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Alinara [238K]
3 years ago
7

If the required direct materials purchases are 24,000 pounds, the direct materials required for production is three times the di

rect materials purchases, and the beginning direct materials are three and a half times the direct materials purchases, what are the desired ending direct materials in pounds?
A. 60,000
B. 12,000
C. 36,000
D. 24,000
Mathematics
1 answer:
ivolga24 [154]3 years ago
6 0

Answer:

explained

Step-by-step explanation:

Desired ending direct material

Particulars                                                                       pounds

Beginning direct material (24000*3.5)                             84,000

Add: purchases                                                                  24,000

Direct material available for production                           1,08,000

Less: Direct material required for production                  72,000

Desired ending direct material                                          36,000

therefore, the desire ending direct material are 36,000 pounds

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The population of a town is 13,000. It decreases at a rate of 5% per year. In about how many years will the population be less t
just olya [345]

This is an exponential decay problem.

Using the equation Y = a *(1-rate)^time

where Y is the future value given as 12,000 and a is the starting value given as 13,000.

The rate is also given as 5%.

The equation becomes:

12,000 = 13,000(1-0.05)^x

12,000 = 13,000(0.95)^x

Divide each side by 13000:

12000/13000 = 0.95^x/13000

12/13 = 0.95^x

Use the natural log function:

x = ln(12/13) / ln(0.95)

x = 1.56 years. ( this will equal 12,000

Round to 2 years it will be less than 12000.

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3 years ago
If there are 10 billion transistors (each of a square shape) uniformly (without space) distributed on a silicon chip with the sq
Karo-lina-s [1.5K]

Answer:

the transistors have L=1 mm of linear size

Step-by-step explanation:

For the silicon chip the area is A=1 cm² and for the transistors the area is At=L² (L=linear size) . Then since N= 10 billion transistors of area At should fit in the area A

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solving for L

L= √(A/N)

Assuming that 1 billion=10⁹ (short scale version of billion), then

L= √(A/N) = √(1 cm²/10⁹) = 1 cm / 10³ = 1 mm

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3 years ago
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