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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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To find : List the sample space and tell whether you think the events are equally likely ?

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b) A family has 3 children; record the number of boys.

Let B denote boy and G denote girl.

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{GGG,GGB,GBG,BGG,BBG,GBB,BGB,BBB}

The possible number of boys are 0,1,2 and 3.

Number of boys      Favorable outcome    Probability

           0                      GGG                        \frac{1}{8}

           1                    GGB,GBG,BGG          \frac{3}{8}

           2                   GBB,BGB,BBG           \frac{3}{8}

           3                       BBB                         \frac{1}{8}

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Getting a head in a trial is dependent on the previous toss.

Similarly getting 3 consecutive tails also dependent on previous toss.

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d) Roll two dice; record the larger number

The sample space of rolling two dice is

(1,1) (1,2) (1,3) (1,4) (1,5) (1,6)

(2,1) (2,2) (2,3) (2,4) (2,5) (2,6)

(3,1) (3,2) (3,3) (3,4) (3,5) (3,6)

(4,1) (4,2) (4,3) (4,4) (4,5) (4,6)

(5,1) (5,2) (5,3) (5,4) (5,5) (5,6)

(6,1) (6,2) (6,3) (6,4) (6,5) (6,6)

Now we form a table that the number of time each number occurs as maximum number then we find probability,

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           1                                   1                     \frac{1}{36}

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           3                                  5                    \frac{5}{36}

           4                                  7                    \frac{7}{36}

           5                                  9                    \frac{9}{36}

           6                                  11                    \frac{11}{36}

Since the probabilities are not the same the events are not equally likely.

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