The ratio of bananas to strawberries is 11:3.
What is ratio?A ratio is a comparison between two amounts that is calculated by dividing one amount by the other. The quotient a/b is referred to as the ratio between a and b if a and b are two values of the same kind and with the same units, such that b is not equal to 0. Ratios are represented by the colon symbol (:). As a result, the ratio a/b has no units and is represented by the notation a: b.
Given:
Number of bananas = 11 ounces,
Number of strawberries = 3 ounces
The ratio of bananas to strawberries
= 11/3
= 11:3.
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An ordinary (fair) die is a cube with the numbers 1 through 6 on the sides (represented by painted spots). Imagine that such a die is rolled twice in succession and that the face values of the two rolls are added together. This sum is recorded as the outcome of a single trial of a random experiment.
Compute the probability of each of the following events:
Event A: The sum is greater than 6. Event B: The sum is divisible by 3 or 5 (or both).
Write your answers as exact fractions.
P(a)
P(b)
The final answers are:
a) P(A) = 7/12
b) P(B) = 11/36
To compute the probabilities of Events A and B, we can use a table to list all possible outcomes and their corresponding sums:
Die 1 Die 2 Sum
1 1 2
1 2 3
1 3 4
1 4 5
1 5 6
1 6 7
2 1 3
2 2 4
2 3 5
2 4 6
2 5 7
2 6 8
3 1 4
3 2 5
3 3 6
3 4 7
3 5 8
3 6 9
4 1 5
4 2 6
4 3 7
4 4 8
4 5 9
4 6 10
5 1 6
5 2 7
5 3 8
5 4 9
5 5 10
5 6 11
6 1 7
6 2 8
6 3 9
6 4 10
6 5 11
6 6 12
a) Event A: The sum is greater than 6. From the table, we can see that there are 21 outcomes where the sum is greater than 6 (highlighted in bold). Thus, the probability of Event A is:
P(A) = 21/36 = 7/12
b) Event B: The sum is divisible by 3 or 5 (or both). From the table, we can see that the sums that are divisible by 3 or 5 (or both) are:
3, 5, 6, 9, 10, 12
There are 11 outcomes with these sums (highlighted in bold). Thus, the probability of Event B is:
P(B) = 11/36
Therefore, the final answers are:
a) P(A) = 7/12
b) P(B) = 11/36
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1.28 Ф Vector operations and units. (Chapter 2) Circle the vector operations below (scalar multiplication, addition, dot-product, etc.) that are defined for a position vector a (with units of m) and a velocity vector b (with units of P. a+b 1.29 Using vector identities to simplify expressions (refer to Homework 1.13). One reason to treat vectors as basis-independent quantities is to simplify vector expressions with- out resolving the vectors into orthogonal x, y, z or i k components. Simplify the following Vector expression (3u-29) x (u + v) vector expressions using various properties of dot-products and cross-products Simplified vector expression Express your results in terms of dot-products - and cross- products x of the arbitrary vectors υ. v. w (ie.. υ. v, w are not orthogonal). 2 1.30 Vector concepts: Solving a vector equation. (Section 2.10.5) Consider the vector equation to the right and the process that follows that solves for θ (a, is a unit vector and v,, θ, R are scalars). This process is a valid way to solve for θ. True/False Explain: vr ax 0
The vector operations defined for a position vector a and a velocity vector b are vector addition and scalar multiplication.
Simplified Vector expression is (3u - 29) x u + (3u - 29) x v.
False, as the process that follows, solving for θ by dividing by the magnitude of v and taking the inverse sine, is only valid if v x a = 0.
1.28: The vector operations defined for a position vector a and a velocity vector b are vector addition and scalar multiplication.
1.29: Using the properties of the cross-product, the simplified expression is:
3u x u + 3u x v - 29 x u - 29 x v
Using the fact that the cross-product is distributive over vector addition, we can write this as:
3u x u + 3u x v - 29 x u - 29 x v = (3u - 29) x u + (3u - 29) x v
So the simplified vector expression is (3u - 29) x u + (3u - 29) x v.
1.30: False.
The given equation,
v x (a x R) = 0,
can be simplified to a scalar triple product,
(v x a) · R = 0.
Since the scalar triple product is equal to the determinant of a matrix formed by the three vectors,
this equation implies that either v x a = 0 or R = 0.
Therefore, the process that follows, solving for θ by dividing by the magnitude of v and taking the inverse sine, is only valid if v x a = 0.
The vector operations defined for a position vector a and a velocity vector b are vector addition and scalar multiplication.
Simplified Vector expression is (3u - 29) x u + (3u - 29) x v.
False as the process that follows, solving for θ by dividing by the magnitude of v and taking the inverse sine, is only valid if v x a = 0.
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if we want to measure the magnitude of the orbital angular momentum and the projection of the angular momentum in
To measure the magnitude of the orbital angular momentum L and the projection of the angular momentum along a particular axis (say, the z-axis), we need to use the mathematical formalism of quantum mechanics.
In quantum mechanics, the orbital angular momentum L of a particle is an operator that acts on the wave function describing the particle's motion. Similarly, the z-component of the angular momentum Lz is also an operator.
