Which fraction is equal to 0.45...? A. 5/11 B. 41/99 C. 41/90 D. 41/9

Mathematics

2 answers:

Butoxors [25]3 years ago

8 0

The answer is A. 5/11

GrogVix [38]3 years ago

6 0

0.45 Isn't Equal To Any Of These Fractions But It Is Equal To 45/100

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Need help with cross multiplying

saw5 [17]

So to solve for this, we need to set up proportional fractions, which I will help show you how to do.
First, if we are given an amount out of a total, we need to put it over x (if we are looking for the total). It looks like this:
12/x, 12 being the given number and x being the total.
If we are given the total but are looking for an amount, put the total at the bottom of the fraction (aka the denominator). It looks like this: x/16, 16 being the total amount and x being the amount out of the total.
We have a total of 40 test problems, so we can put our total at the bottom, x/40.
X is the amount of questions answered correctly (we are looking for x in the question).
We have answered 80% correct, so put 80% over 100 (100 being the total). It should look like this: 80/100.
Now we have our two fractions: x/40 & 80/100.
Set these up as an equation.
x/40 = 80/100.
Now this is where things may get tricky if you don't pay attention.
Multiply the numerator (the top number of a fraction) of x/40 by the denominator (the bottom number of a fraction) of 80/100.
Your product equation should look like this:
x times 100. This will give is 100x. Leave it at that.
Now, multiply the denominator of x/40 (the bottom number of the fraction) by the numerator (the top number of a fraction) of 80/100. It should look like this:
80 x 40. This will give us 3200.
Now set up our products as an equation.
100x = 3200.
To solve for x, divide both sides by 100.
3200/100 = 32.
x = 32.
I hope this helps and has taught you something!

4 0

3 years ago

226Ra has a half-life of 1599 years. How much is left after 1000 years if the initial amount was 10 g?

Ray Of Light [21]

$\bf \textit{Amount for Exponential Decay using Half-Life}\\\\
A=P\left( \frac{1}{2} \right)^{\frac{t}{h}}\qquad 
\begin{cases}
A=\textit{accumulated amount}\\
P=\textit{initial amount}\to &10\\
t=\textit{elapsed time}\to &1000\\
h=\textit{half-life}\to &1599
\end{cases}
\\\\\\
A=10\left( \frac{1}{2} \right)^{\frac{1000}{1599}}$

8 0

3 years ago

Please solve with explanation

Juliette [100K]

Step-by-step explanation:

$4\frac{9}{20} - 1\frac{2}{6} =\\\frac{89}{20} - \frac{8}{6} = \frac{89 - 26.7}{20} = \frac{62.3}{20} \\$

$4\frac{6}{9} - 3\frac{3}{6} =\\\frac{42}{9} - \frac{21}{6} = \frac{252 - 189}{54} = \frac{63}{54} = 1\frac{9}{54}$

$2\frac{2}{4} - 1\frac{1}{3} =\\\frac{10}{4} - \frac{4}{3} = \frac{30 - 16}{12} = \frac{14}{12} = 1\frac{2}{12}$

8 0

3 years ago

Read 2 more answers

Suppose a certain type of fertilizer has an expected yield per acre of mu 1 with variance sigma 2, whereas the expected yield fo

mart [117]

Answer:

See the proof below.

Step-by-step explanation:

For this case we just need to apply properties of expected value. We know that the estimator is given by:

$S^2_p= \frac{(n_1 -1) S^2_1 +(n_2 -1) S^2_2}{n_1 +n_2 -2}$

And we want to proof that $E(S^2_p)= \sigma^2$

So we can begin with this:

$E(S^2_p)= E(\frac{(n_1 -1) S^2_1 +(n_2 -1) S^2_2}{n_1 +n_2 -2})$

And we can distribute the expected value into the temrs like this:

$E(S^2_p)= \frac{(n_1 -1) E(S^2_1) +(n_2 -1) E(S^2_2)}{n_1 +n_2 -2}$

And we know that the expected value for the estimator of the variance s is $\sigma$ , or in other way $E(s) = \sigma$ so if we apply this property here we have: