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3 years ago

In microsoft word, when you highlight existing text you want to replace, you’re in

1 answer:

6 0

Hello <span>Christinamauffp2olwu</span><span>

Answer: In Microsoft Word, when you highlight existing text you want to replace, you're in insert mode.

Hope This Helps :-)
-Chris</span>

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All of the following are methods to improve the communication of your key messages EXCEPT:

uysha [10]

Answer:

Fill the slides with images and animations

Explanation:

these will most likely distract the audience

6 0

2 years ago

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Using the following table as your starting point for each algorithm SJN Given the following information: Job: A B C D E Arrival

victus00 [196]

Answer:

Please check the attachment,

The average turnaround time is 4.8 and average waiting time is 1.

Please note: Exit time is same as completion time.

Explanation:

Please check the attachment.

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2 years ago

What your roblox user >:)

sweet-ann [11.9K]

Answer:

whats yours?

Explanation:

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2 years ago

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What is the common name for the growth rate function N?

kolezko [41]

Answer:

Linear.

Explanation:

The common name for the growth rate function N is linear.N growth rate is directly proportional to the input.So it grows linearly.It does not grow rapidly like other growth functions like quadratic,exponential and cubical.

Constant growth does not grow over time it stays as it is.

Hence the answer to this question is linear.

3 0

3 years ago

Suppose you wish to develop a matrix-multiplication algorithm that is asymptotically faster than Strassen’s algorithm. Your algo

Kazeer [188]

Answer:

The number of subproblems are given as $T(n)=a*T(n/8)+\theta(n^2)$ while the value of T(n) to be less than S(n) is for 342.

Explanation:

The number of subproblems are given as

$T(n)=a*T(n/8)+\theta(n^2)$

Asymptotic running time for Strassen’s algorithm is $S(n)=\theta(n^{log(7)})$

Now, when a increases, number of subproblems determines the asymptotic running time of the problem and case 1 of master theorem applies. So, in worst case, asymptotic running time of the algorithm will be

$T(n)=\theta(n^{logb(a)})=\theta(n^{log8(a)})=\theta(n^{log_{2}(a^{1/3})})$

Now, for T(n) to be smaller than S(n)

$n^{loga^{1/3}}$

So,

$log(a^{(1/3)})$

So,

a=342

5 0

3 years ago