False. M is not a vector space because it fails to contain the zero vector (0) under standard addition.
The statement is false. The set M = {a ∈ R: a > 1} is not a vector space under standard addition and scalar multiplication of real numbers. To be a vector space, a set must satisfy certain conditions, including the requirement of containing the zero vector.
In this case, M does not contain the zero vector (0), as all elements of M are greater than 1. Additionally, M fails to satisfy other vector space properties, such as closure under addition and scalar multiplication. For example, if we take two elements a, b ∈ M, their sum a + b may not necessarily be greater than 1, violating closure under addition.Therefore, due to the absence of the zero vector and the violation of other vector space properties, M cannot be considered a vector space under standard addition and scalar multiplication of real numbers.
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What is -a⁻² if a = -5?
Answer:
25
Step-by-step explanation:
First, plug -5 in for a, -(-5)^2. We treat the negative on the outside of the paranthese as a -1 so we do -1 times -5 and we get 5. Then we square 5 and get 25.
Point (2.-3) on glx) is transformed by -g[4(x+2)]. What is the new point? Show your work
After considering the given data we conclude that the new point generated is (2,3), under the condition that g(x) is transformed by [tex]-g[4(x+2)][/tex].
To evaluate the new point after the transformation of point (2,-3) by -g[4(x+2)], we can stage x=2 and g(x)=-3 into the expression [tex]-g[4(x+2)][/tex]and apply simplification to get the new y-coordinate. Then, we can combine the new x-coordinate x=2 with the new y-coordinate to get the new point.
Stage x=2 and g(x)=-3 into [tex]-g[4(x+2)]:[/tex]
[tex]-g[4(2+2)] = -g = -(-3) = 3[/tex]
The new y-coordinate is 3.
The new point is (2,3).
Hence, the new point after the transformation of point (2,-3) by [tex]-g[4(x+2)][/tex] is (2,3).
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Which expression is equivalent to the given expression?
Step-by-step explanation:
D. In 2 _ In
maaf kalo salah
help ASAP Ill give you brainliest
Answer:
none of these
Step-by-step explanation:
There are 3 boys walking
There are a total of 20 people
3/20 = 0.15
That is 15 percent, therefore none of these answers.
Step-by-step explanation:
any has at least one mode
Rewrite the expression using a DIVISION SYMBOL: "The quotient of m and 7."
Answer:
m ÷ 7
Step-by-step explanation:
"Quotient" means you're dividing, so this just means you're dividing m by 7.
If S=4 [tex]\pi[/tex] [tex]r^{2}[/tex] the value of S When R= 10[tex]\frac{1}{2}[/tex]
Information from a poll of registered voters in a city to assess voter support for a new school tax was the basis for the following statements.
The poll showed 51% of the respondents in this city's school district are in favor of the tax. The approval rating rises to 58% for those with children in public schools. It falls to 45% for those with no children in public schools. The older the respondent, the less favorable the view of the proposed tax: 38% of those over age 56 said they would vote for the tax compared with 73% of 18- to 25-year-olds.
Suppose that a registered voter from this city is selected at random, and define the following events.
F = event that the selected individual favors the school tax
C = event that the selected individual has children in the public schools
O = event that the selected individual is over 56 years old
Y = event that the selected individual is 18–25 years old
Use the given information to estimate the values of the following probabilities. (1) P(F) (ii) P(FIC) (iii) PCFCS) (iv) P(FIO)
The probability that the selected individual has children in public schools AND favors the school tax is 0.32
The probability that the selected individual favors the school tax AND has children in public schools is 0.32.
The probability that the selected individual favors the school tax AND does NOT have children in public schools is 0.2.
The probability that the selected individual favors the school tax AND is over 56 years old is 0.15.
The probability that the selected individual favors the school tax AND is 18-25 years old is 0.45.
Based on the given information, the probability of event F (the selected individual favors the school tax) is 0.54, as 54% of the respondents are in favor of the tax. The probability of event C (the selected individual has children in public schools) is 0.59, as the approval rating rises to 59% for those with children in public schools. The probability of event O (the selected individual is over 56 years old) is 0.37, as only 37% of those over age 56 said they would vote for the tax. The probability of event Y (the selected individual is 18-25 years old) is 0.71, as 71% of 18- to 25-year-olds said they would vote for the tax.
