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Furkat [3]
3 years ago
14

A tree is 7 ft tall when it is planted and this type of tree grows two feet per year. How tall will the tree be in 6 years? How

long will it take for the tree to reach 44ft?
Mathematics
1 answer:
s344n2d4d5 [400]3 years ago
4 0

2x+7=?

X=6

2(6)+7=

12+7=19

In six years, the tree will be 19ft

2x+7=44

Minus seven on both sides

2x=37

Divide two from both sides

X=18.5

Thus, it will take 18 and a half years for the tree to reach 44ft

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2. A solar lease customer built up an excess of 6,500 kilowatt hours (kwh) during the summer using his solar
tia_tia [17]

9514 1404 393

Answer:

  a)  E = 6500 -50d

  b)  5000 kWh

  c)  the excess will last only 130 days, not enough for 5 months

Step-by-step explanation:

<u>Given</u>:

  starting excess (E): 6500 kWh

  usage: 50 kWh/day (d)

<u>Find</u>:

  a) E(d)

  b) E(30)

  c) E(150)

<u>Solution</u>:

a) The exces is linearly decreasing with the number of days, so we have ...

  E(d) = 6500 -50d

__

b) After 30 days, the excess remaining is ...

  E(30) = 6500 -50(30) = 5000 . . . . kWh after 30 days

__

c) After 150 days, the excess remaining would be ...

  E(150) = 6500 -50(150) = 6500 -7500 = -1000 . . . . 150 days is beyond the capacity of the system

The supply is not enough to last for 5 months.

3 0
2 years ago
The number of messages arriving at a multiplexer is a Poisson random variable with mean 15 messages/second. Use the central limi
Mila [183]

Answer:

0.048 is the probability that more than 950 message arrive in one minute.

Step-by-step explanation:

We are given the following information in the question:

The number of messages arriving at a multiplexer is a Poisson random variable with mean 15 messages/second.

Let X be the number of messages arriving at a multiplexer.

Mean = 15

For poison distribution,

Mean = Variance = 15

\text{Standard Deviation} = \sqrt{\text{Variance}} = \sqrt{15} = 3.872

From central limit theorem, we have:

z = \displaystyle\frac{x-n\mu}{\sigma\sqrt{n}}

where n is the sample size.

Here, n = 1 minute = 60 seconds

P(x > 950)

P( x > 950) = P( z > \displaystyle\frac{950 - (60)(15)}{\sqrt{(15)(60)}}) = P(z > 1.667)

= 1 - P(z \leq 1.667)

Calculation the value from standard normal z table, we have,  

P(x > 950) = 1 - 0.952 = 0.048

0.048 is the probability that more than 950 message arrive in one minute.

3 0
3 years ago
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Anika [276]
It won't. Both are increasing by 1.5 every year. If you graph that, the lines will be parallel. Parallel lines have no intercept.
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Answer: 1649

Step-by-step explanation:

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