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2 years ago

How to add and subtract fractions

1 answer:

7 0

To add or subtract fractions , their denominator ( The one beneath the numerator, the one that divides the numerator ).

If they both terms have the same denominators you just have to add or subtract their numerators and stay with your denominator:

example: 3/5 + 4/5 = 7/5
2/3 - 1/3 = 1/3

But if the terms you are adding or subtracting do NOT have the same denominator , you have to make their denominator equal by equivalency :

Example : 2/3 + 1/6
I cant add 2/3 to 1/6 because they have different denominators. But if I multiply 2/3 by 2/2 it wil be 4/6:
4/6 + 1/6 = 5/6 !!
As 2/2 = 1 , I did not change the value but I changed the numerator and denominator of 2/3 :
If I multiply the numerator of a fraction by any number , but I do the same thing to the denominator, It creates a equivalency, where I dont change the actual values of the fraction.

I hope you understood my brief explanation , and please consider marking this awnser as Branliest if you think it deserves it. Thank you ! :)

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Step-by-step explanation:

I said because Exponential describes a very rapid increase. ... Exponential is also a mathematical term, meaning "involving an exponent." When you raise a number to the tenth power, for example, that's an exponential increase in that number. the part involves exponent is the y part same number multiple to it self to get the next answer while the x part one specific number to get the next number which is linear but since theyre asking about the relationship it’s neither because both were supposed to have the same relationship linea for x and y or exponential for x and y but they both are different.

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2 years ago

Verify that the following function is a probability mass function, and determine the requested probabilities. f left-parenthesis

Licemer1 [7]

Answer:

$f(x) = \frac{27}{13} (\frac{1}{3})^x , x=1,2,3$

1) $f(x_i) \geq 0, \forall x_i$

2) $sum_{i=1}^n P(X_i) =1$

We can find the individual probabilities and we got:

$f(1)= \frac{27}{13} (\frac{1}{3})^1 =0.6923$

$f(2)= \frac{27}{13} (\frac{1}{3})^2 =0.2307$

$f(3)= \frac{27}{13} (\frac{1}{3})^3 =0.0770$

And the sum of the 3 values 0.6923+0.2307+0.0770= 1 so then we satisfy all the conditions and we can conclude that f(x) is a probability distribution.

$P(X \leq 1) = P(X=1) =0.6923$

$P(X>1) = P(X=2) +P(X=3) = 0.2307+0.0770=0.3077$

$P(2$

Step-by-step explanation:

For this case we have the following density function:

$f(x) = \frac{27}{13} (\frac{1}{3})^x , x=1,2,3$

In order to satisfty that this function is a probability mass function we need to check two conditions:

1) $f(x_i) \geq 0, \forall x_i$

2) $sum_{i=1}^n P(X_i) =1$

We can find the individual probabilities and we got:

$f(1)= \frac{27}{13} (\frac{1}{3})^1 =0.6923$

$f(2)= \frac{27}{13} (\frac{1}{3})^2 =0.2307$

$f(3)= \frac{27}{13} (\frac{1}{3})^3 =0.0770$

And the sum of the 3 values 0.6923+0.2307+0.0770= 1 so then we satisfy all the conditions and we can conclude that f(x) is a probability distribution.

And if we want to find the following probabilities:

$P(X \leq 1) = P(X=1) =0.6923$

$P(X>1) = P(X=2) +P(X=3) = 0.2307+0.0770=0.3077$

$P(2$

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Step-by-step explanation:

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