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AleksandrR [38]
3 years ago
15

What major financial flop led to the end of the Sega Dreamcast and ultimately caused Sega to stop making game consoles altogethe

r?
1: The founder and CEO of Sega was found to be secretly skimming money off of the top of profits, leading to widespread distrust by the public and a sharp decline in sales until they ultimately had to shut down due to making no profit.

2: A small group of employees found a way to drain all of the Sega Corporation funding accounts and flee the country with all of the money, never to be heard from again.

3: A game called Shenmue, that cost more than $70 million to make, meant that everyone who owned a Dreamcast needed to buy two copies of the game just for Sega to make back the money they had spent to develop it-which didn't happen.

4: A game called Katamari Damacy that carried a virus that would infect any console that it was played on forced Sega to spend millions of dollars in refunds and bankrupted the company.​
Engineering
1 answer:
icang [17]3 years ago
6 0

Answer:

The major financial flop that led to the end of the Sega Dreamcast and ultimately caused Sega to stop making game consoles altogether is:

3: A game called Shenmue, that cost more than $70 million to make, meant that everyone who owned a Dreamcast needed to buy two copies of the game just for Sega to make back the money they had spent to develop it-which didn't happen.

Explanation:

Shenmue, released on December 29, 1999, was created for Dreamcast by Suzuki.  It was widely described as the most expensive video game ever produced.  It had an estimated production and marketing cost of between US$47 and $70 million, according to the latest available data.

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Search up A gardener can increase the number of dahlia plants in an annual garden by either buying new bulbs each year or dividing the existing bulbs to create new plants . The table below shows the expected number of bulbs for each method

Part A
For each method,a function to model the expected number of plants for each year

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Use the Functions to Find the expected number of plants in 10 years for each method.

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What is a magnitute?
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Answer:

Magnitude is a specific type of norm. Magnitude is what is known as the Euclidean norm. There are other norms, such as the Leibniz and the Chebychev norm.

Explanation:

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2 years ago
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Suppose that the time (in hours) required to repair a machine is an exponentially distributed random variable with parameter ???
Andrej [43]

We need to define the variables,

So,

F_x (x) = 1-e^{-\lambda x}\\F_x (x) = 1-e^{-0.5x}

Therefore, the probability that the repair time is more than 4 horus can be calculate as,

P(x>4)=1-P(x4)= 1-F_x(4)\\P(x>4) = 1-e^{-0.5*4}\\P(x>4) = 1-0.98\\P(x>4) = 0.018

The probability that the repair time is more than 4 hours is 0.136

b) The probability that repair time is at least 12 hours given that the repair time is more than 7 hoirs is calculated as,

P(x\geq 12|x>7)=P(X\geq7+5|x>7)\\P(x\geq12|x>7)=P(X\geq5)\\P(x\geq12|x>7)=1-P(x\leq 5)\\P(x\geq12|x>7)=1-e^{-0.5(2)}

P(x\geq 12|x>7)=0.6321

The probability that repair time is at least 12 hours given that the repair time is more than 7 hours is 0.63

3 0
3 years ago
A weighted, frictionless piston-cylinder device initially contains 5.25 kg of R134a as saturated vapor at 500 kPa. The container
kykrilka [37]

Answer:

-6.326 KJ/K

Explanation:

A) the entropy change is defined as:

delta S_{12}=\int\limits^2_1  \, \frac{dQ}{T}

In an isobaric process heat (Q) is defined as:

Q= m*Cp*dT

Replacing in the equation for entropy  

delta S_{12}=\int\limits^2_1 \frac{m*Cp*dT}{T}

m is the mass and Cp is the specific heat of R134a. We can considerer these values as constants so the expression for entropy would be:  

delta S_{12}= m*Cp*\int\limits^2_1 \frac{ dT }{T}  

Solving the integral we get the expression to estimate the entropy change in the system  

delta S_{12}= m*Cp *ln(\frac{T_{2}}{T_{1}})

The mass is 5.25 Kg and Cp for R134a vapor can be consulted in tables, this value is 0.85\frac{kJ}{Kg*K}

We can get the temperature at the beginning knowing that is saturated vapor at 500 KPa. Consulting the thermodynamic tables, we get that temperature of saturation at this pressure is: 288.86 K

The temperature in the final state we can get it from the heat expression, since we know how much heat was lost in the process (-976.71 kJ). By convention when heat is released by the system a negative sign is used to express it.

Q= m*Cp*dT

With dt=T_{2}-T_{1} clearing for T2 we get:

T_{2}=\frac{Q}{m*Cp}+T1= \frac{-976.71kJ}{5.25Kg*0.85\frac{kJ}{Kg*K}}+288.86 K =69.98 K

Now we can estimate the entropy change in the system

delta S_{12}= m*Cp*ln(\frac{T_{2}}{T_{1}})= 5.25Kg*0.85\frac{kJ}{Kg*K}*ln(\frac{69.98}{288.861})= -6.326\frac{kJ}{K}

The entropy change in the system is negative because we are going from a state with a lot of disorder (high temperature) to one more organize (less temperature. This was done increasing the entropy of the surroundings.  

b) see picture.

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Answer: the correct answer is D

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