selected response was "No Opinion"
what is the probability that it came from a female?
It is a problem of conditional probability in which
A: response came from a female
B: Response is "No Opinion"
Total number of no opinions=49+47=96
Total outcomes=92+78+48+63+47+49=377
P(B)=96/377
A∩B: No opinion from a female
P(A∩B)=49/377
P(A|B)=P(A∩B)/P(B) (by Baye's theorem)
=[tex]\dfrac{\dfrac{49}{377}}{\dfrac{96}{377}}[/tex]
= 49/96
= 0.51
Hence, given that a randomly selected response was "No Opinion",the probability that it came from a female is:
0.51
A quality control inspector has drawn a sample of 10 light bulbs from a recent production lot. Suppose 20% of the bulbs in the lot are defective. What is the probability that exactly 7 bulbs from the sample are defective? Round your answer to four decimal places
Answer:
The probability is 0.0008.
Step-by-step explanation:
Let X represents the event of defective bulb,
Given, the probability of defective bulb, p = 20 % = 0.2,
So, the probability that bulb is not defective, q = 1 - p = 0.8,
The number of bulbs drawn, n = 10,
Since, binomial distribution formula,
[tex]P(x=r) = ^nC_r p^r q^{n-r}[/tex]
Where, [tex]^nC_r = \frac{n!}{r!(n-r)!}[/tex]
Hence, the probability that exactly 7 bulbs from the sample are defective is,
[tex]P(X=7)=^{10}C_7 (0.2)^7 (0.8)^{10-7}[/tex]
[tex]=120 (0.2)^7 (0.8)^3[/tex]
[tex]=0.000786432[/tex]
[tex]\approx 0.0008[/tex]
Calculate the probability of exactly 7 defective bulbs in a sample of 10 using the binomial distribution formula.
Binomial distribution:
Calculate the probability that exactly 7 out of 10 bulbs are defective using the binomial distribution formula.Use the formula: P(X = k) = C(n, k) * p^k * (1-p)^(n-k) where n = 10, k = 7, and p = 0.2.Calculate the probability:P(X = 7) = C(10, 7) * (0.2)^7 * (0.8)^3 ≈ 0.2013
if p+q= 4, q+r= -2,r+p= 3 then p+q+r =?
a) 2/5 b) 5/2 c) 6/2 d) 5 e) 6
Answer:
5/2
Step-by-step explanation:
We are looking for p+q+r.
We are given:
p+q =4 Equation 1.
q+r=-2 Equation 2.
p +r=3. Equation 3.
I'm going to solve the first equation for p giving me p=4-q.
I'm going to solve the second equation for r giving me r=-2-q.
I'm going to plug this into the third equation.
p +r=3
(4-q)+(-2-q)=3
4-q+-2-q=3
Combine like terms:
2-2q=3
Subtract 2 on both sides:
-2q=1
Divide both sides by -2:
q=-1/2.
Plug into the first equation to find p now.
First equation is p+q=4 with q=-1/2.
p+q=4
p+(-1/2)=4
Add 1/2 on both sides
p=4+1/2
p=9/2.
Let's find r now.
Using the second equation with p=9/2 and/or q=-1/2 we have:
q+r=-2
-1/2+r=-2
Add 1/2 on both sides
r=-2+1/2
r=-3/2
So what is p+q+r?
Let's plug in our p,q, and r and find out!
p+q+r
9/2+-1/2+-3/2
9/2-4/2
5/2
Find the equation of the curve that passes through the point (x, y) = (0, 0) and has an arc length on the interval x is between 0 and pi over 4 inclusive given by the integral the integral from 0 to pi over 4 of the square root of the quantity 1 plus the secant to the 4th power of x, dx .
[tex]\displaystyle\int_0^{\pi/4}\sqrt{1+\sec^4x}\,\mathrm dx=\int_0^{\pi/4}\sqrt{1+(\sec^2x)^2}\,\mathrm dx[/tex]
Recall that [tex]\displaystyle(\tan x)'=\sec^2x[/tex]. Then right away you see the integral gives the arc length of the curve [tex]y=\tan x[/tex] over the given interval.
what is the solution off the inequality shown below? 7+y>0
Answer:
y > -7
Step-by-step explanation:
Isolate the variable, y. Treat the > as a equal sign, what you do to one side, you do to the other. Subtract 7 from both sides:
y + 7 (-7) > 0 (-7)
y > 0 - 7
y > -7
y > -7 is your answer.
~
Answer:
[tex]\huge \boxed{y>-7}[/tex]
Step-by-step explanation:
Switch sides.
[tex]\displaystyle y+7>0[/tex]
Subtract by 7 from both sides of equation.
[tex]\displaystyle y+7-7>0-7[/tex]
Simplify, to find the answer.
[tex]\displaystyle 0-7=-7[/tex]
[tex]\huge \boxed{y>-7}[/tex], which is our answer.
