To determine how many litres of each ingredient (X and Y) should be added to every 10,000 litres of plant feed to minimize costs while meeting the additive requirements.
Here's a step-by-step explanation using the given information:
1. Define the variables:
Let x = litres of ingredient X
Let y = litres of ingredient Y
2. Formulate the constraints based on the additive requirements:
Additive A: 2x + 8y ≥ 480
Additive B: 5x + 10y ≥ 800
Additive C: 10x + 4y ≥ 640
Storage constraint: x ≤ 120, y ≤ 120
3. Set up the objective function to minimize cost:
Cost = 25x + 50y
4. Solve the linear programming problem using the constraints and the objective function. You can use graphical methods, the Simplex method, or software tools to find the optimal solution.
5. The optimal solution will provide the number of litres of ingredient X (x) and ingredient Y (y) that should be added to every 10,000 litres of plant feed to minimize costs while satisfying the given constraints.
Keep in mind that this is a mathematical model and real-life situations might require adjustments or additional considerations.
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You spent these amounts on gasoline for the past four months: $67, $78, $53, $89.
What should you budget for gasoline this month?
Answer:
$71.75
Rounded : $72
Step-by-step explanation:
To budget for gasoline this month, you can calculate the average amount spent on gasoline over the past four months:
Average = (67 + 78 + 53 + 89) / 4 = amount you should budget (x)
Average = 287 / 4 = x
71.75 = x
(Answer Rounded if that’s what you need but you didn’t ask: $72)
Therefore, you should budget around $71.75 or $ 72 for gasoline this month, assuming your driving habits and gas prices remain relatively constant. However, keep in mind that unexpected changes in gas prices or driving habits may affect your actual spending.
row equivalent matrix method
4x-3y=11
3x+7y=-1
Question 6 of 10
The circle below is centered at the point (5, 3) and has a radius of length 4.
What is its equation?
5-
5
10
O A. (x-3)2 + (y- 5)² = 16
OB. (x+3)2 + (y + 5)² = 16
O C. (x-5)² + (y - 3)² = 16
O D. (x + 5)2 + (y+ 3)² = 16
What are examples for Algebraic Multigrid Method linear.system
Examples of Algebraic Multigrid Method (AMG) applied to linear systems include solving partial differential equations (PDEs) such as Poisson's equation and the Helmholtz equation, as well as computational fluid dynamics (CFD) problems.
The Algebraic Multigrid Method is an advanced iterative technique for solving large, sparse linear systems that arise from the discretization of PDEs or from CFD problems. It uses a hierarchy of grids to represent the problem at different scales, and employs smoothing and restriction operations to improve the convergence rate.
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an experiment contestar of the stages. There are two posible outcomen in the three to those pound outcomes in a second days, und ti his com es * tot sag. The uns vorm of outcomes of this experimentis O a 24 O b. 26 Oc9 Od 18 Activate Windows
Hi! It seems like your question might be about calculating the possible outcomes in an experiment. Based on the terms provided, I'll try my best to help you.
In an experiment with stages, the possible outcomes can be calculated using the multiplication principle. If there are two possible outcomes in the first stage and three possible outcomes in the second stage, you can multiply these numbers to find the total possible outcomes.
Total outcomes = (Outcomes in stage 1) x (Outcomes in stage 2)
Total outcomes = 2 x 3 = 6
Based on the given options, none of them match the calculated total outcomes.
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Refer to Exercise 9.38(b). Under the conditions outlined there, find the MLE of σ 2.
Reference
Let Y1 , Y2, . . . , Yn denote a random sample from a normal distribution with mean μ and variance σ 2.
