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3 years ago

The fraction 3 over 8 is equivalent to what percent?

Mathematics

2 answers:

kirill115 [55]3 years ago

7 0

Answer: 37.5
Steps right it as a decimal:0.375
Calculate: 37.5/100
Solution:37.5

Sladkaya [172]3 years ago

6 0

Answer:

37.5 %

Step-by-step explanation:

$\frac{3}{8}=0.375\\ 0.375 * 100 = 37.5$

$\frac{3}{8}$ as a percent is 37.5%

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Y = 3x - 4 y = 4 + x

Andrei [34K]

Is This Substitution?

Cause if it is, then Y = 3x - 4
Y = 4 + x
= (4, 8)

7 0

3 years ago

Of his 94 compact discs, Raul plans to donate 17 to the thrift store. To the nearest tenth of a percent, what percent is this?

Masteriza [31]

For this case we can solve the problem by means of the following rule of three:
94 --------> 100%
17 --------> x
Clearing x we have:
x = (17/94) * (100)
x = 18,08510638%
To the nearest tenth of a percent:
x = 18.1%
Answer:
This is 18.1% of his compact discs.

6 0

3 years ago

A man sold a fan for Rs. 400. Find the cost price if he incurred a loss of 15%.

statuscvo [17]

Answer:

Cost price of the fan is $\text {Rs. 470.59}$ .

Step-by-step explanation:

Given: A man sold a fan for $\text{Rs.}\:400$ . He incurred a loss of $15\%$ .

To find: The cost price of the fan

Solution:

We have,

Selling price, S.P., of the fan $= \text{Rs.} 400$

Loss $=15\%$

We know that $\text {Loss\%}=\frac{\text{C.P.-S.P.}}{\text{C.P.}}\times100\%$

Now let the C.P. of the fan be $\text{Rs.} \:x$

So, $15=(\frac{x-400}{x})\times 100$

$\implies 15x=100x-40000$

$\implies 100x-15x=40000$

$\implies 85x=40000$

$\implies x=\frac{40000}{85}$

$\implies x= 470.59$

Hence, the cost price of the fan is $\text {Rs. 470.59}$ .

7 0

2 years ago

You are arguing over a cell phone while trailing an unmarked police car by 25 m; both your car and the police car are traveling

BARSIC [14]

Answer:

(a) 15 m

(b) ground speed: 26.14 m/s (94.1 km/h); relative speed 12 m/s

Step-by-step explanation:

We can choose the frame of reference to be the trailing car. We can start counting time from the point the lead car begins deceleration. Then its position (in meters) relative to the trailing car is ...

d(t) = 25 - 5/2t² . . . . . . where 25 m is the initial distance and -5 is the acceleration in m/s².

(a) Then the distance between cars at t=2 is ...

d(2) = 25 - (5/2)·4 = 15 . . . . meters

(b) The relative velocity at 2.4 seconds is ...

d'(t) = -5t

d'(2.4) = -5(2.4) = -12.0 . . . m/s

The closure speed with the police car is 12 m/s.

We need to do a little more work to find the ground speed of the trailing car when the cars bump.

The distance between the cars at t=2.4 seconds is ...

d(2.4) = 25 -(5/2)2.4² = 10.6 . . . . meters

At the closure speed of 12 m/s, it will take an additional ...

10.6 m/(12 m/s) = 0.8833... s = 53/60 s

to close the gap. The speed of the trailing car at that point in time will be the original speed less the deceleration for 0.8833 seconds:

(110 km/h)·(1/3.6 (m/s)/(km/h))-(5 m/s²)(53/60 s)

= 26 5/36 m/s = 94.1 km/h

_____

Comment on following distance

To avoid collision, the trailing car must be trailing by at least the distance it covers in 2.4 seconds, about 73 1/3 meters.

6 0

2 years ago

What is the greatest factor (GCF) of 24 and 40?

bagirrra123 [75]

Answer:

Step-by-step explanation:

that;'s what i put the first time

3 0

3 years ago

Read 2 more answers