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3 years ago

What does the “MIN” and “MAX” functions allow you to do when inserted in a spreadsheet?

Computers and Technology

1 answer:

Aleks04 [339]3 years ago

7 0

Answer:

Please see explanation

Explanation:

A spreadsheet is used to make calculations and payrolls etc. Different functions are used in spreadsheet. Two of them are MIN and MAX.

The MIN function gives the minimum value out of all the cells in the range.

It is written as:

= MIN(number1, number2..)

= MIN(Starting_cell_address:Last_cell_address)

Similarly, MAX function gives the maximum value from the selected range of cells.

= MAX(number1, number2..)

= MAX(Starting_cell_address:Last_cell_address)

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Can someone tell me how to fix the keyboard on ipad?- its in the middle of my screen andd i dont know how to do it

mamaluj [8]

Answer:

Here

Explanation:

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2 years ago

1. How many bits would you need to address a 2M × 32 memory if:

Dominik [7]

Answer:

a) 23 b) 21
a) 43 b) 42
a) 0 b) 0

Explanation:

1) How many bits is needed to address a 2M * 32 memory

2M = 2^1*2^20, while item =32 bit long word

hence ; L = 2^21 ; w = 32

a) when the memory is byte addressable

w = 8; L = ( 2M * 32 ) / 8 = 2M * 4

hence number of bits = log2(2M * 4)= log2 ( 2 * 2^20 * 2^2 ) = 23 bits

b) when the memory is word addressable

W = 32 ; L = ( 2M * 32 )/ 32 = 2M

hence the number of bits = log2 ( 2M ) = Log2 (2 * 2^20 ) = 21 bits

2) How many bits are required to address a 4M × 16 main memory

4M = 4^1*4^20 while item = 16 bit long word

hence L ( length ) = 4^21 ; w = 16

a) when the memory is byte addressable

w = 8 ; L = ( 4M * 16 ) / 8 = 4M * 2

hence number of bits = log 2 ( 4M * 2 ) = log 2 ( 4^1*4^20*2^1 ) ≈ 43 bits

b) when the memory is word addressable

w = 16 ; L = ( 4M * 16 ) / 16 = 4M

hence number of bits = log 2 ( 4M ) = log2 ( 4^1*4^20 ) ≈ 42 bits

3) How many bits are required to address a 1M * 8 main memory

1M = 1^1 * 1^20 , item = 8

L = 1^21 ; w = 8

a) when the memory is byte addressable

w = 8 ; L = ( 1 M * 8 ) / 8 = 1M

hence number of bits = log 2 ( 1M ) = log2 ( 1^1 * 1^20 ) = 0 bit

b) when memory is word addressable

w = 8 ; L = ( 1 M * 8 ) / 8 = 1M

number of bits = 0

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2 years ago

Tracy has a file that contains a list of actors and the movies in which they acted. She wants to know the top 3 ranked actors fr

harina [27]

Answer:

{

private String name;

private String review;

/** Constructs a ProductReview object and initializes the instance variables. */

public ProductReview(String pName, String pReview)

{

name = pName;

review = pReview;

}

/** Returns the name of the product. */

public String getName()

{ return name; }

/** Returns the review of the product. */

public String getReview()

{ return review; }

}

The ReviewCollector class, shown below, is used to represent a collection of reviews to be analyzed.

public class ReviewCollector

{

private ArrayList<ProductReview> reviewList;

private ArrayList<String> productList;

/** Constructs a ReviewCollector object and initializes the instance variables. */

public ReviewCollector()

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reviewList = new ArrayList<ProductReview>();

productList = new ArrayList<String>();

}

/** Adds a new review to the collection of reviews, as described in part (a). */

public void addReview(ProductReview prodReview)

{ /* to be implemented in part (a) */ }

/** Returns the number of good reviews for a given product name, as described in part (b). */

public int getNumGoodReviews(String prodName)

{ /* to be implemented in part (b) */ }

// There may be instance variables, constructors, and methods not shown.

}

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3 years ago

The _________ in an internet are responsible for receiving and forwarding packets through the interconnected set of networks and

raketka [301]

Answer:

routers

Explanation:

<h2>Fill in the blanks</h2>

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3 years ago

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DanielleElmas [232]

Diversity of their audiences

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2 years ago

Read 2 more answers