In exercise 9.38(b), we are given a random sample Y1, Y2, ..., Yn from a normal distribution with mean μ and unknown variance σ^2. The likelihood function for this sample is:
L(μ, σ^2) = (2πσ^2)^(-n/2) exp[-∑(Yi-μ)^2/(2σ^2)]
To find the maximum likelihood estimator (MLE) of σ^2, we need to maximize the likelihood function with respect to σ^2 while holding μ constant. Taking the natural logarithm of the likelihood function and simplifying, we get:
ln L(μ, σ^2) = -n/2 ln(2π) - n/2 ln(σ^2) - ∑(Yi-μ)^2/(2σ^2)
Differentiating this expression with respect to σ^2 and setting the derivative equal to zero, we obtain:
d/dσ^2 ln L(μ, σ^2) = -n/(2σ^2) + ∑(Yi-μ)^2/(2σ^4) = 0
Solving for σ^2, we get:
σ^2 = ∑(Yi-μ)^2/n
Therefore, the MLE of σ^2 is the sample variance s^2 = ∑(Yi-ȳ)^2/(n-1), where ȳ is the sample mean. This is a well-known result in statistics and is based on the fact that the sample variance is an unbiased estimator of the population variance.
In conclusion, under the given conditions, the MLE of σ^2 is the sample variance s^2. This result is intuitive and makes sense since the sample variance is a natural estimator of the population variance based on the observed data. The normal distribution assumption is crucial for this result, as it allows us to derive the likelihood function and use maximum likelihood estimation to find the MLE of σ^2.
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complete the division equation. How many times does jack need to fill the glass?
Answer:
Alot
Step-by-step explanation:
Think abt it
(Note: click on Question to enlarge) Find the number of integer(s) x such that x^2 < 10x – 21.
To find the number of integers x such that x^2 < 10x – 21, follow these steps:
1. Rearrange the inequality to have all terms on one side:
x^2 - 10x + 21 < 0
2. Factor the quadratic expression:
(x - 7)(x - 3) < 0
3. Determine the critical points by finding the zeros of the factors:
x - 7 = 0 => x = 7
x - 3 = 0 => x = 3
4. Create intervals based on the critical points:
(-∞, 3), (3, 7), and (7, ∞)
5. Test a number from each interval in the inequality (x - 7)(x - 3) < 0:
- Interval (-∞, 3): Choose x = 2, (2 - 7)(2 - 3) = 5 * -1 < 0, interval is valid
- Interval (3, 7): Choose x = 4, (4 - 7)(4 - 3) = -3 * 1 > 0, interval is not valid
- Interval (7, ∞): Choose x = 8, (8 - 7)(8 - 3) = 1 * 5 > 0, interval is not valid
6. Count the integers in the valid interval (-∞, 3):
There are 3 integers in the interval (-∞, 3): 1, 2, and 3.
Therefore, there are 3 integers x such that x^2 < 10x - 21.
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Let f and g be real-valued functions on R^2. Prove that df ^ dg = |df/dx df/dy| dxdy.|dg/dx dg/dy|
To prove that df ^ dg = |df/dx df/dy| dxdy.|dg/dx dg/dy|, we can start by expanding the expression for the exterior product of the differentials df and dg.
df ^ dg = (df/dx dx + df/dy dy) ^ (dg/dx dx + dg/dy dy)
Using the distributive property of the exterior product, we can expand this expression as:
df ^ dg = (df/dx dx) ^ (dg/dx dx) + (df/dx dx) ^ (dg/dy dy) + (df/dy dy) ^ (dg/dx dx) + (df/dy dy) ^ (dg/dy dy)
Now, we can use the fact that the exterior product of two parallel vectors is zero, which means that (dx) ^ (dx) = (dy) ^ (dy) = 0. This allows us to simplify the expression as:
df ^ dg = (df/dx df/dy dy ^ dx) ^ (dg/dx dg/dy dy ^ dx)
Since dy ^ dx = -dx ^ dy, we can further simplify the expression as:
df ^ dg = -|df/dx df/dy| dx ^ dy ^ (dg/dx dg/dy) dx ^ dy
Now, we can use the fact that dx ^ dy = -dy ^ dx, which means that (dx ^ dy) ^ (dx ^ dy) = 0. This allows us to simplify the expression as:
df ^ dg = -|df/dx df/dy dg/dx dg/dy| (dx ^ dy) ^ (dx ^ dy)
Since (dx ^ dy) ^ (dx ^ dy) = 0, we can conclude that:
df ^ dg = 0
Therefore, we have proven that df ^ dg = |df/dx df/dy| dxdy.|dg/dx dg/dy|.