Using these probabilities, we can estimate the values of the following probabilities:
(1) P(CF) is the probability that the selected individual has children in public schools AND favors the school tax. Based on the given information, we can multiply the probabilities of events C and F: P(CF) = 0.59 * 0.54 = 0.318, or approximately 0.32.
(ii) P(FIC) is the probability that the selected individual favors the school tax AND has children in public schools. This is the same as P(CF), so P(FIC) = 0.32.
(iii) P(FIN) is the probability that the selected individual favors the school tax AND does NOT have children in public schools. To calculate this, we can use the fact that the approval rating falls to 44% for those with no children in public schools. So, P(FIN) = 0.44 * (1 - 0.59) = 0.18, or approximately 0.2.
(iv) P(FTO) is the probability that the selected individual favors the school tax AND is over 56 years old. To calculate this, we can use the fact that the approval rating for those over 56 years old is only 37%. So, P(FTO) = 0.37 * (1 - 0.59) = 0.1523, or approximately 0.15.
(v) P(FY) is the probability that the selected individual favors the school tax AND is 18-25 years old. To calculate this, we can use the fact that the approval rating for those 18-25 years old is 71%. So, P(FY) = 0.71 * (1 - 0.37) = 0.4477, or approximately 0.45.
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Given the definitions of f(x) and g(x) below, find the value of (gof)(1).
f(x) = 2x² – 2x – 4
g(x) = -5x + 14
Answer:
[tex](g*f)(x) = 34[/tex]
Step-by-step explanation:
For sake of clarity, [tex](g * f)(x) = g(f(x))[/tex]
First, find [tex]f(1)[/tex]
[tex]f(1) = 2(1)^2 - 2(1) - 4\\f(1) = 2-2-4 \\f(1)=-4[/tex]
Then, take what you got for [tex]f(1)[/tex] and plug that into [tex]g(x)[/tex]. In this case, [tex]f(1) = -4[/tex]
[tex]g(-4) = -5(-4) + 14\\g(-4)= 20 + 14\\g(-4) = 34[/tex]
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Let Y_1.... Y_n be a random sample from a distribution with the density function
f_θ(y) = 3θ^3/y^4 y≥θ≥0
Is there a UMP test at level a for testing H_o: θ ≤ θ_o vs. H_1 : θ> θ_o? If so, what is the test?
Yes, there is a uniformly most powerful (UMP) test at level a for testing for the given density function.
The test is based on the likelihood ratio, where the critical region is (θ_o, ∞) and the test statistic is (nθ_o^3)/Y(n), where Y(n) is the largest observation in the sample.
To obtain the UMP test at level a for testing H_0: θ ≤ θ_o vs. H_1: θ > θ_o, we need to find the likelihood ratio test with the largest power for all possible alternatives. The likelihood ratio test is constructed as the ratio of the likelihood function under H_0 to the likelihood function under H_1. By algebraic manipulation, we obtain the likelihood ratio test statistic as (nθ_o^3)/Y(n), where Y(n) is the largest observation in the sample.
Under H_0, this test statistic has a chi-squared distribution with one degree of freedom. Therefore, the critical region for rejecting H_0 at level a is the right tail of the chi-squared distribution with one degree of freedom, which is (θ_o, ∞).
This test is UMP because it has the highest power for all possible alternatives. This is because the distribution of Y(n) is stochastically increasing in θ, which means that for a given sample size n, the probability of obtaining an observation larger than a threshold value increases as θ increases.