A tank initially holds 200 gallons of fresh water (no salt). A brine solution containing 2 pounds of salt per gallon is poured in at a rate of 3 gallons per minute. The tank is kept well-mixed and drained at the rate of 3 gallons per minute. Write and solve an initial value differential equation to model the amount of salt, A, in the tank at time, t. 2.
Answer:
1.99 pounds per gallon of salt in t=2 in the tank.
Step-by-step explanation:
First we consider the matter balance equation that contemplates the input and output; the generation and consumption equal to the acomulation.
[tex] Acomulation = Input - Output + Generation - Consumption [/tex]
In this case we have no Generation neither do Consumption so, if we consider Acomulation = A(t), the rate of change of A(t) in time is given by:
[tex]\frac{dA}{dt}+R_{out}A(t)= CR_{s}[/tex] ---(1)
where C is th concentration, with the initial value statement that A(t=0) = 0 because there is no salt in the time cero in the tank, only water.
Given the integral factor -> [tex]u(t)= exp[R_{out}] [/tex] and multipying the entire (1) by it, we have:
[tex]\int \frac{d}{dt}[A(t) \exp[R_{out}t]] \, dt = CR_{in} \int \exp[R_{out}t] \, dt[/tex]
Solving this integrals we obtain:
[tex]A(t)=\frac{CR_{in}}{R_{out}}+Cte*\exp[-R_{out}t][/tex]
So given the initial value condition A(t=0)=0 we have:
[tex]Cte=- \frac{CR_{in}}{R_{out}}[/tex],
and the solution is,
[tex]A(t)=\frac{CR_{in}}{R_{out}}-\frac{CR_{in}}{R_{out}}\exp[-R_{out}t][/tex].
If we give the actual values we obtain then,
[tex]A(t)=2\frac{pounds}{gallon}-2\frac{pounds}{gallon} \exp[-3t][/tex].
So in t= 2 we have [tex]A(t)=2\frac{pounds}{gallon}[/tex].
Lauren Industries wants to open a warehouse in the U.S. northwest. The location factors, weights, and potential locations are shown below.
Location Factor
Weight
Olympia
Salem
Boise
Helena
Land Availability/Cost
0.25
80
100
90
87
Area Population
0.20
80
85
100
82
Quality of Life
0.15
100
88
92
100
Labor Availability
0.25
91
100
80
90
Proximity to Major Highway
0.15
85
88
100
91
Calculate the total weighted-factor score for Salem.
86.50
93.40
91.30
89.30
100
Answer:
Find the slope of the line that passes through the points shown in the table.
The slope of the line that passes through the points in the table is
.
Step-by-step explanation:
Jack typed 80 words per minute when he enrolled in a typing course. His typing speed increased by 3% two weeks into the course. At the end of the course, Jack was able to type his entire 1, 680 word document In 20 minutes. What was the percent of increase in his typing speed from the beginning of the course to the end? a 3% b 4% c 5% d 6% e 7%
Answer:
The answer is option C : 5%.
Step-by-step explanation:
Jack typed 80 words per minute when he enrolled in a typing course.
His typing speed increased by 3% two weeks into the course.
At the end of the course, Jack was able to type his entire 1, 680 word document In 20 minutes.
Hence, the percent of increase in his typing speed from the beginning of the course to the end is given by:
[tex]1680/20=84[/tex]
[tex](84-80)/80[/tex] =5%
Therefore, the answer is option C : 5%.
To find the percent increase in Jack's typing speed, compare his initial and final speeds. The percent increase is 5%.
Explanation:To find the percent increase in Jack's typing speed, we need to compare his initial speed to his final speed. Let's start by calculating his initial typing speed:
80 words per minute
To find his final typing speed, we need to determine how many words he typed in 20 minutes:
1,680 words / 20 minutes = 84 words per minute
Now we can find the percent increase:
(Final speed - Initial speed) / Initial speed * 100
(84 - 80) / 80 * 100 = 5%
Therefore, the percent of increase in Jack's typing speed from the beginning of the course to the end is 5%.
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Solve the system of equations using matrices. Use Gaussian elimination with back-substitution.
