3 years ago

How do I cancel my subscription

Engineering

2 answers:

Aleksandr [31]3 years ago

7 0

Hey, it depends what you subscribed to, or what platform you are trying to unsubscribe from?

Ksju [112]3 years ago

5 0

Answer:

it just depends on what you subscribed to and what aplication....

Explanation:

You might be interested in

Jade wanted to test the effect of ice on the weathering of rocks. She filled two containers with gypsum and placed a water ballo

Aleks [24]

Answer:

The gypsum block with the water balloon has contracted.

Explanation:

7 0

2 years ago

A computer has a two-level cache. Suppose that 60% of the memory references hit on the first level cache, 35% hit on the second

Andreyy89

Answer:

t=14ns

Explanation:

We make the relation between the specific access time and the memory percentage in each level, so

$60\% \Rightarrow 60/100 = 0.60\\35\% \Rightarrow 35/100 = 0.35\\05\% \Rightarrow 05/100 = 0.05$

$t= 0.6(5) + 0.35(5+15) + 0.05(5+15+60)\\t= 0.6(5) + 0.35(20) + 0.05(80)\\t= 3 + 7 + 4\\t= 14 ns$

Average Access Time is 14 nsec.

4 0

3 years ago

If a population has no predadors and plenty of available resources, how might that population change

ExtremeBDS [4]

Answer:

B (exponential growth )

Explanation: wish u the best bby <33

4 0

2 years ago

When using fall arrest, free fall must be kept at or below how many feet

FromTheMoon [43]

6 feet
Explanation: A few extra feet of free fall can significantly increase the arresting force on the employee, possibly to the point of causing injury. Because of this, the free fall distance should be kept at a minimum, and, as required by the standard, in no case greater than 6 feet (1.8 m).

5 0

2 years ago

In a cellular phone system, a mobile phone must be paged to receive a phone call. However, paging attempts don’t always succeed

Alina [70]

Answer:

The correct response will be "0.992". The further explanation to the following question is given below.

Explanation:

The probability that paging would be beneficial becomes 0.8

Effective paging at the very first attempted is 0.8

On the second attempt the success probability will be:

⇒ $0.2\times 0.8$