The magnitude of the orbital angular momentum L can be calculated from the components of the angular momentum operator using the expression:
L^2 = L₁^2 + L₂^2 + L₃^2
where L₁, L₂, and L₃ are the x, y, and z components of the angular momentum operator, respectively. To measure the magnitude of the orbital angular momentum, we would need to measure the values of L₁ L₂, and L₃ and use them to calculate L^2.
The projection of the angular momentum along a particular axis (say, the z-axis) is given by the operator L₃. To measure the z-component of the angular momentum, we would need to measure the value of L₃ for a particular state of the system.
In practice, these measurements are often carried out using experiments involving the interaction of particles with magnetic fields. The behavior of the particles in the magnetic field allows us to infer information about the angular momentum of the particles.
The measurement of the magnitude and projection of angular momentum is an important part of many areas of physics, including quantum mechanics and solid-state physics.
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The distance traveled in the time interval 0 ≤ t ≤ 6, given the velocity function v(t) = t^2 - 6t - 16, can be determined by calculating the definite integral of the absolute value of the velocity function over the given time interval. The result is 128 units of distance.
To find the distance traveled in the given time interval, we need to integrate the absolute value of the velocity function v(t) = t^2 - 6t - 16 over the interval 0 ≤ t ≤ 6. The reason for taking the absolute value is that distance is a scalar quantity and does not depend on the direction of motion.
Taking the integral of the absolute value of the velocity function, we have:
∫|v(t)| dt = ∫|t^2 - 6t - 16| dt
To evaluate this integral, we need to split it into intervals where the velocity function is positive and negative. The absolute value function essentially removes the negative sign from the expression inside the absolute value brackets.
Next, we find the points where the velocity function changes sign by setting v(t) = 0:
t^2 - 6t - 16 = 0
Solving this quadratic equation, we find t = -2 and t = 8 as the points where the velocity changes sign.
Now, we evaluate the integral over the intervals [0, 2] and [2, 6] separately, considering the absolute value of the velocity function within each interval.
∫|v(t)| dt = ∫(t^2 - 6t - 16) dt over [0, 2] + ∫(-(t^2 - 6t - 16)) dt over [2, 6]
Evaluating the definite integrals, we obtain:
(128/3) + (128/3) = 256/3
Therefore, the distance traveled in the time interval 0 ≤ t ≤ 6 is 256/3 units of distance, which is approximately 85.33 units of distance.
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( 7 x + 5 ) ( 2 x 3 − 4 x 2 + 9 x − 3 ) A. 14 x 4 − 18 x 3 + 43 x 2 + 24 x − 15 B. 9 x 4 − 11 x 3 + 16 x 2 + 11 x − 15 C. 9 x 4 − 11 x 3 + 16 x 2 + 11 x − 8 D.
Expanding the polynomial, (7x + 5)(2x³ − 4x² + 9x − 3), we have: A. 14x^4 - 18x³ + 43x² + 24x - 15.
How to Expand Expressions?By applying the distribution property, we can expand a given polynomial expression like the one given above.
Given, (7x + 5)(2x³ − 4x² + 9x − 3), distribute to eliminate the parentheses:
7x(2x³ − 4x² + 9x − 3) + 5(2x³ − 4x² + 9x − 3)
14x^4 - 28x³ + 63x² - 21x + 10x³ - 20x² + 45x - 15
Combine like terms:
14x^4 - 18x³ + 43x² + 24x - 15
The correct solution is option A.
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PLEASE HELP WILL GIVE 100 PTS AND BRAINLYEST
A sequence of transformations is applied to a polygon. [ The following statements represent a sequence of transformations where the resulting polygon is similar to the original polygon but have a smaller area than the original polygon is that: a dilation about the origin by a scale factor of 2 3 followed by a rotation of 90° counterclockwise about the origin and a translation 5 units left.
In each of the following systems, find conditions on a, b, and c for which the system has solutions: (a)3x + 2y - z=a x + y + 2z = b 5x+4y + 32=c (b) -3x + 2y + 4z = a
- x - 2y + 3z = b
-X -6y + 232 = 0 (C)4x - 2y + 3z = a 2x - 3y – 2z = b 4x - 2y + 32=c
The conditions on a, b, and c for which the Linear equations has solutions are det(A) ≠ 0, Rank[A|B] = Rank(A).
(a) 3x + 2y - z = a
x + y + 2z = b
5x + 4y + 32 = c
For this system to have solutions, the augmented matrix [A|B] must have a unique solution. This is equivalent to determinant of the coefficient matrix A is non-zero, and the system is consistent (the row rank of the augmented matrix is equal to the row rank of the coefficient matrix). Therefore, the conditions on a, b, and c for this system to have solutions are:
det(A) ≠ 0
Rank[A|B] = Rank(A)
(b) -3x + 2y + 4z = a
-x - 2y + 3z = b
-x - 6y + 232 = 0
For this system to have solutions, the conditions are the same as in (a):
det(A) ≠ 0
Rank[A|B] = Rank(A)
(c) 4x - 2y + 3z = a
2x - 3y – 2z = b
4x - 2y + 32 = c
For this system to have solutions, the conditions are the same as in (a) and (b):
det(A) ≠ 0
Rank[A|B] = Rank(A)
For each system of linear equations to have solutions, the number of equations must be equal to the number of variables, and the determinant of the coefficient matrix must be non-zero.
