Please i need to get good grades

The scale on a map is 1: 75 000. (a) Calculate the ACTUAL distance, in km, between two places which are 16 cm apart on the map.

finlep [7]

Answer:

a. 1 200 000 km

b. 0.000 773 cm

c. 67 500 000 000 km²

d. 0.000 000 016 cm²

Step-by-step explanation:

a. Calculate the ACTUAL distance, in km, between two places which are 16 cm apart on the map.

Since the scale is 1 : 75000 that is 1 cm : 75000 km, we want to find the actual distance of 16 cm on the map. Let x be the actual distance in km.

So, 1 cm : 75000 km = 16 cm : x km

So, 1 cm/75000 km = 16 cm/x km

Thus x km = 16 cm × 75000 km/1 cm = 1 200 000 km

(b) Calculate length of a road on the map which is 0.58 km.

Let y be the length of road on the map. So, using our scale

1 cm : 75000 km = y cm : 0.58 km

So, 1 cm/75000 km = y cm/0.58 km

Thus y cm = 0.58 km × 1 cm/75000 km = 0.000 773 cm

(c) Calculate, in km², the ACTUAL area of a lake which is 12 cm2 on the map.

Since the scale is 1 : 75000 , the scale for the area would be the square of this. So, 1² : 75000² = 1 : 5 625 000 000 which is 1 cm² : 5 625 000 000 km²

Let z be the actual area of the lake.

So, 1 cm² : 5 625 000 000 km² = 12 cm² : z km²

So, 1 cm²/5 625 000 000 km² = 12 cm²/z km²

Thus z km² = 12 cm² × 5 625 000 000 km²/1 cm² = 67 500 000 000 km²

(d) Calculate the area of a play field on the map which has has an actual area of 90 km².

Let a represent the area of the play field on the map. Using our scale for the area,

1 cm² : 5 625 000 000 km² = a cm² : 90 km²

So, a cm² = 90 km² × 1/5 625 000 000 km² = 0.000 000 016 cm²

4 0

2 years ago

Dale keeps all of his pairs of shoes in his closet. select the number of shoes that deal could have in his closet. Mark all that

bekas [8.4K]

As it says he keeps 'pairs' of shoes so the answer has to be an even 'number' - ie B, C and E are all possible answers.

6 0

3 years ago

Read 2 more answers

25 POINTS! URGENT! HELP PLS!! 3 PARTS BUT YOU CAN DO HOW EVER MANY U WANT

goldenfox [79]

A pyramid has 7 faces

3 0

2 years ago

Which expression has the fewest significant figures?

yanalaym [24]

Hello!

Here are some rules to determine the number of significant figures.

Numbers that are not zero are significant (45 - all are sigfigs)
Zeros between non-zero digits are significant (3006 → all are sigfigs)
Trailing zeros are not significant (0.067 → the first two zeros are not sigfigs)
Trailing zeros after a decimal point are always significant (1.000 → all are sigfigs)
Trailing zeros in a whole number are not significant (7800 → the last two zeros are not sigfigs)
In scientific notation, the exponential digits are not significant, known as place holders (6.02 x 10² → 10² is not a sigfig)

Now, let's find the number of significant figures in each given number.

A). 296.54

Since these digits are all <em>non-zero</em>, there are 5 significant figures.

B). 5003.1

Since the two <em>zeros are between non-zero digits</em>, they are significant figures. Thus, there are 5 significant figures.

C). 360.01

Again, the two zeros are between non-zero digits. There are 5 significant figures.

D). 18.3

All of these digits are non-zero, hence, there are 3 significant figures.

Therefore, expression D has the fewest number of significant figures being 3.

4 0

3 years ago

Read 2 more answers

Suppose each laptop of a certain type is assigned a series number, which consists of a sequenceof eight symbols: number, letter,

Varvara68 [4.7K]

Answer:

Step-by-step explanation:

There are 5numbers and 3 letters. Each number can be one among the 10 possible.

Each letter is 1 among the 26 available.

a)The number of possible unique serial numbers are

10^5x 26³

The number of 5 number combinations each different (order important) is 5!(10/5)= 30240

The number of 3 letter combinations each different (order important) is 3!(26/3)=15600

The required probability (that all symbols are different if one laptop is picked at random with equal probability) is

15600x3240/ 10^5x 26³=0.2684

b) The first letter must be at least 4 to be largest among the 5 numbers. So the first letter can be one of . 4,5,6,7,8,9..

The number of 5 number combinations each different and starting with 4 (order important) is .

4!=24

The number of 5 number combinations each different and starting with 5 (order important) is .

4!(5/4)= 120

The number of 5 number combinations each different and starting with 6 (order important) is .

4!(6/4)= 360

The number of 5 number combinations each different and starting with 7 (order important) is .

4!(7/4)= 840

The number of 5 number combinations each different and starting with 8 (order important) is

4!(8/4)= 1680

The number of 5 number combinations each different and starting with 9 (order important) is .

4!(9/4)= 3024

Thus, there are

The required probability ( that all symbols are different and the first number is the largest among the numbers) is

15600* 6040/10^5x 26³= 0.0537

5 0

3 years ago