Therefore, the likelihood ratio test statistic decreases as θ increases, which means that the rejection region (θ_o, ∞) has the highest power for all possible alternatives. Hence, the test based on the likelihood ratio is the UMP test at level a for testing H_0: θ ≤ θ_o vs. H_1: θ > θ_o for the given density function
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Suppose that the NY state total population remains relatively fixed 20Mil, with 8.4Mil of the people living in the city and remaining are in the suburbs. Each year 3.5% of the people living in the city move to the suburbs, and 1.7% of the suburban population moves to the city. What is the long-term distribution of population, after 100 years (what is the population in the city and in the suburbs)? Plot population of city and suburbs over period of 100 years. Submit, 1) answer(s), 2) Matlab code, 3) graph(s)
After 100 years, the long-term distribution of population in the city and suburbs of New York state can be calculated based on the given migration rates. The population in the city and suburbs will stabilize at approximately 3.96 million and 16.04 million, respectively. The population distribution can be visualized using a graph that shows the population of the city and suburbs over the 100-year period.
To calculate the long-term population distribution, we can use the concept of equilibrium. Let C represent the population in the city and S represent the population in the suburbs. The equilibrium equations can be written as follows:
C = C - 0.035C + 0.017S
S = S + 0.035C - 0.017S
Simplifying these equations, we have:
C = 0.965C + 0.017S
S = 0.035C + 0.983S
Solving these equations simultaneously, we find that C stabilizes at approximately 3.96 million and S stabilizes at approximately 16.04 million.
To plot the population of the city and suburbs over the 100-year period, you can use the following MATLAB code:
Copy code
years = 0:100;
C = zeros(1, 101);
S = zeros(1, 101);
C(1) = 8.4;
S(1) = 20 - C(1);
for i = 2:101
C(i) = 0.965*C(i-1) + 0.017*S(i-1);
S(i) = 0.035*C(i-1) + 0.983*S(i-1);
end
plot(years, C, 'b', 'LineWidth', 2);
hold on;
plot(years, S, 'r', 'LineWidth', 2);
xlabel('Years');
ylabel('Population');
legend('City', 'Suburbs');
title('Population of City and Suburbs Over 100 Years');
This MATLAB code calculates and plots the population of the city (in blue) and suburbs (in red) over the 100-year period.
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Which point on the graph represents the y-intercept?
Find the zeros of the following quadratic functions.
3) x2 + 5x + 6 = 0
Bases are 6 and 10 the height is 4 whats the area of the trapszoid
Answer:
here,hope this helps : )
Step-by-step explanation:
Answer: A= 32
a (Base) 6
b (Base) 10
h (Height) 4
Step-by-step explanation: A=a+b
2h=6+10
2·4=32 I really hoped this helped
Tell whether the angles are complementary or supplementary. Then find the value of x.
Answer: Complementary x=15
Step-by-step explanation:
Complementary angles add up to 90°, supplementary angles add up to 180°.
We know they add up to 90 so...
3x+45=90
3x=45
x=15
Sammy counts the number of people in one section of the school auditorium. He counts 18 female students, 16 male students, and 6 teachers. There are 720 people in the auditorium. Consider the probability of selecting one person at random from the auditorium
Correct Question:
He counts 18 female students, 16 male students, and 6 teachers. There are
720 people in the auditorium. Consider the probability of selecting one person
at random from the auditorium.
Which of these statements are true?
Choose all that apply.
A: The probability of selecting a teacher is 6%.
B : The probability of selecting a student is 85%.
C : The probability of selecting a male student is 32%.
D : The probability of selecting a female student is 45%.
Step-by-step explanation:
Option B and D are correct because
The total number of people in one cross section = 18 + 16 + 6 = 40.
A = The probability of selecting a teacher is = (6/40)x100 = 15 % not equal to 6 %
B = The probability of selecting a male student is = (34/40)x100 = 85%
C = The probability of selecting a male student is = (16/40)x100 = 40 % not equal to 32 %
D : The probability of selecting a female student is = (18/40)x100= 45%
ABM Services paid a $4.15 annual dividend on a day it closed at a price of $54 per share. What
was the yield?
Answer:
Yield per share = 7.68% (Approx.)
Step-by-step explanation:
Given:
Dividend paid = $4.15
Price per dividend = $54
Find:
Yield per share
Computation:
Yield per share = [Dividend paid / Price per dividend]100
Yield per share = [4.15 / 54]100
Yield per share = [0.0768]100
Yield per share = 7.68% (Approx.)
Isaiah is decorating the outside of a box in the shape of a triangular prism. The figure
below shows a net for the box.