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A total of 14 different people were randomly surveyed and asked how many hours per day they worked the week before Their answers are included below. Construct a box and whisker plot using a TI-83. TI-83 Plus, or TI-84 graphing calculator. Au plots below have the following window settings: Xmin = 7.95. Xmax = 8.25, Xsel = 0.01, Ymin = -0.5, Ymax = 3.5, Ysel = 1, Xres=1 Average Work Hours Per Day 8.08 8.08 8.04 8.12 8.05 8.05 7.97 8.00 8.10 8.22 8.09 8.00 815 D ex 70 6 t y u g LLLL wa Lose Calculate measures of center and spread using Technology Calculator Med=8.08 P1L1 Med=8.065 Med=8.14 P1:11 Med=8.09
The boxplot for the provide data is present in above figure. Median or central tendency of data is equals to 8.08. The lower and upper limits of outliers are 7.928 and 8.248 respectively.
The first task is to compute the median and the quartiles. And, in order to compute the median and the quartiles, the data needs to be put into the ascending order, as shown in the above table. Since the sample size n = 14 is even, we have that (n+1)/2 = (14+1)/2 =7.5, is not an integer value, the median is computing by taking the average of the values at the positions 7ᵗʰ and 8ᵗʰ, as shown below Median = (8.08 + 8.08)/2 = 8.08
Quartiles : The quartiles are computed using the table with the data in the asencending order. For Q₁ we have to compute the following position:
pos(Q) = (n+1)× 25/100 = 15×25/100 = 3.75
Since 3.75 is not an integer number, Q₁ is computed by interpolating between the
values located in the 3ᵗʰ and 4ᵗʰ positions, as shown in the formula above
Q₁ = 8.04+ (3.75 - 3) (8.05 - 8.04) = 8.0475
For Q3 we have to compute the following position:pos(Q₃) = (n + 1)× 75/100
= (14+1)×75/100 = 11.25
Since 11.25 is not an integer number, Q₃ is computed by interpolating between
the values located in the 11ᵗʰ and 12ᵗʰ positions, as shown in the formula below
Q₃ = 8.12 + (11.25 - 11) × (8.15 8.12) = 8.1275
The interquartile range is therefore
IQR = Q₃ - Q₁ = 8.1275 - 8.0475 = 0.08
Now, we can compute the lower and upper limits for outliers:
Lower = Q₁ - 1.5 x IQR = 8.0475 - 1.5×0.08 =7.928
Upper = Q₃ + 1.5 x IQR = 8.1275 + 1.5× 0.08 =8.248
and then, an outcome X is an outlier if X < 7.928, or if X > 8.248.
In this case since all the outcomes X are within the values of Lower = 7.9275 and Upper = 8.2475, then there are no outliers. The above boxplot is obtained.
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An entry-level civil engineer earns an average bi-weekly net pay of $2,250.65. The engineer has created a monthly budget using the following percentages for expenses: Percent Housing 25% Food/Household 20% Savings 10% Transportation 5% Debt 15% Entertainment 5% Medical/Personal Care 5% Giving 5% Clothing 2% Miscellaneous 8% Which balance sheet correctly represents the engineer's income, expenses, and balance?
The engineer's balance sheet would show a monthly income of $9,002.60 ($4,501.30 x 2), total expenses of $4,001.36, and a balance of $499.94.
What is a balance sheet?Balance sheet is a financial statement that contains details of a company's assets or liabilities at a specific point in time.
First we can use the given information to create the following balance sheet for the civil engineer:
INCOME:
Bi-weekly net pay: $2,250.65 x 2 = $4,501.30 (this assumes two pay periods per month)
EXPENSES:
Housing: 25% of $4,501.30 = $1,125.33Food/Household: 20% of $4,501.30 = $900.26Savings: 10% of $4,501.30 = $450.13Transportation: 5% of $4,501.30 = $225.07Debt: 15% of $4,501.30 = $675.20Entertainment: 5% of $4,501.30 = $225.07Medical/Personal Care: 5% of $4,501.30 = $225.07Giving: 5% of $4,501.30 = $225.07Clothing: 2% of $4,501.30 = $90.03Miscellaneous: 8% of $4,501.30 = $360.10TOTAL EXPENSES: $4,001.36
BALANCE: $4,501.30 - $4,001.36 = $499.94
Therefore, the engineer's balance sheet would show a monthly income of $9,002.60 ($4,501.30 x 2), total expenses of $4,001.36, and a balance of $499.94.
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Maximize Z = 7X1 + 5X2
Subject to X1 + 2X2 ≤ 6
4X1 + 3X2 ≤ 12
X1, X2 ≥ 0
The solution is [tex]x_1[/tex] = 3 and [tex]x_2[/tex]= 0.
What is Simplex Method?Simplex method is an approach to solving linear programming models by hand using slack variables, tableaus, and pivot variables as a means to finding the optimal solution of an optimization problem.
