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grandymaker [24]
3 years ago
10

By careful reorganization, a company was able to reduce the cost of operation from $21,840 per month to $17,190 per month. By wh

at percent was the cost of operation reduced?
Mathematics
2 answers:
Ede4ka [16]3 years ago
4 0
Subtract the 17190 from the 21840. you get 4650. 4650 divided by 21840 is 0.21. The percent decrease is 21%
lesya692 [45]3 years ago
4 0

Answer: The cost of operation was reduced by 21.29%.

Step-by-step explanation:

Since we have given that

Cost of operation earlier = $21,840 per month

Cost of operation now = $17,190 per month

Difference between them is given by

$21,840 - $17,190

=$4,650

Percentage by which the cost of operation was reduced is given by

\dfrac{4650}{21840}\times 100\\\\=21.29\%

Hence, the cost of operation was reduced by 21.29%.

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What is the distance between the coordinates (8,-5) and (8,13).
Darina [25.2K]

Answer:

18

Step-by-step explanation:

use the distance forumla to detwemine the distance between the points.

7 0
2 years ago
An angle of measure 120 degrees intersects the unit circle at point (-1/2,√3/2) What is the exact value of cos(120)
notka56 [123]

The exact value of cos120 if the measure 120 degrees intersects the unit circle at point (-1/2,√3/2) is 0.5

<h3>Solving trigonometry identity</h3>

If an angle of measure 120 degrees intersects the unit circle at point (-1/2,√3/2), the measure of cos(120) can be expressed as;

Cos120 = cos(90 + 30)

Using the cosine rule of addition

cos(90 + 30) = cos90cos30 - sin90sin30

cos(90 + 30) = 0(√3/2) - 1(0.5)

cos(90 + 30) = 0 - 0.5

cos(90 + 30) = 0.5

Hence the exact value of cos120 if the measure 120 degrees intersects the unit circle at point (-1/2,√3/2) is 0.5

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7 0
1 year ago
The braking distance, in feet of a car a Travling at v miles per hour is given.
irakobra [83]

The braking distance is the distance the car travels before coming to a stop after the brakes are applied

a. The braking distances are as follows;

  • The braking distance at 25 mph, is approximately <u>63.7 ft.</u>
  • The braking distance at 55 mph,  is approximately <u>298.35 ft.</u>
  • The braking distance at 85 mph,  is approximately <u>708.92 ft.</u>

b. If the car takes 450 feet to brake, it was traveling with a speed of 98.211 ft./s

Reason:

The given function for the braking distance is D = 2.6 + v²/22

a. The braking distance if the car is going 25 mph is therefore;

25 mph = 36.66339 ft./s

D = 2.6 + \dfrac{36.66339^2}{22} = 63.7 \ ft.

At 25 mph, the braking distance is approximately <u>63.7 ft.</u>

At 55 mph, the braking distance is given as follows;

55 mph = 80.65945  ft.s

D = 2.6 + \dfrac{80.65945^2}{22} \approx 298.35 \ ft.

At 55 mph, the braking distance is approximately <u>298.35 ft.</u>

At 85 mph, the braking distance is given as follows;

85 mph = 124.6555 ft.s

D = 2.6 + \dfrac{124.6555^2}{22} \approx 708.92 \ ft.

At 85 mph, the braking distance is approximately <u>708.92 ft.</u>

b. The speed of the car when the braking distance is 450 feet is given as follows;

450 = 2.6 + \dfrac{v^2}{22}

v² = (450 - 2.6) × 22 = 9842.8

v = √(9842.2) ≈ 98.211 ft./s

The car was moving at v ≈ <u>98.211 ft./s</u>

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8 0
2 years ago
Question 12 (1 point) Given P(A) 0.34, P(A and B) 0.27, P(A or B) 0.44, what is P(B)? Answer in decimal form. Round to 2 decimal
Tema [17]

Answer:  The required probability of event B is P(B) = 0.37.

Step-by-step explanation:  For two events A and B, we are given the following probabilities :

P(A) = 0.34,    P(A ∩ B) = 0.27   and   P(A ∪ B) = 0.44.

We are to find the probability of event B, P(B) = ?

From the laws of probability, we have

P(A\cup B)=P(A)+P(B)-P(A\cap B)\\\\\Rightarrow 0.44=0.34+P(B)-0.27\\\\\Rightarrow 0.44=0.07+P(B)\\\\\Rightarrow P(B)=0.44-0.07\\\\\Rightarrow P(B)=0.37.

Thus, the required probability of event B is P(B) = 0.37.

6 0
3 years ago
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