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Alisiya [41]
3 years ago
7

Using the merge method of the Map interface, which statement correctly updates the salesTotalByDept map of type Map to update th

e sales total for a dept by an integer sales value? Question 3 options: salesTotalByDept.merge(dept, sales,(d,s) -> s + 1); sales.merge(salesTotalByDept, dept, (d,s) -> s + sales); dept.merge(salesTotalByDept, sales, (d,s) -> s + sales); salesTotalByDept.merge(dept, sales, (d,s) -> s + sales);
Computers and Technology
1 answer:
Alexxandr [17]3 years ago
7 0

Answer:

salesTotalByDept.merge(dept, sales, (d,s) -> s + sales);

Explanation:

In the above statement merge() method is being used which updates the salesTotalByDept map of type Map to update the sales total for a dept by an integer sales value.

Here dept and sales both are used and passed as a parameter to merge() method because salesTotalByDept can be update if both the information about sales and department are provided.

Map.merge() method is used to place new value under the given key if the key does not correspond to any value, or is used to update existing key with a given value.

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The exercise is about filling in the gaps and is related to the History of the ARPANET.

<h3>What is the History of the ARPANET?</h3>

From the text:

In 1972, earlier designers built the <u>ARPANET </u>connecting major universities. They broke communication into smaller chunks, or <u>packets </u>and sent them on a first-come, first-serve basis. The limit to the number of bytes of data that can be moved is called line capacity, or <u>bandwidth</u>.

When a network is met its capacity the user experiences <u>unwanted pauses</u>. When the network is "slowing down", what is happening is users are waiting for their packet to leave the <u>queue</u>.

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Find the propagation delay for a signal traversing the in a metropolitan area through 200 km, at the speed of light in cable (2.
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Answer:

t= 8.7*10⁻⁴ sec.

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If the signal were able to traverse this distance at an infinite speed, the propagation delay would be zero.

As this is not possible, (the maximum speed of interactions in the universe is equal to the speed of light), there will be a finite propagation delay.

Assuming that the signal propagates at a constant speed, which is equal to 2.3*10⁸ m/s (due to the characteristics of the cable, it is not the same as if it were propagating in vaccum, at 3.0*10⁸ m/s), the time taken to the signal to traverse the 200 km, which is equal to the propagation delay, can be found applying the average velocity definition:

v = \frac{(xf-xo)}{(t-to)}

If we choose x₀ = 0 and t₀ =0, and replace v= 2.3*10⁸ m/s, and xf=2*10⁵ m, we can solve for t:

t =\frac{xf}{v}  =\frac{2e5 m}{2.3e8 m/s} =8.7e-4 sec.

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