Given:
We transfer the row with the resolving element from the previous table into the current table, elementwise dividing its values into the resolving element:
[tex]P_2[/tex] = [tex]P_2[/tex] / [tex]x_{2,1[/tex] = 12 / 4 = 3;
[tex]x_{2,1[/tex] = [tex]x_{2,1[/tex] / [tex]x_{2,1[/tex] = 4 / 4 = 1;
[tex]x_{2,2[/tex] = [tex]x_{2,2[/tex] / [tex]x_{2,1[/tex] = 3 / 4 = 0.75;
[tex]x_{2,3[/tex] = [tex]x_{2,3[/tex] / [tex]x_{2,1[/tex] = 0 / 4 = 0;
[tex]x_{2,4[/tex] = [tex]x_{2,4[/tex] / [tex]x_{2,1[/tex] = 1 / 4 = 0.25;
The remaining empty cells, except for the row of estimates and the column Q, are calculated using the rectangle method, relative to the resolving element:
[tex]P_1[/tex] = ([tex]P_1[/tex] . [tex]x_{2,1[/tex]) - ([tex]x_{1,1[/tex] . [tex]P_2[/tex]) / [tex]x_{2,1[/tex]= ((6 x 4) - (1 x 12)) / 4 = 3
[tex]x_{1,1[/tex]= (([tex]x_{1,1[/tex]. [tex]x_{2,1[/tex]) - ([tex]x_{1,1[/tex] . [tex]x_{2,1[/tex])) / [tex]x_{2,1[/tex]= ((1 x 4) - (1 x 4)) / 4 = 0
[tex]x_{1,3[/tex]= (([tex]x_{1, 3[/tex]. [tex]x_{2,1[/tex]) - ([tex]x_{1,1[/tex] . [tex]x_{2,3[/tex])) / [tex]x_{2,1[/tex]= ((1 x 4) - (1 x 0)) / 4 = 1
[tex]x_{1,4[/tex] = (([tex]x_{1, 4[/tex]. [tex]x_{2,1[/tex]) - ([tex]x_{1,1[/tex]. [tex]x_{2,4[/tex])) / [tex]x_{2,1[/tex]= ((0 x 4) - (1 x 1)) / 4 = -0.25
Objective function value
We calculate the value of the objective function by elementwise multiplying the column [tex]C_b[/tex] by the column P, adding the results of the products.
Max P = (C[tex]b_1[/tex] x [tex]P_{01[/tex]) + (C[tex]b_{11[/tex] x [tex]P_2[/tex])= (0 x 3) + (7 x 3) = 21;
Evaluated Control Variables
We calculate the estimates for each controlled variable, by element-wise multiplying the value from the variable column, by the value from the Cb column, summing up the results of the products, and subtracting the coefficient of the objective function from their sum, with this variable.
Max [tex]x_1[/tex]= ((C[tex]b_1[/tex] . [tex]x_{1,1[/tex]) + (C[tex]b_2[/tex] . [tex]x_{2,1[/tex]) ) - k[tex]x_1[/tex] = ((0 x 0) + (7 x 1) ) - 7 = 0
Max [tex]x_2[/tex]= ((C[tex]b_1[/tex] . [tex]x_{1,2[/tex]) + (C[tex]b_2[/tex] . [tex]x_{2,2[/tex]) ) - k[tex]x_2[/tex] = ((0 x 1.25) + (7 x 0.75) ) - 5 = 0.25
Max [tex]x_3[/tex]= ((C[tex]b_1[/tex] . [tex]x_{1,3[/tex]) + (C[tex]b_2[/tex] . [tex]x_{2,3[/tex]) ) - k[tex]x_3[/tex] = ((0 x 1) + (7 x 0) ) - 0 = 0
Max [tex]x_4[/tex]= ((C[tex]b_1[/tex] . [tex]x_{1,4[/tex]) + (C[tex]b_2[/tex] . [tex]x_{2,4[/tex]) ) - k[tex]x_4[/tex] = ((0 x -0.25) + (7 x 0.25) ) - 0 = 1.75
Since there are no negative values among the estimates of the controlled variables, the current table has an optimal solution.
The value of the objective function:
F = 21
The variables that are present in the basis are equal to the corresponding cells of the column P, all other variables are equal to zero:
[tex]x_1[/tex] = 3;
[tex]x_2[/tex]= 0;
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If ABCD is a parallelogram,
find the value of x.
