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3 years ago

Find the following percentages. a. 15% of 930 b. 6% of 120 c. 4.2% of 50 d. 200% of 145

Mathematics

2 answers:

Mama L [17]3 years ago

4 0

Answer:

a) 139.5

b) 72

c) 21

d) 290

Step-by-step explanation:

a). 15% of 930

15%930 = (15 * 930)/100

= 139.5

b). 6% of 120

6% of 120 = (6 * 120)/100

= 72

c). 4.2% of 50

4.2% of 50 = (42 * 50)/100

= 290

d). 200% of 145

200% of 145 = (200 * 145)/100

= 290

Therefore,

a) 139.5

b) 72

c) 21

d) 290

Tom [10]3 years ago

3 0

These are 4 questions and 4 answers.

Answer:

a. 15% of 930 = 139.5

b. 6% of 120 = 7.2

c. 4.2% of 50 = 2.1

d. 200% of 145 = 290

Explanation:

Percentage is defined as number of parts per 100.

Percentage can be expressed as a ratio (fraction) or decimal.

1% = 1 / 100 = 0.01

10% = 10 / 100 = 0.1

25% = 25 /100 = 1/4 = 0.25

50% = 50 / 100 = 1/2 = 0.5

75% = 75 / 100 = 3/4 = 0.75

You can use either the fraction or the decimal form to find the percentage.

This is how you apply this to questions a through d.

a. 15% of 930

15% × 930 = 15 / 100 × 930 = 0.15 × 930 = 139.5

b. 6% of 120

6% × 120 = 6 / 100 × 120 = 0.06 × 120 = 7.2

c. 4.2% of 50

4.2% × 50 = 4.2 / 100 × 50 = 0.042 × 50 = 2.1

d. 200% of 145

200% × 145 = 200 / 100 × 145 = 2 × 145 = 290

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earnstyle [38]

Answer:

(i). Average speed = 75 km per hour

(ii). Average speed = 75 km per hour

Step-by-step explanation:

This question is incomplete; here is the complete question.

A train traveled from station A to station B at an average speed of 80 kilometers per hour and then from station B to station C at an average of 60 kilometers per hour. If the train did not stop at station b, what was the average speed at which the train traveled from station A to C?

(i). The distance that the train traveled from station A to station B was 4 times the distance that train traveled from station B to station C.

(ii). The amount of time it took to the train to travel from station A to station B is 3 times the amount of time that it took the train to travel from station B to station C.

(i) Let the distance between station B and station C = x km

So the distance between Station A and station B = 4x km

Therefore, time to travel the distance x km with 80 km per hour = $\frac{\text{Distance traveled}}{\text{Speed}}$

= $\frac{4x}{80}$

Similarly time taken to travel between station B and station C with speed 60 km per hour = $\frac{x}{60}$ km per hour

Now average speed between station A and station C = $\frac{\text{Total distance between station A and station C}}{\text{time taken to travel}}$

= $\frac{4x+x}{\frac{4x}{80}+\frac{x}{60}}$

= $\frac{5x}{\frac{4x}{60} }$

= $\frac{5\times 60}{4}$

= 75 km per hour

(ii). Let the train took the time to cover the distance between station B and station C = t hours

Therefore, time taken by the train between station A and B = 3t

Distance between Station A and station B = Speed × time

= 80 × 3t

= 240t km

Similarly distance between station B and station C = 60 × t

= 60t

Now average speed between station A and station C = $\frac{240t+60t}{(3t+t)}$

= $\frac{300t}{4t}$

= 75 km per hour

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alina1380 [7]

Answer:

Options (A) and (D).

Step-by-step explanation:

Given expression is,

4x³ - 7x² - $\frac{30}{y}$ + 15

Characteristics of the give expression ,

Option (A).

Third term $-\frac{30}{y}$ of the expression is in the form a ratio.

True.

Option (B).

Entire expression is a difference.

False.

Option (C).

There are three terms.

False.

Option (D).

There are four terms in the given expression.

True.

Therefore, Options (A) and (D) will be the correct options.

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---------
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