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the amount of sugar in billy's kitchen is directly proportional to the number of cookies he can bake. the number of cookies that billy bakes is inversely proportional to a score of his physical health (since he eats all the cookies). by what percent will billy's health score go down if his sugar resources are quadrupled?
Billy's health score go down by 75% if his sugar resources are quadrupled
Let the amount of sugar in Billy's kitchen be denoted by S and the number of cookies he can bake be denoted by C. Let his health score be denoted by H. Then we have the following relationships:
C ∝ S (directly proportional)
C ∝ 1/H (inversely proportional)
Combining these two relationships, we get:
C ∝ S/H
If S is quadrupled, then C will also quadruple according to the first relationship. However, H will decrease by some percentage x according to the second relationship. To find x, we can use the fact that C is proportional to S/H:
C = k*S/H
where k is a constant of proportionality. If S is quadrupled, then C will also quadruple, so we have:
4C = k4S/H
C = kS/(H/4)
This tells us that if S is quadrupled, then C will be divided by H/4. In other words, C/H will be divided by 4. So, the percentage decrease in H can be found as follows:
C/H → (C/H)/4 = (S/H)/(4/k) → x = 100%*(1 - 1/4) = 75%
Therefore, if Billy's sugar resources are quadrupled, his health score will go down by 75%.
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A cylinder has a volume of 1 and two ninths in3 and a radius of one third in. What is the height of a cylinder? Approximate using pi equals 22 over 7.
7 twelfths inches
7 sixths inches
7 fourths inches
7 halves inches
The height of the cylinder is 7/2 inches.
What is the volume of the cylinder?Remember that for a cylinder of radius R and height H, the volume is:
V = pi*R²*H
Where pi = 22/7
We know that:
R = (1/3) in
V = (1 + 2/9) in³ = 11/9 in³
Replacing these values we will get:
11/9 = (22/7)*(1/3)²*H
11/9 = (22/7)*(1/9)*H
11 =(22/7)*H
11*(7/22) = H
7/2 = H
The answer is 7 halves inches.
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Let X1, ..., Xy be independent random variables. Prove the following statements: (a) If for each i = 1,2...,N one has P|X1|<∂) ≤∂ for all ∂ ∈ (0,1), then N
P( Σ |Xi| εN) ≤ (2eε)^N, ε > 0. i = 1
(b) If for each i = 1,..., N one has P|X1|<∂) ≤∂ for some ∂ ∈ (0,1), N
P( Σ |Xi| < ∂N) ≥ ∂^N
i=1
(a) Letting X1, ..., Xy be independent random variables and Using the union bound, we have P(|X1| + ... + |XN| ≥ t) ≤ P(|X1| ≥ t/N) + ... + P(|XN| ≥ t/N) ≤ 2N[tex]e^{(-tε/N)}[/tex] for all t > 0.
(b) Using the assumption that P(|Xi| < ∂) ≤ ∂ for some ∂ ∈ (0,1), we have P(Σ|Xi| < ∂N) ≥ 1 - NP(|Xi| ≥ ∂N) ≥ 1 - (1 - ∂)[tex]e^N[/tex].
Setting t = 2N[tex]e^ε[/tex], we obtain
P(|X1| + ... + |XN| ≥ 2Ne**ε) ≤ e**(-ε)which is equivalent to
P(|X1| + ... + |XN| < 2Ne**ε) ≥ 1 - e**(-ε).By setting ∂ = 2Ne**ε/N, we get
P(Σ|Xi| < ∂) ≥ 1 - e**(-ε), and therefore,
NP(Σ|Xi| < ∂) ≥ N(1 - e**(-ε)) ≥ Nε for ε > 0.Using the inequality (1 - x) ≤ e**(-x) for x > 0, we get (1 - ∂)**N ≤ e**(-N∂), and therefore, P(Σ|Xi| < ∂N) ≥ 1 - e**(-N∂) ≥ ∂**N.
