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3 years ago

7

Depreciation is the decrease in value of an item. Three years ago, Aaron bought a computer for $930. The computer is now worth $

570. Assuming the value of the computer depreciates linearly, what is its yearly depreciation?

1 answer:

Umnica [9.8K]3 years ago

7 0

Answer:

$120

Step-by-step explanation:

930-570=360 (Price difference in 3 years)

360 divided by 3= 120 (Price difference for each of the 3 years)

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businessText message users receive or send an average of 62.7 text messages per day. How many text messages does a text message

KiRa [710]

Answer:

(a) The probability that a text message user receives or sends three messages per hour is 0.2180.

(b) The probability that a text message user receives or sends more than three messages per hour is 0.2667.

Step-by-step explanation:

Let <em>X</em> = number of text messages receive or send in an hour.

The random variable <em>X</em> follows a Poisson distribution with parameter <em>λ</em>.

It is provided that users receive or send 62.7 text messages in 24 hours.

Then the average number of text messages received or sent in an hour is: $\lambda=\frac{62.7}{24}= 2.6125$ .

The probability of a random variable can be computed using the formula:

$P(X=x)=\frac{e^{-\lambda}\lambda^{x}}{x!} ;\ x=0, 1, 2, 3, ...$

(a)

Compute the probability that a text message user receives or sends three messages per hour as follows:

$P(X=3)=\frac{e^{-2.6125}(2.6125)^{3}}{3!} =0.21798\approx0.2180$

Thus, the probability that a text message user receives or sends three messages per hour is 0.2180.

(b)

Compute the probability that a text message user receives or sends more than three messages per hour as follows:

P (X > 3) = 1 - P (X ≤ 3)

= 1 - P (X = 0) - P (X = 1) - P (X = 2) - P (X = 3)

$=1-\frac{e^{-2.6125}(2.6125)^{0}}{0!}-\frac{e^{-2.6125}(2.6125)^{1}}{1!}-\frac{e^{-2.6125}(2.6125)^{2}}{2!}-\frac{e^{-2.6125}(2.6125)^{3}}{3!}\\=1-0.0734-0.1916-0.2503-0.2180\\=0.2667$

Thus, the probability that a text message user receives or sends more than three messages per hour is 0.2667.

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3 years ago

The graph illustrates a normal distribution for the prices paid for a particular model of HD television. The mean price paid is

marysya [2.9K]

Answer:

(a) 0.14%

(b) 2.28%

(c) 48%

(d) 68%

(e) 34%

(f) 50%

Step-by-step explanation:

Let <em>X</em> be a random variable representing the prices paid for a particular model of HD television.

It is provided that <em>X</em> follows a normal distribution with mean, <em>μ</em> = $1600 and standard deviation, <em>σ</em> = $100.

(a)

Compute the probability of buyers who paid more than $1900 as follows:

$P(X>1900)=P(\frac{X-\mu}{\sigma}>\frac{1900-1600}{100})$

$=P(Z>3)\\=1-P(Z$

*Use a <em>z</em>-table.

Thus, the approximate percentage of buyers who paid more than $1900 is 0.14%.

(b)

Compute the probability of buyers who paid less than $1400 as follows:

$P(X$

$=P(Z$

*Use a <em>z</em>-table.

Thus, the approximate percentage of buyers who paid less than $1400 is 2.28%.

(c)

Compute the probability of buyers who paid between $1400 and $1600 as follows:

$P(1400$

$=P(-2$

*Use a <em>z</em>-table.

Thus, the approximate percentage of buyers who paid between $1400 and $1600 is 48%.

(d)

Compute the probability of buyers who paid between $1500 and $1700 as follows:

$P(1500$

$=P(-1$

*Use a <em>z</em>-table.

Thus, the approximate percentage of buyers who paid between $1500 and $1700 is 68%.

(e)

Compute the probability of buyers who paid between $1600 and $1700 as follows:

$P(1600$

$=P(0$

*Use a <em>z</em>-table.

Thus, the approximate percentage of buyers who paid between $1600 and $1700 is 34%.

(f)

Compute the probability of buyers who paid between $1600 and $1900 as follows:

$P(1600$

$=P(0$

*Use a <em>z</em>-table.

Thus, the approximate percentage of buyers who paid between $1600 and $1900 is 50%.

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Answer:

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