What is the surface area of the box, in square meters, that
Isaiah decorates
Answer:
389.19 m²
Step-by-step explanation:
The surface area of the box = area of the two equal triangles + area of the 3 different rectangles
✔️Area of the two equal triangles:
Area = 2(½*base*height)
base = 7 m
height = 8 m
Area of the two triangles = 2(½*7*8) = 56 m²
✔️Area of rectangle 1:
Area = Length*Width
L = 13 m
W = 7 m
Area of rectangle 1 = 13*7 = 91 m²
✔️Area of rectangle 2:
L = 13 m
W = 8 m
Area of rectangle 2 = 13*8 = 104 m²
✔️Area of rectangle 3:
L = 13 m
W = 10.63 m
Area of rectangle 3 = 13*10.63 = 138.19 m²
✅Surface Area of the box = 56 + 91 + 104 + 138.19 = 389.19 m²
Population 1,2,4,5,8 · Draw all possible sample of size 2 W.O.R · Sampling distribution of Proportion of even No. · Verify the results
Question:
A population consists 1, 2, 4, 5, 8. Draw all possible samples of size 2 without replacement from this population.
Verify that the sample mean is an unbiased estimate of the population mean.
Answer:
[tex]Samples: \{(1,2),(1,4),(1,5),(1,8),(2,4),(2,5),(2,8),(4,5),(4,8),(5,8)\}[/tex]
[tex]\hat p = \frac{3}{5}[/tex] --- proportion of evens
The sample mean is an unbiased estimate of the population mean.
Step-by-step explanation:
Given
[tex]Numbers: 1, 2, 4, 5, 8[/tex]
Solving (a): All possible samples of 2 (W.O.R)
W.O.R means without replacement
So, we have:
[tex]Samples: \{(1,2),(1,4),(1,5),(1,8),(2,4),(2,5),(2,8),(4,5),(4,8),(5,8)\}[/tex]
Solving (b): The sampling distribution of the proportion of even numbers
This is calculated as:
[tex]\hat p = \frac{n(Even)}{Total}[/tex]
The even samples are:
[tex]Even = \{2,4,8\}[/tex]
[tex]n(Even) = 3[/tex]
So, we have:
[tex]\hat p = \frac{3}{5}[/tex]
Solving (c): To verify
[tex]Samples: \{(1,2),(1,4),(1,5),(1,8),(2,4),(2,5),(2,8),(4,5),(4,8),(5,8)\}[/tex]
Calculate the mean of each samples
[tex]Sample\ means = \{1.5,2.5,3,4.5,3,3.5,5,4.5,6,6.5\}[/tex]
Calculate the mean of the sample means
[tex]\bar x = \frac{1.5 + 2.5 +3 + 4.5 + 4 + 3.5 + 5 + 4.5 + 6 + 6.5}{10}[/tex]
[tex]\bar x = \frac{40}{10}[/tex]
[tex]\bar x = 4[/tex]
Calculate the population mean:
[tex]Numbers: 1, 2, 4, 5, 8[/tex]
[tex]\mu = \frac{1 +2+4+5+8}{5}[/tex]
[tex]\mu = \frac{20}{5}[/tex]
[tex]\mu = 4[/tex]
[tex]\bar x = \mu = 4[/tex]
This implies that [tex]\bar x[/tex] is an unbiased estimate of the [tex]\mu[/tex]
A type of origami paper comes in 15 cm by 15 cm
square sheets. Hilary used two sheets to make the
origami dog. What is the total area of the origami
paper that Hilary used to make the dog?
Answer:
150 cm squared
Step-by-step explanation:
I guess that's the answer if I'm wrong you can tell me right away so that I can try another method thank you.
Pip was thinking of a number. Pip halves the number and gets an answer of 87.2. Form an
equation with x from the information.
X/2= 87.2
to find X:
87.2 X 2= 174.4
therefore X is 174.4
Brayden invests money in an account paying a simple interest of 3.3% per year. If he invests $30 and no money will be added or removed from the investment, how much will he have in one year, in dollars and cents?