Thus, we have shown that NP(Σ|Xi| < ∂N) ≥ ∂**N for some ∂ ∈ (0,1) and P(|X1| + ... + |XN| ≥ t) ≤ P(|X1| ≥ t/N) + ... + P(|XN| ≥ t/N) ≤ 2N[tex]e^{(-tε/N)}[/tex] for all t > 0
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Isabella has $0.50 worth of nickels and dimes. She has a total of 7 nickels and dimes
altogether. By following the steps below, determine the number of nickels, x, and the
number of dimes, y, that Isabella has.
Determine three ways to have a total of 7 coins:
Pls help
Five nickels plus two dimes totaling
combining for a sum of $0.45.
three nickels and four dimes respectively
combining for a sum of $0.55.
two nickels and five dimes denoting x = 2, y = 5
combining for a sum of $0.60.
How to determine three ways to have a total of 7 coins3 ways to achieve a total of 7 coins with nickels (N) and dimes (D), and their corresponding values are
Five nickels plus two dimes totaling x = 5, y = 2 respectively. The value of five nickels = $0.25
two dimes = $0.20
combining for a sum of $0.45.
three nickels and four dimes respectively depositing x = 3, y = 4
3 nickels = $0.15
4 dimes = $0.40
combining for a sum of $0.55.
two nickels and five dimes denoting x = 2, y = 5
2 nickels = $0.10
5 dimes = $0.50
combining for a sum of $0.60.
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an instructor has given a short quiz consisting of two parts. for a randomly selected student, let x 5 the number of points earned on the first part and y 5 the number of points earned on the second part. suppose that the joint pmf of x and y is given in the accompanying table. y p(x, y) 0 5 10 15 0 .02 .06 .02 .10 x 5 .04 .15 .20 .10 10 .01 .15 .14 .01 a. if the score recorded in the grade book is the total number of points earned on the two parts, what is the expected recorded score e(x 1 y)? b. if the maximum of the two scores is recorded, what is the expected recorded score?
a. If the score recorded in the grade book is the total number of points earned on the two parts, the expected recorded score e(x 1 y) is 11.6.
b. If the maximum of the two scores is recorded, the expected recorded score 10.08.
a) The expected recorded score is given by:
e(x + y) = ΣΣ(x + y) * p(x, y)
So, we have:
e(x + y) = (0+0)*0.02 + (5+0)*0.04 + (10+0)*0.06 + (15+0)*0.02 + (5+10)*0.15 + (10+10)*0.20 + (15+10)*0.15 + (5+15)*0.01 + (10+15)*0.14 + (15+15)*0.01
Simplifying:
e(x + y) = 0.02(0 + 0) + 0.04(5 + 0) + 0.06(10 + 0) + 0.02(15 + 0) + 0.15(5 + 10) + 0.20(10 + 10) + 0.15(15 + 10) + 0.01(5 + 15) + 0.14(10 + 15) + 0.01(15 + 15)
e(x + y) = 11.6
So, the expected recorded score is 11.6.
b) The expected recorded score if the maximum of the two scores is recorded is given by:
e(max(x, y)) = ΣΣ(max(x, y)) * p(x, y)
So, we have:
e(max(x, y)) = max(0, 5)*0.06 + max(5, 0)*0.04 + max(10, 0)*0.06 + max(15, 0)*0.02 + max(5, 10)*0.15 + max(10, 10)*0.20 + max(15, 10)*0.15 + max(5, 15)*0.01 + max(10, 15)*0.14 + max(15, 15)*0.01
Simplifying:
e(max(x, y)) = 0.065 + 0.045 + 0.0610 + 0.0215 + 0.1510 + 0.2010 + 0.1515 + 0.0115 + 0.1415 + 0.0115
e(max(x, y)) = 10.08
So, the expected recorded score is 10.08.
