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MrMuchimi
2 years ago
6

Based on the given information, determine which company has a lower minimum and find the minimum value. g(x) at (10, 855) f(x) a

t (16, 536) g(x) at (16, 536) f(x) at (10, 855)
SAT
1 answer:
Gekata [30.6K]2 years ago
7 0

The minimum production cost of company 2 is greater than the minimum production cost of company 1. We arrived at this value by comparing the production cost of both companies.

<h3>What is meant by minimum production cost?</h3>

The overall cost incurred by a company to manufacture a product or provide services is known as the cost of production.

The objective of every company is to keep this cost at minimum, hence the minimum production cost.

<h3>How do find minimum Production Cost?</h3>

Recall that the production function is given as:

f(x) = 0.25x² - 8x + 600

Inserting the values given by the schedule we have

  1. f(6) = 0.25(6²) - 8(6) + 600 = 561
  2. f(8) = 0.25(8²) - 8(8) + 600 = 552
  3. f(10) = 0.25(10²) - 8(10) + 600 = 545
  4. f(12) = 0.25(12²) - 8(12) + 600 = 540
  5. f(14) = 0.25(14²) - 8(14) + 600 = 537

For company 2, we are given the various production costs as;

x   - g(x)

6  - 862.2

8  - 856.8

10 -  855

12 -  856.8

14 - 862.2

Juxtaposing the above, we can infer that the minimum production cost of company 2 is greater than the minimum production cost of company 1.

Learn more about minimum production cost at:
brainly.com/question/9871118
#SPJ1

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A bat emits a sound at a frequency of 3. 00 × 104 hz as it approaches a wall. The bat detects beats such that the frequency of t
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Taking into account the Doppler efect, the speed of the bat is closest to 5.02 \frac{m}{s}.

The Doppler effect is defined as the apparent frequency change of a wave produced by the relative movement of the source with respect to its observer. In other words, this effect is the change in the perceived frequency of any wave motion when the sender and receiver, or observer, move relative to each other.

The following expression is considered the general case of the Doppler effect:

f1=f\frac{v+-vR}{v+-vE}

Where:

  1. f1, f: Frequency perceived by the receiver and frequency emitted by the issuer respectively. Its unit of measurement in the International System (S.I.) is the hertz (Hz), which is the inverse unit of the second (1 Hz = 1 s⁻¹)
  2. v: Velocity of propagation of the wave in the medium. It is constant and depends on the characteristics of the medium. In this case, the speed of sound in air is considered to be 340 m/s
  3. vR, vE: Speed ​​of the receiver and the sender respectively. Its unit of measure in the S.I. is the m/s
  4. ±, ∓:
  • We will use the + sign:

In the numerator if the receiver approaches the sender

In the denominator if the sender moves away from the receiver

  • We will use the sign -:

In the numerator if the receiver moves away from the emitter

In the denominator if the sender approaches the receiver

In this case, a bat emits a sound at a frequency of 3.00×10⁴ Hz= 30,000 Hz as it approaches a wall.

The bat detects beats such that the frequency of the echo is 900 hz higher than the frequency the bat is emitting.

So, you know:

  • f1= 30,000 Hz + 900 Hz= 30,900 Hz
  • f= 30,000 Hz
  • v= 340 m/s
  • vE=vR= vbat
  • Receiver approaches the sender

Replacing in the expression is considered the general case of the Doppler effect:

30900 Hz=30000 Hz\frac{340 \frac{m}{s} +vbat}{340\frac{m}{s} -vbat}

Solving:

\frac{30900 Hz}{30000 Hz} =\frac{340 \frac{m}{s} +vbat}{340\frac{m}{s} -vbat}

1.03=\frac{340 \frac{m}{s} +vbat}{340\frac{m}{s} -vbat}

1.03×( 340\frac{m}{s} -vbat)= 340 \frac{m}{s} + vbat

1.03×340\frac{m}{s} - 1.03×vbat= 340 \frac{m}{s} + vbat

350.2\frac{m}{s} - 1.03×vbat= 340 \frac{m}{s} + vbat

350.2\frac{m}{s} - 340 \frac{m}{s}= vbat + 1.03×vbat

10.2 \frac{m}{s}= 2.03×vbat

10.2 \frac{m}{s}÷ 2.03 = vbat

<u><em>5.02 </em></u>\frac{m}{s}<u><em>= vbat</em></u>

In summary, the speed of the bat is closest to 5.02 \frac{m}{s}.

Learn more about the Doppler effect:

  • brainly.com/question/17107808
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