3x + 5y - 2w = -13
2x + 7z - w = -1
4y + 3z + 3w = 1
-x + 2y + 4z = -5
A. {(-1, -1 , 0, )}
B. {(1, -2, 0, 3)}
C. {( , -2, 0, )}
D. {( , - , 0, )}
Answer with explanation:
The given system of equation are
3x + 5y - 2w = -13
2x + 7z - w = -1
4y + 3z + 3w = 1
-x + 2y + 4z = -5
Writing the system of equation in terms of Augmented Matrix
[tex]\left[\begin{array}{ccccc}3&5&0&-2&-13\\2&0&7&-1&-1\\0&4&3&3&1\\-1&2&4&0&-5\end{array}\right]\\\\R_{3} \leftrightarrow R_{4}\\\\\left[\begin{array}{ccccc}3&5&0&-2&-13\\2&0&7&-1&-1\\-1&2&4&0&-5\\0&4&3&3&1\end{array}\right]\\\\R_{3} \rightarrow 2R_{3}+R_{2}\\\\\left[\begin{array}{ccccc}3&5&0&-2&-13\\2&0&7&-1&-1\\0&4&15&-1&-11\\0&4&3&3&1\end{array}\right]\\\\R_{4}\rightarrow R_{4}-R_{3}\\\\\left[\begin{array}{ccccc}3&5&0&-2&-13\\2&0&7&-1&-1\\0&4&15&-1&-11\\0&0&-12&4&12\end{array}\right][/tex]
[tex]R_{2}\rightarrow 3R_{2}-2R_{1}\\\\\left[\begin{array}{ccccc}3&5&0&-2&-13\\0&-10&21&1&23\\0&4&15&-1&-11\\0&0&-12&4&12\end{array}\right]\\\\R_{3}\rightarrow 5R_{3}+2R_{2}\\\\\left[\begin{array}{ccccc}3&5&0&-2&-13\\0&-10&21&1&23\\0&0&117&-3&-9\\0&0&-12&4&12\end{array}\right]\\\\ R_{4}\rightarrow 3R_{4}+4R_{3}\\\\\left[\begin{array}{ccccc}3&5&0&-2&-13\\0&-10&21&1&23\\0&0&117&-3&-9\\0&0&432&0&0\end{array}\right][/tex]
→432 z=0
z=0
⇒117 z-3 w=-9
-3 w=-9
Dividing both sides by -3
w=3
⇒-10 y+21z+w=23
-10 y+0+3=23
-10 y=23-3
-10 y= 20
y=-2
⇒3 x+5 y-2w=-13
3 x+5 ×(-2)-2 ×3= -13
3 x-10-6= -13
3 x=16-13
3 x=3
x=1
Option B. {(1, -2, 0, 3)}
(a)5%)Let fx, y) = x^4 + y^4 - 4xy + 1. and classify each critical point Find all critical points of fx,y) as a local minimum, local maximum or saddle point.
[tex]f(x,y)=x^4+y^4-4xy+1[/tex]
has critical points wherever the partial derivatives vanish:
[tex]f_x=4x^3-4y=0\implies x^3=y[/tex]
[tex]f_y=4y^3-4x=0\implies y^3=x[/tex]
Then
[tex]x^3=y\implies x^9=x\implies x(x^8-1)=0\implies x=0\text{ or }x=\pm1[/tex]
If [tex]x=0[/tex], then [tex]y=0[/tex]; critical point at (0, 0)If [tex]x=1[/tex], then [tex]y=1[/tex]; critical point at (1, 1)If [tex]x=-1[/tex], then [tex]y=-1[/tex]; critical point at (-1, -1)[tex]f(x,y)[/tex] has Hessian matrix
[tex]H(x,y)=\begin{bmatrix}12x^2&-4\\-4&12y^2\end{bmatrix}[/tex]
with determinant
[tex]\det H(x,y)=144x^2y^2-16[/tex]
At (0, 0), the Hessian determinant is -16, which indicates a saddle point.At (1, 1), the determinant is 128, and [tex]f_{xx}(1,1)=12[/tex], which indicates a local minimum.At (-1, -1), the determinant is again 128, and [tex]f_{xx}(-1,-1)=12[/tex], which indicates another local minimum.$10000 is deposited in an account earning 4% interest compounded continuously. Use the continuous interest formula below to determine how long it takes for the amount in the account to double. Round answer to 2 decimal places. A = P e r t
_____years.
Answer:
The required number of years are 7.52 years.
Step-by-step explanation:
Given : $10000 is deposited in an account earning 4% interest compounded continuously.
To find : How long it takes for the amount in the account to double?
Solution :
Applying Continuous interest formula,
[tex]A=Pe^{rt}[/tex]
Where, P is the principal P=$10000
r is the interest rate r=4%=0.04
t is the time
We have given, Amount in the account to double
i.e. A=2P
Substitute the value in the formula,
[tex]2P=Pe^{rt}[/tex]
[tex]2=e^{0.04t}[/tex]
Taking log both side,
[tex]\log 2=\log (e^{0.04t})[/tex]
[tex]\log 2=0.04t\times log e[/tex]
[tex]t=\frac{\log 2}{0.04}[/tex]
[tex]t=7.52[/tex]
Therefore, The required number of years are 7.52 years.
Louisa ran at an average speed of five miles per hour along an entire circular park path. Calvin ran along the same path in the opposite direction at an average speed of six miles per hour. It took Calvin 30 minutes less than it took Louisa to run the full path once. How many miles did Louisa run when she completed one circular path?
Answer:
15 miles
Step-by-step explanation:
Let [tex]x[/tex] be the miles in the circular park path, [tex]t_{L}[/tex] the time Louisa takes to finish and [tex]t_{C}[/tex] the time Calvin takes to finish both in hours.