7x + 2
9x-28
contracts for two construction jobs are randomly assigned to one or more of three firms, a, b, and c. the joint distribution of y1, the number of contracts awarded to firm a, and y2, the number of contracts awarded to firm b, is given by the entries in the following table. y1 y2 0 1 2 0 1 9 2 9 1 9 1 2 9 2 9 0 2 1 9 0 0 find cov(y1, y2). (round your answer to three decimal places.) cov(y1, y2)
The value of covariance cov(y1, y2) is -0.222
y1
y2 0 1 2 Total
0 1/9 2/9 1/9 4/9
1 2/9 2/9 0 4/9
2 1/9 0 0 1/9
Total 4/9 4/9 1/9 1
marginal distribution of y2:
y2 P(y2) y2P(y2) y2^2P(y2)
0 4/9 0.0000 0.0000
1 4/9 0.4444 0.4444
2 1/9 0.2222 0.4444
total 1 0.66667 0.88889
E(y2) = 0.6667
E(y2^2) = 0.8889
Var(y2) = E(y2^2)-(E(y2))^2=0.4444
marginal distribution of y1
y1 P(y1) y1P(y1) y1^2P(y1)
0 4/9 0.0000 0.0000
1 4/9 0.4444 0.4444
2 1/9 0.2222 0.4444
total 1.00 0.6667 0.8889
E(y1) = 0.6667
E(y1^2) = 0.8889
Var(y1)= σy = E(y1^2)-(E(y1))^2 = 0.4444
E(XY) =∑xyP(x,y)=0.2222
Cov(y1,y2) = -0.222
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please help. . . . . . . .
Answer:
3. [tex]y=(x-3)(x-2)[/tex]
4. [tex]y = (x + 2)^2[/tex]
Step-by-step explanation:
In this problem, we are asked to find the equations in their factored form of the graphed parabolas.
3. We can see that the parabola's vertex is at (3, -4). We can plug the coordinates of that point into the vertex form equation:
[tex]y=(x-a)^2 + b[/tex]
where [tex](a,b)[/tex] is the vertex of the parabola.
[tex]y=(x-3)^2 -4[/tex]
Then, we can expand the right side of the equation to an unfactored form.
[tex]y=x^2-6x+9 -4[/tex]
[tex]y=x^2-6x-5[/tex]
Finally, we can factor the right side of the equation.
[tex]\boxed{y=(x-3)(x-2)}[/tex]
4. First, input the vertex's coordinates into the vertex form equation.
[tex]y=(x - (-2))^2 + 0[/tex]
Then, simplify.
(Remember that subtracting a negative is the same as adding the positive)
[tex]\boxed{y=(x+2)^2}[/tex]
Question 5.4° A school with 20 professors forms 10 committees, each containing 6 profes- sors, such that every professor is on exactly 3 committees. Prove that it is possible to select a distinct representative from each committee.
Since there are 20 professors and each professor is on exactly 3 committees, there must be a total of 60 committee positions.
Let's assume that it is not possible to select a distinct representative from each committee. This means that there must be some committee that does not have a distinct representative.
Let's consider the professors on this committee. Since each professor is on exactly 3 committees, each of the professors on this committee must also be on 2 other committees.
Let's consider the 2 other committees for each of the professors on this committee. Since there are 6 professors on the committee, there are a total of 12 other committees to consider.
Each of these 12 committees must have a representative chosen from the remaining 14 professors, since we have assumed that it is not possible to select a distinct representative from the original 10 committees.
However, since there are only 14 professors remaining, and each professor can only be on 3 committees, it is not possible to choose a representative from all 12 of these committees without choosing at least one professor to be on 4 committees. This is a contradiction, since we have assumed that each professor is on exactly 3 committees.
Therefore, our assumption that it is not possible to select a distinct representative from each committee must be false, and it is indeed possible to do so.
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The surface of a table to be built will be in the shape shown below. The distance from the center of the shape to the center of each side is 10.4 inches and the length of each side is 12 inches.
A hexagon labeled ABCDEF is shown will all 6 sides equal in length. ED is labeled as 12 inches. A perpendicular is drawn from the center of the hexagon to the side ED. This perpendicular is labeled as 10.4 inches.
Part A: Describe how you can decompose this shape into triangles. (2 points)
Part B: What would be the area of each triangle? Show every step of your work. (5 points)
Part C: Using your answers above, determine the area of the table's surface. Show every step of your work.
A) We can draw a line perpendicular to side ED from the center of the hexagon, which will bisect side ED and create two additional triangles.
B) Area of each triangle = 31.16 square inches (rounded to two decimal places)
C) Total surface area = 375.87 square inches (rounded to two decimal places)
What is Surface Area ?Any geometric shape with three dimensions can have its surface area determined. The area or region that an object's surface occupies is known as its surface area.
Now in the given question,
Part A: To decompose this shape into triangles, we can draw lines connecting the center of the hexagon to each of its vertices, creating six triangles. We can also draw a line perpendicular to side ED from the center of the hexagon, which will bisect side ED and create two additional triangles.