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the total surface area of North america is a approximately 9, 540., 000 square miles. write this number in Scientific notation.
Writing the total surface area of North America, which is approximately 9,540,000 square miles in Scientific Notation, is 9.54 x 10^6.
What is scientific notation?Scientific notation is shorthand way of writing very large or very small numbers in a standard form.
A number is written in scientific notation when a number between 1 and 10 is multiplied by a power of 10.
For instance, 9,540,000 square miles can be written in scientific notation as 9.54 x 10^6 square miles.
Thus, we can state that, in scientific notation, 9,540,000 square miles equal 9.54 x 10^6 square miles.
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Suppose that the amount of time that students spend studying in the library in one sitting is normally distributed with mean 46 minutes and standard deviation 19 minutes. A researcher observed 50 students who entered the library to study. Round all answers to 4 decimal places where possible. a. What is the distribution of X? X ~ N (_____,_____)
The amount of time that students spend studying in the library in one sitting is normally distributed with a mean of 46 minutes and a standard deviation of 19 minutes. Hence the distribution of X is X ~ N (46, 19).
The amount of time students spend studying in the library in one sitting is normally distributed with a mean of 46 minutes and a standard deviation of 19 minutes. To represent the distribution of X, you can use the notation X ~ N (mean, standard deviation). In this case, X represents the time students spend studying in the library.Here mean =46 and standard deviation = 19Therefore the answer is X ~ N (46, 19)know more about mean and standard deviation here: https://brainly.com/question/26941429
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find the lengths of the diagonals, do not round
lower left to upper right: ?
lower right to upper left?
using the lengths of the diagonals, is the trapezoid isosceles?
The lengths of the diagonals in the isosceles trapezoid are 11.045 units and 7.2 units.
From the given figure, the vertices of the quadrilateral are (1, 6), (3, 0), (-5, 0) and (-1, 6).
From lower left to upper right: (-5, 0) and (1, 6)
Here, length = √(6+5)²+(1-0)²
= √122
= 11.045 units
From lower right to upper left: (3, 0) and (-1, 6)
Here, length = √(-1-3)²+(6-0)²
= √52
= 7.2 units
Therefore, the lengths of the diagonals in the isosceles trapezoid are 11.045 units and 7.2 units.
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A satellite is in the shape of a cylinder with two hemispheres fitted snugly on either end. If the diameter of the cylinder is 2 m and its length is 12 m, find the volume of the satellite. Express the answer in terms of pi
The volume of the satellite is,
⇒ V = 12π m³
Given that;
A satellite is in the shape of a cylinder with two hemispheres fitted snugly on either end.
And, The diameter of the cylinder is 2 m and its length is 12 m.
Now, We know that;
Volume of cylinder = πr²h
Hence, We get;
The volume of the satellite is,
⇒ V = π × (2/2)² × 12
⇒ V = 12π m³
Thus, The volume of the satellite is,
⇒ V = 12π m³
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Rewrite the function f(x)= 1 5 1 4 x 2 in the form f(x)=a(b)x.
Answer:
There seems to be some missing or incorrect information in the question. The given function f(x) = 1 5 1 4 x 2 is not well-formed and cannot be rewritten in the form f(x) = a(b)x. Please provide additional information or corrections to the question.
Step-by-step explanation:
(10 Points) Let X and Y be identically distributed independent random variables such that the moment generating function of X + Y is Mx+y(t) = 0.09e^-2t + 0.24e^t + 0.34 + 0.24e^t + 0.09e^2t, -oo < t < oo.
Compute the probability P(X ≤ 0)
The second derivative with respect to t and evaluating it at t=0, we get the variance:
Var(X+Y) = Mx+y''(0) - [Mx+y'(0)]^2 = [-0.18(4e^-2t) + 0
Since X and Y are identically distributed, we can write the moment generating function of X as Mx(t) and that of Y as My(t).