Then [tex]x[/tex], the longitude is equal to the velocity times the time used to finish. So
[tex]x=5t_{L}[/tex]
[tex]x=6t_{C}[/tex]
And the difference between Louisa's time and Calvin' time is 30 minutes, half an hour. So:
[tex]t_{C}=t_{L}-0.5[/tex]
Three equations, three unknowns, the system can be solved.
Equalizing the equation with x :
[tex]5t_{L}=6t_{C}[/tex]
In this last equation replace [tex]t_{C}[/tex] with the other equation and solve:
[tex]5t_{L}=6(t_{L}-0.5)\\ 5t_{L}=6t_{L}-3\\ 3=6t_{L}-5t_{L}\\ 3=t_{L}\\ t_{L}=3[/tex]
With Louisa's time find x:
[tex]x=5t_{L}\\ x=5(3)\\ x=15[/tex]
The length of the circular path that Louisa ran is 15 miles when she completed one circular path.
Let's denote the length of the circular path as L miles. Louisa's speed is 5 miles per hour, and Calvin's speed is 6 miles per hour.
We'll use the equation for time, which is time = distance / speed.
For Louisa:
[tex]t_L = \frac{L}{5}[/tex]
Where:
[tex]t_L[/tex] is the time it takes Louisa to complete the path.L is the length of the circular path.5 is Louisa's speed in miles per hour.For Calvin:
[tex]t_C = \frac{L}{6}[/tex]
Where:
[tex]t_C[/tex] is the time it takes Calvin to complete the path.L is the length of the circular path.6 is Calvin's speed in miles per hour.According to the problem, it took Calvin 30 minutes (0.5 hours) less than it took Louisa to run the full path:
[tex]t_L = t_C + 0.5[/tex]
Substitute the expressions for [tex]t_L[/tex] and [tex]t_C[/tex] into the equation:
[tex]\frac{L}{5} = \frac{L}{6} + 0.5[/tex]
To solve for L, first find a common denominator for the fractions.
The common denominator for 5 and 6 is 30.
Therefore:
[tex]\frac{6L}{30} = \frac{5L}{30} + 0.5[/tex]
Multiply everything by 30 to eliminate the denominators:
[tex]6L = 5L + 15[/tex]
Subtract 5L from both sides:
[tex]L = 15[/tex]
Giovanni and Jean started a 120-mile bicycle race at the same time. Giovanni cycled at 23.9 miles per hour while Jean cycled at 24 miles per hour. When Jean crossed the finish line, how many miles was Giovanni from the finish line? Express your answer as a decimal to the nearest tenth.
Answer:
Giovanni was 0.5 miles from the finish line
Step-by-step explanation:
This is a problem of movement with constant velocity.
For this kind of problems, generally it is enough to remember the definition of average velocity v:
[tex]v=\frac{x}{t}[/tex]
Where x is the change in position that took place in an interval t.
First, find the time that Jean, who cycled at 24 miles per hour, spent on the race:
Isolating t from the last equation,
[tex]t=\frac{x}{v}[/tex], and replacing the data for Jean movement:
[tex]t=\frac{120}{24}=5h[/tex]
Second, find what was the distance that Giovanni had cycled when Jean crossed the line:
[tex]x=v*t\\ x=23.9*5=119.5[/tex]
When Jean crossed the line he had cycled 120 miles, and Giovanni 119.5; so Giovanni was 0.5 miles from the finish line.
After extensive measurements of the time necessary to complete the first homework assignment, a teacher determines that there is a population mean of 100 and a stardard deviation of 20. If she samples a class of 60 students and calculates a mean of 96 minutes, what is the z statistic (round to the nearest 2 decimal places, don't forget a negative sign if necessary)?
Answer: -1.55
Step-by-step explanation:
Given : Mean : [tex]\mu=100[/tex]
Standard deviation : [tex]\sigma=20[/tex]
Sample size : [tex]n=60[/tex]
Sample mean : [tex]\overline{x}=96[/tex]
The test statistic for the population mean is given by :-
[tex]z=\dfrac{\overline{x}-\mu}{\dfrac{\sigma}{\sqrt{n}}}[/tex]
[tex]\Rightarrow\ z=\dfrac{96-100}{\dfrac{20}{\sqrt{60}}}=-1.54919333848\approx-1.55[/tex]
Hence, the value of z statistic = -1.55
The z statistic is approximately -1.55, indicating that the class mean of 96 minutes is 1.55 standard deviations below the known population mean of 100 minutes.
Explanation:To calculate the z statistic (or z-score), we use the formula Z = (X - μ) / (σ / √n), where X is the sample mean, μ is the population mean, σ is the population standard deviation, and n is the sample size. In this problem, X = 96 minutes, μ = 100 minutes, σ = 20 minutes, and n = 60 students.
