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Yakvenalex [24]
3 years ago
7

Suppose a company did $2,000,000 in annual maintenance in 2013 and expects 80% of those to renew for 2014. Suppose that product

sales for 2013 were 2,000,000(which include free maintenance in 2013) and 75% of those were expected to pay annual maintenance of 10% of the purchase price in 2014. What will be the annual maintenance collected in 2014?
Please help me. I'll mark the best answer brainiest- just HALPPP
Mathematics
1 answer:
Paha777 [63]3 years ago
6 0

Answer:

$1,750,000

Step-by-step explanation:

The company did $2,000,000 in annual maintenance in 2013 and expects 80% of those to renew for 2014.

That will result in annual maintenance of:

(80 / 100) * 2000000 = $1,600,000

The product sales for 2013 were 2,000,000 (which include free maintenance in 2013) and 75% of those were expected to pay annual maintenance of 10% of the purchase price in 2014.

That will result in annual maintenance of:

(75 / 100) * (10 / 100) * 2000000 = $150,000

Therefore, the total annual maintenance will be:

$1600000 + $150000 = $1,750,000

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12 = 6(n – 7)<br> Solve 2 -step equation
madam [21]

Answer:

9 = n

Step-by-step explanation:

12 = 6(n – 7)

Divide by 6 on both sides

12/6 = 6/6(n – 7)

2 = n-7

Add 7 to each side

2+7 = n-7+7

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4 years ago
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The radius of the base of a cylinder is increasing at a rate of 7 millimeters per hour. The height of the cylinder is fixed at 1
Ilya [14]

Answer:

The rate of change of the volume of the cylinder at that instant = 791.28\ mm^3/hr

Step-by-step explanation:

Given:

Rate of increase of base of radius of base of cylinder = 7 mm/hr

Height of cylinder = 1.5 mm

Radius at a certain instant = 12 mm

To find rate of change of volume of cylinder at that instant.

Solution:

Let r represent radius of base of cylinder at any instant.

Rate of increase of base of radius of base of cylinder can be given as:

\frac{dr}{dt}=7\ mm/hr

Volume of cylinder is given by:

V=\pi\ r^2h

Finding derivative of the Volume with respect to time.

\frac{dV}{dt}=\pi\ h\ 2r\frac{dr}{dt}

Plugging in the values given:

\frac{dV}{dt}=\pi\ (1.5)\ 2(12)(7)

\frac{dV}{dt}=252\pi

Using \pi=3.14

\frac{dV}{dt}=252(3.14)

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Thus rate of change of the volume of the cylinder at that instant = 791.28\ mm^3/hr

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3 years ago
Jaz was 43 inches tall. Eighteen months later she was 52 inches tall. Find the constant rate of change of jaz's height
notka56 [123]

Answer:

.5 inches a month


Step-by-step explanation:

52 - 43 = 9 inches


9/18 = .5



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