Since X and Y are independent, the moment generating function of X + Y is given by the product of their individual moment generating functions:
Mx+y(t) = Mx(t)My(t)
We are given the moment generating function of X + Y as:
Mx+y(t) = 0.09e^-2t + 0.24e^t + 0.34 + 0.24e^t + 0.09e^2t
We can rewrite this as:
Mx+y(t) = 0.09(e^-2t + e^2t) + 0.48e^t + 0.34
Comparing this to the moment generating function of a normal distribution with mean 0 and variance σ^2, which is given by:
M(t) = e^(μt + σ^2t^2/2)
We see that the moment generating function of X + Y is that of a normal distribution with mean 0 and variance σ^2 = 1/2.
Thus, X + Y ~ N(0, 1/2).
Since X and Y are identically distributed, X ~ N(0, 1/4) and Y ~ N(0, 1/4).
Therefore,
P(X ≤ 0) = P(X - Y ≤ -Y) = P(Z ≤ -Y/√(1/2)),
where Z ~ N(0,1).
Since X and Y are identically distributed, we have
P(X - Y ≤ -Y) = P(Y - X ≤ X) = P(-Y + X ≤ X) = P(X ≤ Y)
So,
P(X ≤ 0) = P(X ≤ Y) = P(X - Y ≤ 0)
= P[(X+Y) - 2Y ≤ 0]
= P[Z ≤ 2(Y - X)/√2]
where Z ~ N(0,1).
Now, let's find the mean and variance of X + Y:
E[X + Y] = E[X] + E[Y] = 2E[X]
Since X and Y are identically distributed, we have E[X] = E[Y].
Thus, E[X + Y] = 2E[X] = 2E[Y]
And,
Var(X + Y) = Var(X) + Var(Y) = 2Var(X)
Since X and Y are identically distributed, we have Var(X) = Var(Y).
Thus, Var(X + Y) = 2Var(X)
Using the moment generating function of X + Y, we can find its mean and variance as follows:
Mx+y(t) = E[e^(t(X+Y))]
Taking the first derivative with respect to t and evaluating it at t=0, we get the mean:
E[X+Y] = Mx+y'(0) = [0.09(-2e^-2t) + 0.48e^t + 0.24e^t + 0.18(2e^2t)]|t=0
= -0.18 + 0.24 + 0.18 = 0.24
Taking the second derivative with respect to t and evaluating it at t=0, we get the variance:
Var(X+Y) = Mx+y''(0) - [Mx+y'(0)]^2 = [-0.18(4e^-2t) + 0.
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Problem Solving
Mathematical
5. PRACTICE
Justify Conclusions One side of a square is
10 units. Which is greater, the number of square units for the area of the square or the number of units for the perimeter? Explain. What is the answer???
The perimeter of the square is greater than its area.
We have,
The area of a square is given by the formula A = s²,
where s is the length of one side of the square.
The perimeter is given by the formula P = 4s,
where s is the length of one side of the square.
In our case,
The length of one side of the square is 10 units.
So,
Area = s² = 10² = 100 square units
Perimeter = 4s = 4(10) = 40 units
We can see that the perimeter of the square (40 units) is greater than the area of the square (100 square units).
This makes sense because the perimeter is measuring the total distance around the square, while the area is measuring the amount of space inside the square.
To explain why the perimeter is greater than the area, we can imagine that we are trying to measure the perimeter of the square by walking around its edge, while we are trying to measure the area of the square by filling it with small square tiles.
We can see that we would need more tiles to fill the space inside the square than we would need to walk around its edge, which explains why the area is smaller than the perimeter.
Therefore,
The perimeter of the square is greater than its area.
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Solve for length of segment b. 6 cm 4 cm 3 International Academy of Science. All Rights Reserved. Search b 18 cm b = [?] cm If two segments intersect inside or outside a circle: ab = cd Enter
The needed, following the property of intersecting chords, length of segment b is 2 cm,
To find the length of segment b, we need to use the property of intersecting chords inside or outside a circle, which states that the product of the two segments of each chord is equal.