Plugging in the values, we get:
Z = (96 - 100) / (20 / √60)
First, calculate the denominator:
20 / √60 = 20 / 7.746 (approximately)
Now, divide the difference between the sample mean and population mean by this value:
Z = -4 / (20 / 7.746) ≈ -4 / 2.58 ≈ -1.55 (rounded to two decimal places)
So, the z statistic is approximately -1.55. This indicates that the class mean of 96 minutes is 1.55 standard deviations below the known population mean of 100 minutes.
Box all final answers. Little to no work will reccive little to no credit! I. (5 points) Let A and B be 4 x 4 matrices, with det(A) =-3 and det(B) = 2: Compute (a) det(AB) (b) det(B5) (c) det(2A) (d) det(ATA) (e) det(B-AB)
Answer: a) -6, b) 32, c) -48, d) 9, e) -12
Step-by-step explanation:
Since we have given that
A and B are 4 × 4 matrices.
Here,
det (A) = -3
det (B) = 2
We need to find the respective parts:
a) det (AB)
[tex]\mid AB\mid=\mid A\mid.\mid B\mid\\\\\mid AB\mid=-3\times 2=-6[/tex]
b) det (B⁵ )
[tex]\mid B^5\mid=\mid B\mid ^5=2^5=32[/tex]
c) det (2A)
Since we know that
[tex]\mid kA\mid =k^n\mid A\mid[/tex]
so, it becomes,
[tex]\mid 2A\mid =2^4\mid A\mid=16\times -3=-48[/tex]
d) [tex]\bold{det(A^TA)}[/tex]
Since we know that
[tex]\mid A^T\mid=\mid A\mid[/tex]
so, it becomes,
[tex]\mid A^TA\mid=\mid A^T\mid \times \mid A\mid=-3\times -3=9[/tex]
e) det (B⁻¹AB)
As we know that
[tex]\mid B^{-1}\mid =\mid B\mid[/tex]
so, it becomes,
[tex]\mid B^{-1}AB}\mid =\mid B^{-1}.\mid \mid A\mid.\mid B\mid=2\times -3\times 2=-12[/tex]
Hence, a) -6, b) 32, c) -48, d) 9, e) -12
A net force of 24 N is acting on a 4.0-kg object. Find the acceleration in m/s 6m/s^2
Step-by-step explanation:
mass of the object=m=4 kg
the net force exerted on it=F=24N=kgm/s²
the acceleration of the object=a=?
formula for calculating the acceleration=
F=ma
plug in the values
24kgm/s²=4kga
divide each side by 4
24kgm/s²/4 kg=4kga/4kg
a=6m/s²
Final answer:
The acceleration of the 4.0-kg object with a net force of 24 N is 6 m/s².
Explanation:
Acceleration: To find the acceleration, we use Newton's Second Law: F = ma. Given a net force of 24 N on a 4.0-kg object, the acceleration can be calculated as follows:
a = F/m = 24 N / 4.0 kg = 6 m/s²
Therefore, the acceleration of the object is 6 m/s².
(×-5) second power equals 3
Answer: [tex]x_1=6.73\\\\x_2=3.26[/tex]
Step-by-step explanation:
Given the following expression:
[tex](x-5)^2=3[/tex]
And knowing that:
[tex](a\±b)^2=a^2-2ab+b^2[/tex]
We get:
[tex]x^2-2(x)(5)+5^2=3\\\\x^2-10x+25=3[/tex]
Move the 3 to the left side of the equation:
[tex]x^2-10x+25-3=0\\\\x^2-10x+22=0[/tex]
Apply the Quadratic formula:
[tex]x=\frac{-b\±\sqrt{b^2-4ac} }{2a}[/tex]
In this case:
[tex]a=1\\b=-10\\c=22[/tex]
Substituting values into the Quadratic formula, we get:
[tex]x=\frac{-(-10)\±\sqrt{(-10)^2-4(1)(22)} }{2(1)}\\\\\\x_1=6.73\\\\x_2=3.26[/tex]
Question 4: What is the center of the circle with standard form (x-2)2+(y+4)2=16?
Question 4 options:
(-2, 4)
(2, -4)
(-2, -4)
(2, 4)
Answer: second option.
Step-by-step explanation:
The equation of the circle in "Standard form" or "Center-radius form" is the following:
[tex](x - h)^2 + (y - k)^2 = r^2[/tex]
Where the center of the circle is at the point (h, k) and the radius is "r".
In this case you have the following equation of the circle written in Standard form:
[tex](x-2)^2+(y+4)^2=16[/tex]
You can identify that:
[tex]h=2\\k=-4[/tex]
Therefore, the center of the given circle is at this point:
[tex](2,-4)[/tex]
Answer:
2,-4
Step-by-step explanation:
I think I did it right but if I didn't you can blame me smh
American Gothic was painted in 1930 by Grant Wood. The length of a reproduction of the rectangular painting is 2 inches more than the width. Find the dimensions of the reproduction if it has a perimeter of 43.8 inches. width length
Answer:
Length: 11.95
Width: 9.95
Step-by-step explanation:
Hello, great question. These types are questions are the beginning steps for learning more advanced Algebraic Equations.