Part B: Each of the six triangles created by connecting the center of the hexagon to its vertices is an equilateral triangle, because all sides of the hexagon are equal in length. The area of an equilateral triangle can be calculated using the formula:
Area = (√3 / 4) x side²
where side is the length of one side of the equilateral triangle. In this case, the length of each side of the equilateral triangle is also 12 inches, so we have:
Area = (√3 / 4) x 12²
Area = (√3 / 4) x 144
Area = 36√3 square inches
Each of the two triangles created by drawing a perpendicular from the center of the hexagon to side ED is a right triangle, because one of its angles is 90 degrees. We can use the Pythagorean theorem to calculate the length of the other two sides of the right triangle. We know that one side has length 10.4 inches and the hypotenuse (which is also a side of the hexagon) has length 12 inches. Let x be the length of the other side of the right triangle. Then we have:
x² + 10.4² = 12²
x²+ 108.16 = 144
x² = 35.84
x = √35.84
x = 5.99 inches (rounded to two decimal places)
The area of each right triangle can be calculated using the formula:
Area = (1/2) x base x height
where the base is 5.99 inches (the length of the side opposite the 90 degree angle) and the height is 10.4 inches (the length of the side adjacent to the 90 degree angle). We have:
Area = (1/2) x 5.99 x 10.4
Area = 31.16 square inches (rounded to two decimal places)
Part C: To determine the area of the table's surface, we need to add up the areas of all eight triangles. There are six equilateral triangles, each with an area of 36sqrt(3) square inches, and two right triangles, each with an area of 31.16 square inches. Therefore, the total area of the table's surface is:
Total area = 6 x 36√3 + 2 x 31.16
Total area = 216√3 + 62.32
Total area = 375.87 square inches (rounded to two decimal places)
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Please solve within 30 minutes! Thanks!
Please solve for the variable indicated.
20.) A=1/2bh, solve for b
If you could break it down step by step that would be super helpful! I’m very confused. Thank you!
Step-by-step explanation:
A = 1/2 bh
divide by 1/2
2A = bh
divide by h
2A/h = b
Hope this helps.
Answer:
[tex] b = \dfrac{2A}{h} [/tex]
Step-by-step explanation:
[tex] A = \dfrac{1}{2}bh [/tex]
[tex] 2A = 2 \times \dfrac{1}{2}bh [/tex]
[tex] 2A = bh [/tex]
[tex] \dfrac{2A}{h} = \dfrac{bh}{h} [/tex]
[tex] \dfrac{2A}{h} = b [/tex]
[tex] b = \dfrac{2A}{h} [/tex]
Deduct 20% for taxes from the monthly gross income to get the net income (the amount made after taxes).
What is Andrew's monthly net income?
What is Lisa's monthly net income?
What is their total monthly net income?
Andrew's monthly net income is $65280.
Lisa's monthly net income is $60960.
The total monthly net income is $126240.
What is a percentage?The percentage is calculated by dividing the required value by the total value and multiplying by 100.
Example:
Required percentage value = a
total value = b
Percentage = a/b x 100
Example:
50% = 50/100 = 1/2
25% = 25/100 = 1/4
20% = 20/100 = 1/5
10% = 10/100 = 1/10
We have,
Andrew's monthly gross income.
= $81, 600
Lisa's monthly gross income.
= $76,200
Now,
Tax = 20%
So,
Andrew's monthly net income.
= 80/100 x 81, 600
= $65280
Lisa's monthly net income.
= 80/100 x 76,200
= $60960
Now,
Total monthly net income.
= 65280 + 60960
= $126240
Thus,
Andrew's monthly net income = $65280
Lisa's monthly net income = $60960
Total monthly net income = $126240
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A young family is looking to buy a house and decides they need at least a 3/4-acre lot so there’s room for the kids and their dog Moose to play. Their agent plans to show them a house with a 30,000-sq-ft lot. Will she be wasting her time?
Answer: We can calculate the size of the lot in acres and then compare it to the minimum size of 3/4 of an acre that the family is looking for.
1 acre = 43,560 square feet
So, a 30,000 sq-ft lot is equivalent to:
30,000 sq-ft / 43,560 sq-ft/acre = 0.6864 acres
Since 0.6864 acres is less than 3/4 of an acre, the family's agent will be wasting her time by showing them the house with a 30,000 sq-ft lot. The lot is not big enough for the family's needs.
Step-by-step explanation:
I NEED HELP PLEASE HELP ME
For 3y-2x=-18 determine the value of y when x = 0, and the value of x when y = 0
Answer question on the picture attached
The average rate of change over the interval (-2,0) is -2.
What is the average rate of change?The average rate of change describes how quickly one quantity changes in comparison to another. It indicates how much the function changed per unit during the specified interval.
Given that the interval is (-2,0). The coordinate of the point from the interval -2,0 is ( -2 , 4 ).
The average rate of change will be calculated as:-
Rate of change = ΔY / ΔX
Rate of change = ( 4 - 0 ) ( -2 - 0 )
Rate of change = -2
Therefore, the average rate of change will be -2.
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For each situation, determine why the situation cannot be modelled after a binomial distribution. (2 pts each)a. A bowl of candy contains 23 Pluto Bars, 21 Jolly Farmers, 14 Husky's Kisses, 21 Finnish Fishes, and 18 HairHeads. Kristina decides to pull out 77 pieces of candy, without putting any back, and count only the number of Jolly Farmers that she grabs.b. A bowl of candy contains 23 Pluto Bars, 21 Jolly Farmers, 14 Husky's Kisses, 21 Finnish Fishes, and 18 HairHeads. Ian decides to pull out 70 pieces of candy, with replacement, and record each type of candy the he pulls.c. A bowl of candy contains 23 Pluto Bars, 21 Jolly Farmers, 14 Husky's Kisses, 21 Finnish Fishes, and 18 HairHeads. Marissa decides to pull out pieces of candy, with replacement, until a Husky's Kiss is pulled.