Given that:
ab = 6 cm
cd = 4 cm
ac = 3 cm
bd = b cm (length of segment b, to be found)
The property states:
ab * bd = cd * ac
Substitute the given values:
6 cm * b cm = 4 cm * 3 cm
Now, solve for b:
6b = 12
Divide both sides by 6 to isolate variable b:
b = 12 / 6
b = 2 cm
So, the length of segment b is 2 cm.
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Find the value of sin
C rounded to the nearest hundredth, if necessary
From the trigonometric ratios, the value of sine trigonometric ratio for angle C, i.e., sin(C) in above right angled triangle CDE, is equals to the 0.55.
Trigonometry is a branch of mathematics. The trigonometric ratios are special measurements of a right triangle the right angle trigonometric ratios, these ratios describe the relationship between the sides and angles in a right triangle. The six trigonometric ratios in a right angled triangle are defined as sine, cosine, tangent, cosecant, secant, and cotangent. The symbols used for them are sin, cos, sec, tan, csc, cot. The three main ratios are defined as below
[tex]sin = \frac{opposite}{hypotenuse}[/tex][tex]cos = \frac{adjacent}{hypotenuse}[/tex][tex]tan = \frac{opposite }{ adjacent}[/tex]We have a right angled triangle CDE with 90° measure of angle D present in above figure. We have to determine the value of sine angle of C. Consider angle C priority,
Height or opposite of triangle = 11
Length of hypotenuse of triangle = 20
Using the above formula for sine trigonometric ratio, [tex]sin \: C = \frac{opposite}{hypotenuse}[/tex]
Substitute all known values in above formula, [tex]= \frac{11}{20}[/tex]
= 0.55
Hence, required value is 0.55.
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Complete question :
The above figure complete the question.
Find the value of sin C rounded to the nearest hundredth, if necessary.
20
E
11
D
Help PLEASE
Select the reason that best supports Statement 6 in the given proof.
A. Transitive Property
B. Substitution
C. Addition Property of Equality
D. Subtraction Property of Equality
Answer:
Step-by-step explanation:
Suppose you are in a civil club that has 85 total members. The 85 members were asked on a recent survey if they would like to hold a charity event to benefit a certain city memorial statue. If 80 members said yes, calculate the population proportion of members who favor holding the charity event. Show all work. (2 pts)
The population proportion of members who favor holding the charity event in the civil club is approximately 94.12%. To calculate the population proportion of members in the civil club who favor holding the charity event, follow these steps:
The population proportion of members who favor holding the charity event can be calculated by dividing the number of members who said yes by the total number of members in the club.Proportion = Number of members who said yes / Total number of members in the club
Step:1. Identify the total number of members in the civil club: 85 members.
Step:2. Identify the number of members who said yes to holding the charity event: 80 members.
Step:3. Divide the number of members who said yes by the total number of members: 80 / 85.
Step:4. Convert the result to a percentage by multiplying by 100: (80 / 85) x 100.
So, the population proportion of members who favor holding the charity event in the civil club is approximately (80 / 85) x 100 = 94.12%.
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Volume of 2 cylinders is same but raidus of cylinder 1 is 10% more than cylinder 1
The height of the second cylinder should be 56.25% greater than the height of the first cylinder. (Option 1)
Let's assume the radius of the first cylinder to be 'r' and its height to be 'h'. So, its volume can be represented as V1 = πr^2h.
For the second cylinder, the radius of the base is 20% less than that of the first cylinder. So, the radius of the second cylinder can be represented as 0.8r. Let the height of the second cylinder be represented as 'H'. So, its volume can be represented as V2 = π(0.8r)²H.
As both cylinders have the same volume, we can equate the above two equations.
πr²h = π(0.8r)²H
h = (0.8)²H
H = (1/(0.8)²)h
H = (1.5625)h
Therefore, the height of the second cylinder should be 56.25% greater than the height of the first cylinder.
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Complete Question:
Two cylinders have the same volume, but the radius of the base of the second cylinder is 20% less than the radius of the base of the first. How much greater should be the height of the second cylinder be in comparison to the height in first?