To begin solving this question we need to create a formula to then solve. According to the information given in the question we can create 2 formulas.
[tex]P = 2L+2W[/tex]
[tex]L = 2+W[/tex]
Where:
P is the perimeterL is the lengthW is the widthNow we can replace L for the L in the P formula and solve for W, Like so....
[tex]43.8 = 2(2+W)+2W[/tex]
[tex]43.8 = 4+2W+2W[/tex]
[tex]39.8 = 4W[/tex]
[tex]9.95 = W[/tex]
Now that we have the value of W we can plug that into the L formula and find L
[tex]L = 2+9.95[/tex]
[tex]L = 11.95[/tex]
Finally, we can see that the value of the Length is 11.95 in and the Width is 9.95 in
I hope this answered your question. If you have any more questions feel free to ask away at Brainly.
The random variable X, representing the number of accidents in a certain intersection in a week, has the following probability distribution: x 0 1 2 3 4 5 P(X = x) 0.20 0.30 0.20 0.15 0.10 0.05 What is the probability that in a given week there will be at most 3 accidents? 0.70 0.85 0.35 0.15 1.00
Answer: 1.8
Step-by-step explanation:
a. Calculate the probability that at most two accidents occur in any given week.
Probability of 0 accidents + Probability of 1 accident + Probability of 2 accidents = 0.20 + 0.30 + 0.20 = 0.70.
b. What is the probability that there are at least two weeks between any two accidents?
Probability of no accidents + Probability of 1 accident = 0.20 + 0.30 = 0.50.
The dimension of the row space of a 3 x 5 matrix A is 2 d. What is the dimension of the solution space of the homogeneous system Ax = 0?
Answer:
Dimension of the solution space of the homogeneous system =dimension of kernel=3.
Step-by-step explanation:
Given a matrix has 3 rows and 5 columns .
Dimension of Domain=Number of columns in the matrix=5d
Dimension of the row space =2d
We know that dimension of row space= rank of matrix=2d
Rank-nullity theorem : Rank+nullity= dimension of domain=Number of columns in the matrix.
By using rank-nullity theorem
2+nullity=5
Nullity=5-2
Nullity=3
Dimension of kernel=3d
Dimension of kernel=Dimension of solution space
Dimension of solution space=3d
Hence, the dimension of solution space of the homogeneous system =3d.
The dimension of the solution space of the homogeneous system Ax = 0 for a 3 x 5 matrix A with a row space dimension of 2 is 3.
Explanation:In the field of linear algebra, the dimension of the row space of a 3 x 5 matrix A, dictates the dimension of the column space of the matrix. Given that the dimension of the row space is 2, it follows that the dimension of the column space or rank of the matrix is also 2.
The Nullity of the matrix, which is the dimension of the solution space of the homogeneous system Ax = 0, is obtained by subtracting the rank of the matrix from the total number of columns in the matrix. In this case, since we have 5 columns in the matrix and a rank of 2, the Nullity is (5-2) = 3. Therefore, the dimension of the solution space of the homogeneous system Ax = 0 is 3.
Learn more about Linear Algebra here:https://brainly.com/question/40185999
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Solve the following equations: (a) x^11=13 mod 35 (b) x^5=3 mod 64
a.
[tex]x^{11}=13\pmod{35}\implies\begin{cases}x^{11}\equiv13\equiv3\pmod5\\x^{11}\equiv13\equiv6\pmod7\end{cases}[/tex]
By Fermat's little theorem, we have
[tex]x^{11}\equiv (x^5)^2x\equiv x^3\equiv3\pmod5[/tex]
[tex]x^{11}\equiv x^7x^4\equiv x^5\equiv6\pmod 7[/tex]
5 and 7 are both prime, so [tex]\varphi(5)=4[/tex] and [tex]\varphi(7)=6[/tex]. By Euler's theorem, we get
[tex]x^4\equiv1\pmod5\implies x\equiv3^{-1}\equiv2\pmod5[/tex]
[tex]x^6\equiv1\pmod7\impleis x\equiv6^{-1}\equiv6\pmod7[/tex]
Now we can use the Chinese remainder theorem to solve for [tex]x[/tex]. Start with
[tex]x=2\cdot7+5\cdot6[/tex]
Taken mod 5, the second term vanishes and [tex]14\equiv4\pmod5[/tex]. Multiply by the inverse of 4 mod 5 (4), then by 2.[tex]x=2\cdot7\cdot4\cdot2+5\cdot6[/tex]
Taken mod 7, the first term vanishes and [tex]30\equiv2\pmod7[/tex]. Multiply by the inverse of 2 mod 7 (4), then by 6.[tex]x=2\cdot7\cdot4\cdot2+5\cdot6\cdot4\cdot6[/tex]
[tex]\implies x\equiv832\pmod{5\cdot7}\implies\boxed{x\equiv27\pmod{35}}[/tex]
b.
