Problem 6. (i) suppose that a rank-k approximation, for some k, gives an acceptable approximation. how much data is needed to represent the rank-k approximation? your answer should be an expression in terms of k, m and n. hint: you need k columns of u, k columns of v , and k singular values of s. (ii) the ratio of the amount of data used for the approximation (which you found in part (i)) and the amount of data of the (original format of the) picture is the compression rate. find the compression rate for the value of the rank you determined in problem 4. what does the compression rate represent? hint: after finding the compression rate for the value of the rank you determined in problem 4, you may want to present this number as a percentage.

answer: i'm not sure

step-by-step explanation:

54 km / hours

step-by-step explanation:

we need to convert seconds to hours

60 seconds = 1 minutes

60 minutes = 1hour

15 m         60 second   60 minutes

  * *         = 54000 meters/ hour

second     1 minute             1 hour

now we need to convert meters to km

1000 m = 1 km

54000 m       1 km

*   = 54 km / hours

hour               1000m

the answer is:

the correct option:

b.

to solve the problem, we need to remember how to operate to simplify exponential expressions using the logarithms.

we are given the equation:

then, using placing logarith with base 7 on both sides of the equation, we have:

so, rounding the obtained result to two decimal places (nearest hundreth), we have:

hence, we have that the correct option:

b.

have a nice day!