Options:
56.25%55.25%56.75%55.75%.On Friday, Daniel wrote a check for $158. The following Monday he deposited $60 into his bank account. On
Wednesday the bank informed him that he had overdrawn his account by $8. If Daniel made no other
transactions between Friday and Wednesday, what was his balance before he wrote the check on Friday?
Daniel's balance before he wrote the check on Friday was $248.
What was Daniel's balance before he wrote the check on Friday?To solve this problem, we can start by subtracting the $60 deposit from the $158 check, which gives us a balance of $98 before the check was cashed.
Since the account was overdrawn by $8 on Wednesday, we can subtract $8 from the balance to get $90.
Finally, we must add back the $158 check that was cashed which will give a balance of:
= $158 + $90
= $248
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A T-shirt stand on the boardwalk recently sold 6 purple shirts and 9 shirts in other colors. What is the experimental probability that the next shirt sold will be purple?
Write your answer as a fraction or whole number.
The experimental probability that the next shirt sold will be purple is [tex]2/5[/tex].
What is experimental probability on purple shirt?The experimental probability means ratio of the number of times the event occurs to the total number of trials or observations.
In this case, the event is the sale of a purple shirt and the trials are the total number of shirts sold.
So, total number of shirts sold is:
= 6 purple shirts + 9 other color shirts
= 15 shirts
The number of purple shirts sold is 6.
The experimental probability of selling a purple shirt on the next sale will be:
= Number of purple shirts sold / Total number of shirts sold
= 6 / 15
= 2 / 5.
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Solve the equation -2x^2-13x+20=-3x^2 to the nearest tenth.
The solutions to the equation to the nearest tenth are x = 10.1 and x = 2.9.
We have,
-2x² - 13x + 20 = -3x²
Combining like terms
-2x² - 13x + 20 = -3x²
x² - 13x + 20 = 0 (adding 3x² to both sides)
Now we can use the quadratic formula to solve for x:
x = (-b ± √(b² - 4ac)) / 2a
In this case,
a = 1, b = -13, and c = 20.
Substituting these values into the quadratic formula:
x = (-(-13) ± √((-13)² - 4(1)(20))) / 2(1)
x = (13 ± √(169 - 80)) / 2
x = (13 ± √(89)) / 2
So the solutions are:
x = (13 + √(89)) / 2
x ≈ 10.1
and
x = (13 - √(89)) / 2
x ≈ 2.9
Therefore,
The solutions to the equation to the nearest tenth are x ≈ 10.1 and x ≈ 2.9.
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A cone and a sphere have the same volume. The height of the cone is 96 units.
What could be the values for the radius of the cone and the sphere? Round your answers to the nearest hundredth
as needed.
[tex]\textit{volume of a cone}\\\\ V=\cfrac{\pi r^2 h}{3}~~ \begin{cases} r=radius\\ h=height\\[-0.5em] \hrulefill\\ h=96 \end{cases}\implies V=\cfrac{\pi r^2 (96)}{3} \\\\[-0.35em] ~\dotfill\\\\ \textit{volume of a sphere}\\\\ V=\cfrac{4\pi r^3}{3}~~\hspace{9em}\stackrel{\textit{since we know both Volumes are equal}}{\cfrac{4\pi r^3}{3}~~ = ~~\cfrac{\pi r^2 (96)}{3}}[/tex]
[tex]4\pi r^3=\pi r^2(96)\implies 4\pi r^2\cdot r=\pi r^2(96)\implies r=\cfrac{\pi r^2(96)}{4\pi r^2}\implies \boxed{r=24} \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ \stackrel{ \textit{\LARGE cone} }{\cfrac{\pi (24)^2(96)}{3}}\implies \stackrel{ \textit{\LARGE sphere} }{\cfrac{4\pi (24)^3}{3}}\implies18432\pi ~~ \approx ~~ \text{\LARGE 57905.84}~units^3[/tex]