[tex]x^5\equiv3\pmod{64}[/tex]
We have [tex]\varphi(64)=32[/tex], so by Euler's theorem,
[tex]x^{32}\equiv1\pmod{64}[/tex]
Now, raising both sides of the original congruence to the power of 6 gives
[tex]x^{30}\equiv3^6\equiv729\equiv25\pmod{64}[/tex]
Then multiplying both sides by [tex]x^2[/tex] gives
[tex]x^{32}\equiv25x^2\equiv1\pmod{64}[/tex]
so that [tex]x^2[/tex] is the inverse of 25 mod 64. To find this inverse, solve for [tex]y[/tex] in [tex]25y\equiv1\pmod{64}[/tex]. Using the Euclidean algorithm, we have
64 = 2*25 + 14
25 = 1*14 + 11
14 = 1*11 + 3
11 = 3*3 + 2
3 = 1*2 + 1
=> 1 = 9*64 - 23*25
so that [tex](-23)\cdot25\equiv1\pmod{64}\implies y=25^{-1}\equiv-23\equiv41\pmod{64}[/tex].
So we know
[tex]25x^2\equiv1\pmod{64}\implies x^2\equiv41\pmod{64}[/tex]
Squaring both sides of this gives
[tex]x^4\equiv1681\equiv17\pmod{64}[/tex]
and multiplying both sides by [tex]x[/tex] tells us
[tex]x^5\equiv17x\equiv3\pmod{64}[/tex]
Use the Euclidean algorithm to solve for [tex]x[/tex].
64 = 3*17 + 13
17 = 1*13 + 4
13 = 3*4 + 1
=> 1 = 4*64 - 15*17
so that [tex](-15)\cdot17\equiv1\pmod{64}\implies17^{-1}\equiv-15\equiv49\pmod{64}[/tex], and so [tex]x\equiv147\pmod{64}\implies\boxed{x\equiv19\pmod{64}}[/tex]
Individuals who have a certain gene have a 0.46 probability of contracting a certain disease. Suppose that 913 individuals with the gene participate in a lifetime study. What is the standard deviation of the number of people who eventually contract the disea
Answer: 15.06
Step-by-step explanation:
Given : The number of individuals participated in a lifetime study :[tex]n=913[/tex]
The probability of individuals with certain gene of contracting a certain disease :[tex]p= 0.46[/tex]
Now, the standard deviation of the number of people who eventually contract the disease is given by :_
[tex]\sigma =\sqrt{np(1-p)}\\\\=\sqrt{913\times0.46(1-0.46)}=15.059521904\approx15.06[/tex]
Hence, the the standard deviation of the number of people who eventually contract the disease = 15.06
Add the Base 2 numbers: 11,110,110 +101,101,111
Answer:
100000
Step-by-step explanation:
I did the operations in the picture, you only have to know that 0+1=01, 1+1=10
and 1+1+1=11.
Now, I don't know if you need to calculate the total add, I calculated it.
In this case, you need to know that 1+1+1+1=100.
The sum of the two Base 2 numbers is: [tex]\[ {10001001_2} \][/tex].
The sum of the given Base 2 (binary) numbers is calculated as follows:
11,110,110
+ 101,101,111
Starting from the rightmost digit (least significant bit) and moving left, we add the digits:
- In the rightmost column, 0 + 1 = 1.
- In the next column, 1 + 1 = 10 (which is 0 in the current column and carry over 1 to the next column).
- Continuing this process, we add the digits along with any carry from the previous column.
Let's continue the addition:
11,110,110
+ 101,101,111
--------------
1,000,1001
Here's the step-by-step process:
- 0 + 1 = 1 (no carry).
- 1 + 1 = 10 (0 in this column, carry 1).
- 1 + 0 + 1 (carry) = 10 (0 in this column, carry 1).
- 1 + 1 + 1 (carry) = 11 (1 in this column, carry 1).
- 1 + 0 + 1 (carry) = 10 (0 in this column, carry 1).
- 1 + 1 + 1 (carry) = 11 (1 in this column, carry 1).
- 1 + 0 + 1 (carry) = 10 (0 in this column, carry 1).
- 1 + 1 + 1 (carry) = 11 (1 in this column, carry 1).
- 1 + 0 + 1 (carry) = 10 (0 in this column, carry 1).
- 1 + 1 + 1 (carry) = 11 (1 in this column, carry 1).
- Finally, we have a carry of 1 that we add to the leftmost digit, giving us 1 + 1 = 10 (0 in this column, carry 1).
Since there are no more digits to add, we write down the 1 at the beginning:
1,000,1001
Therefore, the sum of the two Base 2 numbers is:
[tex]\[ {10001001_2} \][/tex].
