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aev [14]
2 years ago
15

Ryan has just baked 27 cookies. He wants to put them in paper bags in groups of 6. How many groups of 6 can he make? How many he

have remaining? Number of groups of 6 he can make = Number of cookies remaining =​
Mathematics
2 answers:
lesantik [10]2 years ago
4 0

Answer:

4 and 3 remaining

Step-by-step explanation:

24 is a multiple of 6 27 is not

rodikova [14]2 years ago
3 0

Answer:

He can make 4 groups of 6

He has 24 remaining

Number of cookies remaining are 2  

Step-by-step explanation:

i hope its right

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Andrew, who operates a laundry business, incurred the following expenses during the year. ∙ Parking ticket of $250 for one of hi
marysya [2.9K]

Answer:

Option d: $0

Step-by-step explanation:

Parking penalties arent deductible from expenses, so Andrew cant deduct neither the 250 for his delivery van nor the 75 for the rock concert. Dui tickets definetely arent deductable from expenses, so Andrew cant deduct the 500 from the ticket either. That gives us only 2 options: Andrew can deduct $600 (if he is able to deduct from attoeney's fee), else he cant deduct anything.

However, the attorney's fee shoudn't be deductible, because as a general rule, personal legal fees are not deductible. So the answer is option d, $0

7 0
2 years ago
Imagine you have some workers and some handheld computers that you can use to take inventory at a warehouse. There are diminishi
Svet_ta [14]

Answer:

a. $1.03

b. $0.93

c. 0.98

d. 2 workers

Step-by-step explanation:

a. Given that:

  • 1 computer : 1 worker : inventory 150 items per hour
  • 1 computer : 2 workers : inventory 200 items per hour
  • 1 computer : 3 workers : inventory 220 items per hour
  • 1 computer : 4+ workers : fewer than 235 items per hour
  • Cost: $100 per computer ; $25 per worker

The fixed production factor in the warehouse is the computer used:

-One computer used, but the number of users is varied to inventory a specified number of items.

-The variable production factor is the number of workers assigned per one computer.

#The cost of inventorying a single item by one worker is:

Cost=\frac{C_{pc}+Wage}{Items} \ , C_{pc}=\$125\\\\Cost_1=\frac{125+30}{150}\\\\\\=1.03

Hence, the cost of inventorying a single item is $1.03

b. Using the information provided above, the cost of inventorying a single item when two workers are assigned is :

Cost=\frac{C_{pc}+Wage}{Items} \ , C_{pc}=\$125\\\\Cost_2=\frac{125+2\times30}{200}\\\\\\=0.925

Hence, the cost of inventorying a single item is $0.93

c.Using the information provided above, the cost of inventorying a single item when three workers are assigned is :

Cost=\frac{C_{pc}+Wage}{Items} \ , C_{pc}=\$125\\\\Cost_3=\frac{125+3\times30}{220}\\\\\\=0.98

Hence, the cost of inventorying a single item is $0.98

d. To determine the most cost-effective job assignment, we calculate the cost of 4+ workers.

Take any number less than 235(say 234) as the inventory units:

Cost=\frac{C_{pc}+Wage}{Items} \ , C_{pc}=\$125\\\\Cost_4=\frac{125+4\times30}{234}\\\\\\=1.05

From our calculations, it's clear that two workers per computer costs the least amount($0.93) per unit item. Hence, it is best to assign two workers per computer.

4 0
3 years ago
HELP PLEASE! WILL GIVE BRAINLIEST TO THE BEST ANSWER!
RSB [31]
A:7 hope with helps :)
7 0
2 years ago
Select the correct answer.
vivado [14]

Answer:

B will be the answer...

Step-by-step explanation:

The second equation in system B is only in terms of y, so we need to use elimination to eliminate the x term from the second equation in system A.

To do that, we need to multiply the first equation by 5.

5 (-x − 2y = 7)

-5x − 10y = 35

Add to the second equation.  Notice the x terms cancel out.

(-5x − 10y) + (5x − 6y) = 35 + (-3)

-16y = 32

Combining this new equation with the first equation from system A will get us system B.

-x − 2y = 7

-16y = 32

4 0
1 year ago
I need help ASAP!
Readme [11.4K]

Answer:

i) Equation can have exactly 2 zeroes.

ii) Both the zeroes will be real and distinctive.

Step-by-step explanation:

x^{2} - 7x + 3 is the given equation.

It is of the form of quadratic equation a^{2} + bx + c and highest degree of the polynomial is 2.

Now, FUNDAMENTAL THEOREM OF ALGEBRA

If P(x) is a polynomial of degree n ≥ 1, then P(x) = 0 has exactly n roots, including multiple and complex roots.

So, the equation can have exact 2 zeroes (roots).

Also, find discriminant D = b^{2}  - 4ac = (-7)^{2}  - 4(1)(3) = 49 - 12 = 37

⇒ D = 37

Here, since D > 0, So both the roots will be real and distinctive.

7 0
2 years ago
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