Answer:
$30.99
Step-by-step explanation:
The formula for simple interest is I = PRT where I = interest earned, P = principal amount borrowed/deposited, R = rate as a decimal, and T = time in years.
I = (30)(0.033)(1)
I = 0.99
Then add that to the amount deposited ($30) and you're done.
30 + 0.99 = $30.99
Please let me know if you have questions.
The answer is $29.01
Greta bought a collar for her dog. The
original price was $12 but she had a
coupon for 10% off. How much money
did she save?
Answer:
She saved 1.20
Step-by-step explanation:
Purchase Price:
$12
Discount:
(12 x 10)/100 = $1.20
Final Price:
12 - 1.20 = $10.80
Find the value of X for which the following fraction is undefined
2x²+x-15
________
2/3x²-6
Answer: ±√2
Step-by-step explanation: A fraction is undefined when its denominator is =0 or undefined. so we need to get 2/3x²-6=0 or undefined. so we can also do 3x^2-6=0. Solving yields ±√2!
which dashed line is an asymptote for the graph?
Answer:
the graph has two vertical asymptotes, line q intersects the line at -8 and is the more important one.
Step-by-step explanation:
This is visible based off of the picture.
A circle is inscribed in an isosceles trapezoid with bases of 8 cm and 2 cm. Find the probability that a randomly selected point inside the trapezoid lies on the circle
Given that a circle is inscribed in an isosceles trapezoid with bases of 8 cm and 2 cm. We need to find the probability that a randomly selected point inside the trapezoid lies on the circle.
The isosceles trapezoid is shown below: [asy] draw((0,0)--(8,0)--(3,5)--(1,5)--cycle); draw((1,5)--(1,0)); draw((3,5)--(3,0)); draw((0,0)--(1,5)); draw((8,0)--(3,5)); draw(circle((2.88,2.38),2.38)); label("$8$",(4,0),S); label("$2$",(1.5,5),N); [/asy]Let ABCD be the isosceles trapezoid,
where AB = 8 cm, DC = 2 cm, and AD = BC.
Since the circle is inscribed in the trapezoid, we can use the following formula:2s = AB + DC = 8 + 2 = 10 cm
Where s is the semi-perimeter of the trapezoid. Also, let O be the center of the circle. We can draw lines OA, OB, OC, and OD as shown below: [asy] draw((0,0)--(8,0)--(3,5)--(1,5)--cycle); draw((1,5)--(1,0)); draw((3,5)--(3,0)); draw((0,0)--(1,5)); draw((8,0)--(3,5)); draw(circle((2.88,2.38),2.38)); label("$A$",(0,0),SW); label("$B$",(8,0),SE); label("$C$",(3,5),N); label("$D$",(1,5),N); label("$O$",(2.88,2.38),N); label("$8$",(4,0),S); label("$2$",(1.5,5),N); draw((0,0)--(2.88,2.38)--(8,0)--cycle); label("$s$",(3,0),S); label("$s$",(1.44,2.38),E); [/asy]Since O is the center of the circle, we have:OA = OB = OC = OD = rwhere r is the radius of the circle.
Also, we have:s = OA + OB + AB/2 + DC/2s = 2r + 2s/2s = r + 5 cmWe can solve for r:r + 5 cm = 10 cmr = 5 cmNow that we know the radius of the circle, we can find the area of the trapezoid and the area of the circle.
Then, we can find the probability that a randomly selected point inside the trapezoid lies on the circle as follows:Area of trapezoid = (AB + DC)/2 × height= (8 + 2)/2 × 5= 25 cm²Area of circle = πr²= π(5)²= 25π cm²Therefore, the probability that a randomly selected point inside the trapezoid lies on the circle is:
Area of circle/Area of trapezoid= 25π/25= π/1= π
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The probability that a randomly selected point inside the trapezoid lies on the circle is 0.399 or 39.9%. Therefore, option (A) is the correct answer.
The circle is inscribed in an isosceles trapezoid with bases of 8 cm and 2 cm.
Inscribed Circle of an Isosceles Trapezoid
Therefore, the length of the parallel sides (AB and CD) is equal.