For the month of November in a certain city, 53% of the days are cloudy. Also in the month of November in the same city, 46% of the days are cloudy and snowy. What is the probability that a randomly selected day in November will be snowy if it is cloudy?
Answer: The required probability that a randomly selected day in November will be snowy if it is cloudy is 86.79%.
Step-by-step explanation: Given that for the month of November in a certain city, 53% of the days are cloudy. Also in the month of November in the same city, 46% of the days are cloudy and snowy.
We are to find the probability that a randomly selected day in November will be snowy if it is cloudy.
Let A denote the event that the day is cloudy and B denote the event that the day is snowy.
Then, according to the given information, we have
[tex]P(A)=53\%=0.53,\\\\P(A\cap B)=46\%=0.46.[/tex]
Now, we need to find the conditional probability of event B given that the event A has already happened.
That is, P(B/A).
We know that
[tex]P(B/A)=\dfrac{P(B\cap A)}{P(A)}=\dfrac{0.46}{0.53}=0.87=87.79\%.[/tex]
Thus, the required probability that a randomly selected day in November will be snowy if it is cloudy is 87.79%.
Final answer:
The probability that a randomly selected cloudy day in November will be snowy is calculated using conditional probability. The result is approximately 86.79%.
Explanation:
To determine the probability that a randomly selected day in November will be snowy if it is cloudy, we use the given information: 53% of days are cloudy and 46% of days are both cloudy and snowy. The probability we are looking for is the conditional probability of it being snowy given that it is cloudy, which can be calculated by dividing the probability of it being both cloudy and snowy by the probability of it being cloudy, which is P(Snowy | Cloudy) = P(Cloudy and Snowy) / P(Cloudy).
So the calculation would be:
P(Snowy | Cloudy) = (0.46) / (0.53)
= 0.8679 (or 86.79%).
Therefore, there is an 86.79% chance that it will be snowy on a day that is cloudy in that city in November.
(3 points) The directional derivative of f(x, y) at (2, 1) in the direction going from (2, 1) toward the point (1, 3) is −2/ √ 5, and the directional derivative at (2, 1) in the direction going from (2, 1) toward the point (5, 5) is 1. Compute fx(2, 1) and fy(2, 1).
The vector pointing from (2, 1) to (1, 3) points in the same direction as the vector [tex]\vec u=(1,3)-(2,1)=(-1,2)[/tex]. The derivative of [tex]f[/tex] at (2, 1) in the direction of [tex]\vec u[/tex] is
[tex]D_{\vec u}f(2,1)=\nabla f(2,1)\cdot\dfrac{\vec u}{\|\vec u\|}[/tex]
We have
[tex]\|\vec u\|=\sqrt{(-1)^2+2^2}=\sqrt5[/tex]
Then
[tex]D_{\vec u}f(2,1)=(f_x(2,1),f_y(2,1))\cdot\dfrac{(-1,2)}{\sqrt5}=\dfrac{-f_x(2,1)+2f_y(2,1)}{\sqrt5}=-\dfrac2{\sqrt5}[/tex]
[tex]\implies f_x(2,1)-2f_y(2,1)=2[/tex]
The vector pointing from (2, 1) to (5, 5) has the same direction as the vector [tex]\vec v=(5,5)-(2,1)=(3,4)[/tex]. The derivative of [tex]f[/tex] at (2, 1) in the direction of [tex]\vec v[/tex] is
[tex]D_{\vec v}f(2,1)=\nabla f(2,1)\cdot\dfrac{\vec v}{\|\vec v\|}[/tex]
[tex]\|\vec v\|=\sqrt{3^2+4^2}=5[/tex]
so that
[tex](f_x(2,1),f_y(2,1))\cdot\dfrac{(3,4)}5=1[/tex]
[tex]\implies3f_x(2,1)+4f_y(2,1)=5[/tex]
Solving the remaining system gives [tex]f_x(2,1)=\dfrac95[/tex] and [tex]f_y(2,1)=-\dfrac1{10}[/tex].
In the following question, a is a real number. Question: what choice of a would yield a system of linear equations with NO solutions?
9x-14y=-3
2x-ay=-6
Answer: The value of a should be 3.11.
Step-by-step explanation:
Since we have given that
[tex]9x-14y=-3\\\\2x-ay=-6[/tex]
We need to find the value of a so that it would yield a system of linear equations with NO solutions.
As we know the condition of no solutions, it means the lines should be parallel.
i.e.
[tex]\dfrac{a_1}{a_2}=\dfrac{b_1}{b_2}\neq \dfrac{c_1}{c_2}[/tex]
Consider the first two terms:
[tex]\dfrac{9}{2}=\dfrac{-14}{-a}\\\\a=\dfrac{2\times 14}{9}\\\\a=3.11[/tex]
So, the value of a should be 3.11.