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Harman [31]
3 years ago
6

A company purchased a weaving machine for $190,000. The machine has a useful life of 8 years and a residual value of $10,000. It

is estimated that the machine could produce 75,000 bolts of woven fabric over its useful life. In the first year, 15,000 bolts were produced. In the second year, production increased to 19,000 units. Using the units-of-production method, what is the amount of accumulated depreciation at the end of the second year?
Mathematics
1 answer:
Dafna1 [17]3 years ago
5 0

Answer:

$45,600

Step-by-step explanation:

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Landen Company uses a single plantwide overhead rate of $100 per direct labor hour (DLH). Landen has two products as follows: ba
KATRIN_1 [288]

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$5,500

Step-by-step explanation:

Since Landen Company uses a single overhead rate of $100 per direct labor hour, the total amount allocated to the deluxe and basic chairs is given by the sum of the DLH used up for both products multiplied by the overhead rate:

A =(20+35)*\$100\\A=\$5,500

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8 0
3 years ago
Suppose the number of radios in a household has a binomial distribution with parameters n=11 and p=40%. Find the probability of
Ivenika [448]

Answer:

Step-by-step explanation:

The formula for binomial distribution is expressed as

P(x=r) = nCr × q^(n-r) × p^r

From the information given,

n = 11

p = 40% = 40/100 = 0.4

q = 1 - 0.4 = 0.6

x represent the number of radios

a) P( x = 1) or P(x = 9)

P(x = 1) = 11C1 × 0.6^(11-1) × 0.4^1

P(x = 1) = 0.027

P(x = 9) = 11C9 × 0.6^(11-9) × 0.4^9

P(x = 9) = 0.0052

P( x = 1) or P(x = 9) = 0.027 + 0.0052 = 0.0322

b) P(x lesser than or equal to 7) = P(x = 0) + P(x = 1) + P(x = 2) + P(x = 3) + P(x = 4) + P(x = 5) + P(x = 6)

P(x = 0) = 11C0 × 0.6^(11-0) × 0.4^0 = 0.004

P(x = 1) = 0.027

P(x = 2) = 11C2 × 0.6^(11-2) × 0.4^2 = 0.089

P(x = 3) = 11C3 × 0.6^(11-3) × 0.4^3 = 0.177

P(x = 4) = 11C4 × 0.6^(11-4) × 0.4^4 = 0.24

P(x = 5) = 11C5 × 0.6^(11-5) × 0.4^5 = 0.22

P(x = 6) = 11C6 × 0.6^(11-6) × 0.4^6 = 0.15

P(x = 7) = 11C7 × 0.6^(11-7) × 0.4^7 = 0.15

P(x lesser than or equal to 7) = 0.004 + 0.027 + 0.089 + 0.177 + 0.24 + 0.22 + 0.15 + 0.07 = 0.977

c) P(x greater than or equal to 5) = 1 - P(x lesser than or equal to 4) = 1 - (0.004 + 0.027 + 0.089 + 0.177 + 0.24) = 1 - 0.537 = 0.463

d) P(x lesser than 9) = P(x lesser than or equal to 7) + P(x = 8)

P(x = 8) = 11C8 × 0.6^(11-8) × 0.4^8 = 0.02

P(x lesser than 9) = 0.977 + 0.02 = 0.997

e. P(x greater than 7) = 1 - P(x lesser than or equal to 7) = 1 - 0.977 = 0.023

5 0
3 years ago
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