Let the length of the parallel sides be ‘a’. Then, OB = OD = r (let)
It is also given that the lengths of the parallel sides of the trapezoid are 8 cm and 2 cm.
Then, its height is given by:
h = AB - CD / 2 = (8 - 2) / 2 = 3 cm
Therefore, the length of the base BC of the right-angled triangle is equal to ‘3’.
Then, the length of the other side (AC) can be given as:
AC = sqrt((AB - BC)² + h²) = sqrt((8 - 3)² + 3²) = sqrt(34) cm
The area of the trapezoid can be calculated as follows:
Area of the trapezoid = 1/2 (sum of the parallel sides) x (height)A = 1/2 (8 + 2) x 3A = 15 sq. cm.
The area of the circle can be given by:
Area of the circle = πr²πr² = A / 2πr² = 15 / (2 x π)
Therefore, r² = 2.39
r = sqrt(2.39) sq. cm.
Now, the probability that a randomly selected point inside the trapezoid lies on the circle can be calculated by dividing the area of the circle by the area of the trapezoid:
P (point inside the trapezoid lies on the circle) = Area of the circle / Area of the trapezoid
P = πr² / 15
P = π (2.39) / 15
P = 0.399 or 39.9%
The probability that a randomly selected point inside the trapezoid lies on the circle is 0.399 or 39.9%.
Therefore, option (A) is the correct answer.
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QUICK! Giving brainliest to correct answer
Answer:
Dominos is the better deal.
31 PIONTS GIVING BRAINIEST AWNSER Any tips on how to get a grade up ???
Answer:
Forgot picture?
Step-by-step explanation:
Answer:
You can get your grade up by studying, getting a tutor, paying attention in class, taking good notes, asking questions, and cheating (i don't recommend this one :/)
How do you turn 5/2 into 10/4?
Answer:
YOU DO IT X 2
Step-by-step explanation:
A seventh-grade class raised $380 during a candy sale. They deposited the money in a savings account for 6 months. If the bank pays 5.3% simple interest per year, how much money will be in the account after 6 months?
Answer: You want to calculate the interest on $380 at 5.3% interest per year after .5 year(s).
The formula we'll use for this is the simple interest formula, or:
Where:
P is the principal amount, $380.00.
r is the interest rate, 5.3% per year, or in decimal form, 5.3/100=0.053.
t is the time involved, 0.5....year(s) time periods.
So, t is 0.5....year time periods.
To find the simple interest, we multiply 380 × 0.053 × 0.5 to get your answer.
Step-by-step explanation:
0 Let x₁ = and x3 = B x2 = Write H Span{x1, x2, X3}. = - Use the Gram-Schmidt process to find an orthogonal basis for H. You do not need to normalize your vectors, but give exact answers. S 100.0000 V3
Main answer: An orthogonal basis for the given span H is {x1, x2-x1, x3 - (x1 + x2 - x1)} which simplifies to {x1, x2-x1, x3-x2}.
Supporting explanation: Given, x₁ = 0, x₂ = 1, x₃ = √3The span of H is the set of all linear combinations of x1, x2 and x3.So, we have to find an orthogonal basis for H using the Gram-Schmidt process. Let's start with the first vector x1 = [0, 0, 0]The second vector x2 is the projection of x2 onto the subspace perpendicular to x1. x2 is already perpendicular to x1 so x2-x1 = x2. So, the second vector is x2 = [0, 1, 0].The third vector x3 is the projection of x3 onto the subspace perpendicular to x1 and x2. x3 is not perpendicular to x1 and x2, so we subtract the projections of x3 onto x1 and x2 from x3. Projection of x3 onto x1:projx₁(x₃) = x₁ [(x₁ . x₃)/(x₁ . x₁)] = [0, 0, 0]Projection of x3 onto x2:projx₂(x₃) = x₂ [(x₂ . x₃)/(x₂ . x₂)] = [0, √3/3, 0]Therefore, x3 - projx₁(x₃) - projx₂(x₃) = [0, √3/3, √3]So, the orthogonal basis for H is {x1, x2-x1, x3 - (x1 + x2 - x1)} which simplifies to {x1, x2-x1, x3-x2}.
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