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irina [24]
2 years ago
11

A week before an election, 1,500 people were asked who they planned to vote for. Of the people asked, 45% said they planned to v

ote for candidate X, and 38% said they planned to vote for candidate Y. The rest said they had not yet decided. How many of the people that were asked have not yet decided who they plan to vote for?
Mathematics
1 answer:
Ivahew [28]2 years ago
5 0
By difference, the percentage of people who are not sure who to vote is 100% - 45% - 38% = 17%. The number of people who have not yet decided to whom their votes goes is the product of 1,500 and 17% in its decimal equivalent.
                                (1,500) x (0.17) = 255
Therefore, 255 have not yet decided who to vote. 

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Solve the linear programming problem.
larisa [96]

Answer:

1. Objective function is a maximum at (16,0), Z = 4x+4y = 4(16) + 4(0) = 64

2. Objective function is at a maximum at (5,3), Z=3x+2y=3(5)+2(3)=21

Step-by-step explanation:

1. Maximize: P = 4x +4y

Subject to: 2x + y ≤ 20

x + 2y ≤ 16

x, y ≥ 0

Plot the constraints and the objective function Z, or P=4x+4y)

Push the objective function to the limit permitted by the feasible region to find the maximum.

Answer: Objective function is a maximum at (16,0),

              Z = 4x+4y = 4(16) + 4(0) = 64

2. Maximize P = 3x + 2y

Subject to x + y ≤ 8

2x + y ≤ 13

x ≥ 0, y ≥ 0

Plot the constraints and the objective function Z, or P=3x+2y.

Push the objective function to the limit in the increase + direction permitted by the feasible region to find the maximum intersection.

Answer: Objective function is at a maximum at (5,3),

              Z = 3x+2y = 3(5)+2(3) = 21

7 0
3 years ago
Rewrite 11/12 and 14/15 so that they have a common denominator
djyliett [7]
The least common denomonator is 60
8 0
3 years ago
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PLEASE HELP! 20 POINTS 1) A ball is thrown starting at a time of 0 and a height of 2 meters. The height of the ball follows the
drek231 [11]

Answer:

1) The height of the ball from 0 to 5  seconds are;

At t = 0 second, height = 2

At t = 1 second, height h = 22.1

At t = 2 seconds, height h = 32.4

At t = 3 seconds, height h = 32.9

At t = 4 seconds, height h = 23.6

At t = 5 seconds, height h = 4.5

2)  The correct option is;

D. -16·t² + 25·t + 1

Step-by-step explanation:

1) The equation of motion of the ball is given as follows;

H(t) = -4.9·t² + 25·t + 2

The height of the ball from 0 to 5 seconds are;

H(0) = -4.9×(0)² + 25×(0) + 2 = 2

H(1) = -4.9×(1)² + 25×(1) + 2 = 22.1

H(2) = -4.9×(2)² + 25×(2) + 2 = 32.4

H(3) = -4.9×(3)² + 25×(3) + 2 = 32.9

H(4) = -4.9×(4)² + 25×(4) + 2 = 23.6

H(5) = -4.9×(5)² + 25×(5) + 2 = 4.5

Therefore, we have;

The height of the ball are

At t = 0 second, height = 2

At t = 1 second, height h = 22.1

At t = 2 seconds, height h = 32.4

At t = 3 seconds, height h = 32.9

At t = 4 seconds, height h = 23.6

At t = 5 seconds, height h = 4.5

2) Given that the equation of the ball is that of a projectile motion, such as follows;

h = h₀ + v₀·sin(θ₀)·t - 1/2·g·t² which is equivalent to h = -1/2·g·t²+ h₀+v₀·sin(θ₀)·t

it is best represented by the quadratic equation of an upside down parabola which is option D. -16·t² + 25·t + 1

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Reduce all of the mutiple choices

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C. 9/16. Cannot reduce the fraction.

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