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2 years ago

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A printer needs 12,000 good copies of a flyer for a particular job. Previous experience has shown that she can expect no more th

an a 5% spoilage rate on this type of job. How many should she print to ensure 12,000 clean copies?

1 answer:

PilotLPTM [1.2K]2 years ago

8 0

Answer:

12600

Step-by-step explanation:

5% of 12000 is 600 so just add them

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In a class 26 of students, 17 have a brother and 7 have a sister. There are 4 students who do not have any siblings. What is the

Anni [7]

Number of students = 26
Number of students without siblings = 4
⇒ Number of students with at least one siblings = 26 - 4 = 22

Number of students with brothers = 17
Number of students with sisters = 7
⇒ Number of students with both a brother and a sister = 17 + 7 - 22 = 2

P(student has a brother, given that they have a sister) = 2/7

Answer: 2/7

6 0

2 years ago

A. Mr. Gordon works with Ms. Patel from the Example. Mr. Gordon writes the expression 3x + 4x to represent the cost of ordering

Leviafan [203]

Answer:

$162

Step-by-step explanation:

5r=20 10p=100 14t=42

100+42+20=162

5r+10p+14t=$162

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2 years ago

Solve these linear equations in the form y=yn+yp with yn=y(0)e^at.

WINSTONCH [101]

Answer:

a) $y(t) = y_{0}e^{4t} + 2$ . It does not have a steady state

b) $y(t) = y_{0}e^{-4t} + 2$ . It has a steady state.

Step-by-step explanation:

a) $y' -4y = -8$

The first step is finding $y_{n}(t)$ . So:

$y' - 4y = 0$

We have to find the eigenvalues of this differential equation, which are the roots of this equation:

$r - 4 = 0$

$r = 4$

So:

$y_{n}(t) = y_{0}e^{4t}$

Since this differential equation has a positive eigenvalue, it does not have a steady state.

Now as for the particular solution.

Since the differential equation is equaled to a constant, the particular solution is going to have the following format:

$y_{p}(t) = C$

So

$(y_{p})' -4(y_{p}) = -8$

$(C)' - 4C = -8$

C is a constant, so (C)' = 0.

$-4C = -8$

$4C = 8$

$C = 2$

The solution in the form is

$y(t) = y_{n}(t) + y_{p}(t)$

$y(t) = y_{0}e^{4t} + 2$

b) $y' +4y = 8$

The first step is finding $y_{n}(t)$ . So:

$y' + 4y = 0$

We have to find the eigenvalues of this differential equation, which are the roots of this equation:

$r + 4 =$

$r = -4$

So:

$y_{n}(t) = y_{0}e^{-4t}$

Since this differential equation does not have a positive eigenvalue, it has a steady state.

Now as for the particular solution.

Since the differential equation is equaled to a constant, the particular solution is going to have the following format:

$y_{p}(t) = C$

So

$(y_{p})' +4(y_{p}) = 8$

$(C)' + 4C = 8$

C is a constant, so (C)' = 0.

$4C = 8$

$C = 2$

The solution in the form is

$y(t) = y_{n}(t) + y_{p}(t)$

$y(t) = y_{0}e^{-4t} + 2$

6 0

3 years ago

A basketball player scored 26 points in one game. In basketball, some baskets are worth 3 points, some are worth 2 points, and f

Vsevolod [243]

Answer:

3 points = 2

2 points = 6

1 point = 8

Step-by-step explanation:

Given that:

Total point scored = 26

Let number of 3 point basket = x

Number of 2 point basket = x + 4

Number of free throws = x + x + 4 = 2x + 4

Hence,

Total 3 points = 3x

Total 2 points 2(x + 4)

Total free throw points = 2x + 4

3x + 2(x + 4) + 2x + 4 = 26

3x + 2x + 8 + 2x + 4 = 26

7x + 12 = 26

7x = 26 - 12

7x = 14

x = 2

Number of 3 points = x = 2

2 points = (x+4) = 2+4 = 6

1 point = (2x+ 4) = 2(2) + 4 = 8

5 0

2 years ago

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Solve-<br>29-(20-18)×(58×74)

serious [3.7K]

Answer: -4790

Step-by-step explanation:

$= 29 - (20 - 18) \times (58 \times 74) \\ = 29 - 2 \times 4592 \\ = 29 - 4819 \\ \ = - 4790$

8 0

2 years ago

Read 2 more answers