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3 years ago

Solve for X and round to the nearest tenth

Mathematics

1 answer:

monitta3 years ago

5 0

Answer:

5.4702≈5.5

Step-by-step explanation:

cos67 =0.3907

x/14= 0.3907

x= 0.3907*14 = 5.4702≈5.5

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40% of the students in a class are girls and the number of boys is 24 find the total number of students and the number of girls

navik [9.2K]

Answer:

Girls = 16

Total students = 40

Step-by-step explanation:

40% of the students are girls, this means 60% of the students are boys.

To find the number of girls, setup a proportion using the percentages and actual number of students in the class. I am going to make the numerator the numbers that go with the girls and the denominator will be the numbers that go with the boys.

Also let's convert the percentages to decimals by moving the decimal two places left.

40% = 0.4 and 60% = 0.6

$\frac{.4}{.6} =\frac{g}{24} \\.6g=9.6\\\frac{.6g}{.6} =\frac{9.6}{.6} \\g=16$

Now let's find the total number of students by adding the number of girls to the number of boys.

16 + 24 = 40

7 0

2 years ago

I’ll give brainliest if correct

Tems11 [23]

Answer:

Step-by-step explanation:

3 0

3 years ago

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Better Products, Inc., manufactures three products on two machines. In a typical week, 40 hours are available on each machine. T

Kaylis [27]

Answer:

z (max) = 1250 $

x₁ = 25 x₂ = 0 x₃ = 25

Step-by-step explanation:

Profit $ mach. 1 mach. 2

Product 1 ( x₁ ) 30 0.5 1

Product 2 ( x₂ ) 50 2 1

Product 3 ( x₃ ) 20 0.75 0.5

Machinne 1 require 2 operators

Machine 2 require 1 operator

Amaximum of 100 hours of labor available

Then Objective Function:

z = 30*x₁ + 50*x₂ + 20*x₃ to maximize

Constraints:

1.-Machine 1 hours available 40

In machine 1 L-H we will need

0.5*x₁ + 2*x₂ + 0.75*x₃ ≤ 40

2.-Machine 2 hours available 40

1*x₁ + 1*x₂ + 0.5*x₃ ≤ 40

3.-Labor-hours available 100

Machine 1 2*( 0.5*x₁ + 2*x₂ + 0.75*x₃ )

Machine 2 x₁ + x₂ + 0.5*x₃

Total labor-hours :

2*x₁ + 5*x₂ + 2*x₃ ≤ 100

4.- Production requirement:

x₁ ≤ 0.5 *( x₁ + x₂ + x₃ ) or 0.5*x₁ - 0.5*x₂ - 0.5*x₃ ≤ 0

5.-Production requirement:

x₃ ≥ 0,2 * ( x₁ + x₂ + x₃ ) or -0.2*x₁ - 0.2*x₂ + 0.8*x₃ ≥ 0

General constraints:

x₁ ≥ 0 x₂ ≥ 0 x₃ ≥ 0 all integers

The model is:

z = 30*x₁ + 50*x₂ + 20*x₃ to maximize

Subject to:

0.5*x₁ + 2*x₂ + 0.75*x₃ ≤ 40

1*x₁ + 1*x₂ + 0.5*x₃ ≤ 40

2*x₁ + 5*x₂ + 2*x₃ ≤ 100

0.5*x₁ - 0.5*x₂ - 0.5*x₃ ≤ 0

-0.2*x₁ - 0.2*x₂ + 0.8*x₃ ≥ 0

x₁ ≥ 0 x₂ ≥ 0 x₃ ≥ 0 all integers

After 6 iterations with the help of the on-line solver AtomZmaths we find

z (max) = 1250 $

x₁ = 25 x₂ = 0 x₃ = 25

6 0

3 years ago

Mary is melting a rectangular block of wax that measures 10 cm×9 cm×20 cm. Determine how many candles she can make of the follow

Rashid [163]

Answer:

Step-by-step explanation:

https://www.numerade.com/questions/reasoning-in-exercises-35-and-36-you-are-melting-a-rectangular-block-of-wax-to-make-candles-how-ma-2/

This link lets you access a video for the correct answer

4 0

3 years ago

Two numbers have a sum of 24<br> and a product of 143. What is the<br> smaller number?

jok3333 [9.3K]

Answer:

The smaller number is 11.

Step-by-step explanation:

11 + 13 = 24

11 x 13 = 143

Hope this helps!

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3 years ago

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