The question is geometric distribution. We are concerned with just the first success. That means there is no fixed number of trials
How to solve this
Before a situation can model binomial distribution.
1) There should be a fixed number of trial2) Each trial should have two possible outcomes (success or failure)3) The probability of success is the same for each trial.4) The trials are independent.For question a.
The probability of success is not the same for each trial. Since we are not putting any back. So it cannot be modeled as binomial distribution
For question b.
This has many outcomes not just success and failure.
Lan needs to record Pluto bar, jolly farm, husky kisses, finish fish, and hair heads.
For question c.
The question is geometric distribution. We are concerned with just the first success. That means there is no fixed number of trials
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Please help !! I need help I can’t figure it out
Answer:
i cant see
Step-by-step explanation:
it, its to tiny
What is the value of X and Y?
The value of x and y is 31° and 101°. Value is the measure of what something is worth, either financially or in terms of importance.
What is meant by value?The value describes the value of each digit in relation to its position in the number. The place value and face value of the digit are multiplied to arrive at the answer. Value equals Location Value minus Face Value.Value is concerned with how much something is worth, whether in monetary terms or in terms of significance. It can mean "identify how much something is worth" like a reward valued at $200 or it can indicate "hold something in high esteem" like "I value our friendship."The things that someone values are known as their values. In other terms, values are what a person or a group of people see as "important." Examples include valor, integrity, freedom, and creativity, among others.[tex]$\angle \mathrm{PRT}+\angle \mathrm{RTP}+\angle \mathrm{TPR}=180^{\circ} \text { (angle sum property of triangle) } \\[/tex]
[tex]& \Rightarrow \mathrm{x}+\left(180^{\circ}-\angle \mathrm{RTQ}\right)+60^{\circ}=180^{\circ} \text { (linear pair) } \\[/tex]
[tex]& \Rightarrow \mathrm{x}+\left(180^{\circ}-97^{\circ}\right)+60^{\circ}=180^{\circ} \\[/tex]
We get,
[tex]& \Rightarrow \mathrm{x}=31^{\circ} \\[/tex]
[tex]& \text { Now } \angle \mathrm{PRT}+\angle \mathrm{TRQ}+\angle \mathrm{QRS}=180^{\circ} \text { (angle of straight line) } \\[/tex]
[tex]& \Rightarrow \mathrm{x}+48^{\circ}+\mathrm{y}=180^{\circ} \\[/tex]
Simplifying,
[tex]& \Rightarrow 31^{\circ}+48^{\circ}+\mathrm{y}=180^{\circ} \\[/tex]
Then we get,
[tex]& \Rightarrow \mathrm{y}=101^{\circ}[/tex]
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Consider a triangle ABC like the one below. Suppose that C-96°, a-33, and b=39. (The figure is not drawn to scale.) Solve the triangle.
Carry your intermediate computations to at least four decimal places, and round your answers to the nearest tenth.
If there is more than one solution, use the button labeled "or".
DSD
X
16
No
solution
5
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The value of C is 108°
How to solve for thisGiven [tex]a=33,\ \ b=26, \ \ \angle B=31^0 .[/tex]
We have to find [tex]\angle C, \ \angle A \ and\ c[/tex]
We can use the cosine formula to find these:
[tex]cosB=\frac{a^2+c^2-b^2}{2ac}\\cos31^o=\frac{33^2+c^2-26^2}{2(33)(c)}\\0.8572=\frac{413+c^2}{66c}\\56.5752c=413+c^2\\i.e. c^2-56.5752c+413=0\\\Rightarrow c=47.9647, \ 8.6104\\So, \mathbf{c=48 \ or \ c=9}\\Now if c=48:\\cosA=\frac{b^2+c^2-a^2}{2bc}\\=\frac{26^2+48^2-33^2}{2(26)(48)}\\[/tex]
[tex]=\frac{1891}{2496}\\=0.75761\\\therefore A=cos^{-1}(0.75761)=40.746^0\approx 41^0\\i.e. \mathbf{\angle A= 41^0}\\cosC=\frac{a^2+b^2-c^2}{2ab}\\=\frac{33^2+26^2-48^2}{2(33)(26)}\\=\frac{-539}{1716}\\=-0.3141\\\therefore C=cos^{-1}(-0.3141)=108.306^0\approx 108^0[/tex]
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What is the equation of the line that passes through the point (-8, 6) and has a slope
of -1/4?
Answer:
= 4y + x - 16 = 0.
Step-by-step explanation:
Point are (-8,6)
where
x1 = -8 and y1 = 6
the equation o the line is given by the formula
[tex] = \frac{y - y1}{x - x1} = m[/tex]
where (m) is the gradient
the gradient (m) = -1/4
therefore the equation of the line
[tex] \frac{y - y1}{x - x1} = m \\ = \frac{y - 6}{x - ( - 8)} = \frac{ - 1}{4} \\ = \frac{y - 6}{x + 8} = \frac{ - 1}{4} \\ = 4(y - 6) = - 1(x + 8) \\ = 4y - 24 = - x - 8 \\ = 4y = - x - 8 + 24 \\ = 4y = - x + 16 \\ = 4y + x - 16 = 0.[/tex]
therefore the equation of the line is = 4y + x - 16 = 0.
note// y and x where not given any value.
