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kow [346]
4 years ago
12

What is the area of a circle with a radius of 13 millimeters? (Use 3.14 for Pi.)

Mathematics
1 answer:
ludmilkaskok [199]4 years ago
5 0
The area of a circle with radius of 13 millimeters is A≈530.93A=πr2=π·132≈530.92916
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Consider a manufacturing process called as turning (a type of machining process) that is used to manufacture cylindrical metal s
Sav [38]

Answer: 86.64%

Step-by-step explanation:

Let x be a random variable that represents the diameter of metal samples.

Given : Population mean : \mu=10

Standard deviation: s=0.50

Specified tolerance on the diameter is 0.75 mm.

i.e. range of diameter = 10-0.75< x <10+0.75 = 9.25< x< 10.75

Formula to find the z-score corresponds to x: z=\dfrac{x-\mu}{s}

At x= 0.75,  z=\dfrac{9.25-10}{0.50}=-1.5

z=\dfrac{9.25-10}{0.50}=1.5

Using standard normal table for z-value,

P-value : p(-1.5

∴ Percentage of samples manufactured using this process satisfy the tolerance specification = 86.64%

6 0
3 years ago
Hardcover book sales for $24 at the bookmark men days of total of $25.02 for the book what is the sales tax rate
Ber [7]
25.02-24=1.02
1.02/24=0.0425
The sales tax rate is 4.25%.
3 0
3 years ago
Which inequalities are true? Select the four correct answers.
tiny-mole [99]

The correct anwer is option C and option E which are √4<√5<√√55 and √8<2.9<√√9.

<h3>What is inequality?</h3>

Inequality is the relationship between two expressions that are not equal, employing a sign such as ≠ “not equal to,” > “greater than,” or < “less than.”.

The two inequalities √4<√5<√√55 and √8<2.9<√√9. are true.

√4<√5<√√55  → 2 < 2.23 < 7.41

√8<2.9<√√9    → 2.82 < 2.9 < 3

Therefore correct answers is option C and option E which are √4<√5<√√55 and √8<2.9<√√9.

To know more about inequality follow

brainly.com/question/24372553

#SPJ1

6 0
2 years ago
The ratio of runners to swimmers was 3:2. When 28 runners dropped out of the competition, the ratio became 5:6. How many total s
xz_007 [3.2K]

The total of 42 swimmers are competing

Step-by-step explanation:

The given is:

  • The ratio of runners to swimmers was 3 : 2
  • When 28 runners dropped out of the competition, the ratio became 5 : 6

We need to find how many total swimmers are competing

∵ The ratio of runners to swimmers was 3 : 2

- Multiply each term of the ratio by x

∴ There are 3x runners and 2x swimmers

∵ 28 runners dropped out of the competition

∴ The number of runners who complete = 3x - 28

∵ The ratio became 5 : 6

- Equate the ratio between the new number of runners and the

   number of swimmers by 5 : 6

∵ The new number of runners = 3x - 28

∵ The number of swimmers = 2x

∴ \frac{3x-28}{2x}=\frac{5}{6}

- By using cross multiplication

∴ 6(3x - 28) = 5(2x)

- Simplify both sides

∴ 18x - 168 = 10x

- Add 168 to both sides

∴ 18x = 10x + 168

- Subtract 10x from both sides

∴ 8x = 168

- Divide both sides by 8

∴ x = 21

∵ The number of swimmers is 2x

∵ x = 21

∴ The number of swimmers = 2(21) = 42

The total of 42 swimmers are competing

Learn more:

You can learn more about the ratio in brainly.com/question/2707032

#LearnwithBrainly

7 0
3 years ago
Cost, Revenue, and Profit A company invests $98,000 for equipment to produce a new product. Each unit of the product costs $12.2
Vesnalui [34]

Given:

Investment on equipment = $98,000

Cost of each unit = $12.20

Selling price of each unit = $16.98.

To find:

(a) The total cost C as a function of x.

(b) The revenue R as a function of x.

(c) The profit P as a function of x.

Solution:

Let x be the number of units produced and sold.

We have,

Fixed cost = $98,000

Variable cost = $12.20x

Total cost = Fixed cost + Variable cost

C(x)=98000+12.20x

Therefore, the cost function is C(x)=98000+12.20x.

Selling price of each unit = Revenue from each unit =  $16.98.

Total revenue = Revenue from x units

R(x)=16.98x

Therefore, the revenue function is R(x)=16.98x.

Profit = Revenue - Cost

P(x)=R(x)-C(x)

P(x)=16.98x-(98000+12.20x)

P(x)=16.98x-98000-12.20x

P(x)=4.78x-98000

Therefore, the profit function is P(x)=4.78x-98000.

6 0
4 years ago
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