Convert 2.5 to a fraction in simplest form

1 answer:

3 0

1. You times 2.5 by 10 (because there is 1 decimal point) which then equals 25/10.
2. Then you reduce 25/10 by 5, so it will be in simples form. And your answer is 5/2!

Hope this helped and Good Luck! :D

You might be interested in

In a survey, four out of five people surveyed preferred creamy over chunky peanut butter. 120 people are shopping at the grocery

elixir [45]

Answer: Around 96 people would prefer creamy peanut butter.

Step-by-step explanation:

In the survey was found that 4 out of 5 preferred creamy over chunky peanut butter.

Then the relative frequency is:

4/5 = 0.8 prefer creamy peanut butter.

And we can assume that:

1/5 = 0.2 prefer chunky peanut butter.

We also could multiply these numbers by 100% to obtain the percentage forms:

0.8*100% = 80%

0.2*100% = 20%

Now, if there are 120 people in the grocery store, we can expect that the 80% will prefer creamy peanut butter.

Then we can expect that the number of people that prefers creamy peanut butter is:

N = (80%/100%)*120 = 0.8*120 = 96

Around 96 people would prefer creamy peanut butter.

3 0

2 years ago

Simplify -3x^2-5x^2(4x^3-x^2)2

PSYCHO15rus [73]

Let's simplify step-by-step.

−3x2−5x2(4x3−x2)(2)

Distribute:

=−3x2+−40x5+10x4

Answer:

=−40x5+10x4−3x2

8 0

3 years ago

Suppose after 2500 years an initial amount of 1000 grams of a radioactive substance has decayed to 75 grams. What is the half-li

krok68 [10]

Answer:

The correct answer is:

Between 600 and 700 years (B)

Step-by-step explanation:

At a constant decay rate, the half-life of a radioactive substance is the time taken for the substance to decay to half of its original mass. The formula for radioactive exponential decay is given by:

$A(t) = A_0 e^{(kt)}\\where:\\A(t) = Amount\ left\ at\ time\ (t) = 75\ grams\\A_0 = initial\ amount = 1000\ grams\\k = decay\ constant\\t = time\ of\ decay = 2500\ years$

First, let us calculate the decay constant (k)

$75 = 1000 e^{(k2500)}\\dividing\ both\ sides\ by\ 1000\\0.075 = e^{(2500k)}\\taking\ natural\ logarithm\ of\ both\ sides\\In 0.075 = In (e^{2500k})\\In 0.075 = 2500k\\k = \frac{In0.075}{2500}\\ k = \frac{-2.5903}{2500} \\k = - 0.001036$

Next, let us calculate the half-life as follows:

$\frac{1}{2} A_0 = A_0 e^{(-0.001036t)}\\Dividing\ both\ sides\ by\ A_0\\ \frac{1}{2} = e^{-0.001036t}\\taking\ natural\ logarithm\ of\ both\ sides\\In(0.5) = In (e^{-0.001036t})\\-0.6931 = -0.001036t\\t = \frac{-0.6931}{-0.001036} \\t = 669.02 years\\\therefore t\frac{1}{2} \approx 669\ years$