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solniwko [45]
3 years ago
9

A machine can produce 198 bobbles every 5 days. If an order comes in for 2200, how many days will that take to complete?

Mathematics
2 answers:
Ede4ka [16]3 years ago
5 0

Answer:

55.5 (56 if Rounded)

Step-by-step explanation:

198/5 (39.6) = 1 Day

2200/39.6 = 55.5 Days

nadya68 [22]3 years ago
3 0

Answer:

Not sure use a math word problem solver online

Step-by-step explanation:

......

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irinina [24]
What is the amount of cans so i can answer
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3 years ago
Find the product: (x² - x + 1)(2x² + 3x + 2).
Tresset [83]

Answer:

(2x^3-2x^2-12x) is the required product.

Step-by-step explanation:

The given equation is:

2x(x-3)(x+2)

Solving the above given equation, we get

=(2x^2-6x)(x+2)

which can be written as:

=(2x^3-6x^2+4x^2-12x)

Solving the like terms, we get

=(2x^3-2x^2-12x)

which is the required product of the given expression.

3 0
4 years ago
4. Kevin has $24 to buy a gift for his cousin.
Roman55 [17]

Answer:  Yes, he would have enough money. In total he would spend  $23.10

Step-by-step explanation

Add 5% to the $24

4 0
3 years ago
If $15,000 is invested at an interest rate of 2% per year, compounded semiannually, find the value of the investment after the g
Gennadij [26K]

Answer:

Step-by-step explanation:

If $10,000 is invested at an interest rate of 2% per year, compounded semiannually, find the value of the investment after the given number of years. (Round your answers to the nearest cent.)

a)6 years

b)12 years

c)18 years

***

compound interest formula: A=P(1+r/n)^nt, P=initial investment, r=interest rate, n=number of compounding periods per year, A=amt after t-years.

For given problem:

P=10000

r=.02

n=2

t=6, 12, 18

..

A(6)=10000(1+.02/2)^2*6

A(6)=10000(1.01)^12=11,268.25

A(12)=10000(1.01)^24=12,697.35

A(18)=10000(1.01)^36=14,307.69

3 0
3 years ago
Please can somebody help me with this question please...!
olga55 [171]
Since you are given that the student registered early, the total number you deal with is all students who registered early.

211 + 329 = 540

number of undergraduates who registered early = 211

Among students who registered early:

p(undergraduate) = 211/540
4 0
4 years ago
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