Classify the following functions on R as even, odd or neither: (Problem 1.4A) (a) fi(t)=I (b) f2(t) = t3-2t
For each of the functions in problem (1), find the even and odd parts of the function. (Prob- lem 1.4C)
a) f1(t) = 1 is a even function, the even part is simply 1/2, and the odd part is 0.
b) f2(t) = t^3 - 2t is odd function, the even part is t^3 - t, and the odd part is -t.
Part (a) asks us to classify the function f1(t) = I as even, odd, or neither. Here, I represents a constant, so f1(t) is a constant function.
A function f(t) is even if f(-t) = f(t) for all t in the domain of f. In other words, the function has symmetry about the y-axis.
A function f(t) is odd if f(-t) = -f(t) for all t in the domain of f. In other words, the function has rotational symmetry about the origin.
For a constant function, we have f(-t) = I = f(t) for all values of t. Therefore, f1(t) is even.
Part (b) asks us to classify the function f2(t) = t^3 - 2t as even, odd, or neither.
To determine whether f2(t) is even, odd, or neither, we need to apply the definitions of even and odd functions.
If f2(t) is even, then we have f2(-t) = f2(t) for all t.
If f2(t) is odd, then we have f2(-t) = -f2(t) for all t.
We can check these conditions as follows:
f2(-t) = (-t)^3 - 2(-t) = -t^3 + 2t
Since f2(-t) is not equal to f2(t) for all t, we know that f2(t) is not even.
We can also check if f2(t) is odd:
f2(-t) = (-t)^3 - 2(-t) = -t^3 + 2t = -f2(t)
Since f2(-t) = -f2(t) for all t, we know that f2(t) is odd.
To find the even and odd parts of a function, we use the following formulas:
even part of f(t) = (f(t) + f(-t))/2
odd part of f(t) = (f(t) - f(-t))/2
For f1(t) = I, the even part is (I + I)/2 = I/2, and the odd part is (I - I)/2 = 0.
For f2(t) = t^3 - 2t, the even part is (t^3 - 2t + (-t)^3 - 2(-t))/2 = t^3 - t, and the odd part is (t^3 - 2t - (-t)^3 + 2(-t))/2 = -t.
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Cecily purchases a box of 100 paper clips. She puts 37 /100 of the paper clips in a jar on her desk and puts another 6 /10 in her drawer at home. Shade a grid that shows how many of the paper clips are in Cecily's jar and drawer, then write the fraction tbe grid represents.
The fraction of the total of paper clips Cecily put in the jar and drawer is [tex]\frac{97}{100}[/tex] (see the attachment below).
How many clips did Cecily put in the jar and drawer?Paper clips in the jar: 37/100, which means Cecily put 37 clips in the jar.
Paper clips in the drawer: 6/10, now let's find out the number of clips this fraction is equivalent to:
100 (total clips) / 10 x 6 = 60
Total clips: 60 +37= 97. This number of clips can be expressed as [tex]\frac{97}{100}[/tex] .
How to represent this in a grid?To represent this in a grid color a total of 97 squares in a grid with 100 squares as it is shown below.
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indicate whether the following statements regarding the variance and standard deviation are true or false. a) the population variance and sd use n in the denominator instead of n-1. false b) the sample variance and sd inherit the strengths and weaknesses of the sample mean. _______
c) if we did not square the distances in the numerator, we would end up with 0. ______
d) the sample variance and sd are robust to outliers and/or extreme points.________
a) the population variance and sd use n in the denominator instead of n-1. - false
b) the sample variance and sd inherit the strengths and weaknesses of the sample mean. - True
c) if we did not square the distances in the numerator, we would end up with 0. -false
d) the sample variance and sd are robust to outliers and/or extreme points.-false
a) False. The population variance and standard deviation use N in the denominator, where N is the size of the population. The sample variance and standard deviation use n-1 in the denominator, where n is the size of the sample.
b) True. The sample variance and standard deviation inherit the strengths and weaknesses of the sample mean, since they are based on the deviations of the sample values from the sample mean.
c) False. If we did not square the distances in the numerator, we would end up with the mean deviation, which is a valid measure of dispersion but not the variance.
d) False. The sample variance and standard deviation are not robust to outliers and/or extreme points, as they are sensitive to the values of the data points and can be significantly influenced by the presence of outliers.
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Steven sprinted 14 1/3 laps and then took a break by jogging 4 laps. How much farther did Steven sprint than jog?
help please see photo
The solution of the inequality using the coordinate pair is 5.
What is Linear Inequality?Linear inequalities are those expressions which are connected by inequality signs like >, <, ≤, ≥ and ≠ and the value of the exponent of the variable is 1.
Given linear inequality is,
y ≤ -2x + 3
We have a coordinate pair (-1, 5).
So substitute x = -1 and y = 5 in the given inequality.
5 ≤ (-2 × -1) + 3
5 ≤ 2 + 3
5 ≤ 5
Hence the missing numbers are found.
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