All categories

2 years ago

What is the fraction 25/30 in simplest form

Mathematics

1 answer:

Anuta_ua [19.1K]2 years ago

5 0

$\frac{5}{6}$

To simplify the fraction, divide the numerator / denominator ( cancel ) by the highest common factor of 25 and 30, that is 5

$\frac{25}{30}$ = $\frac{5}{6}$ in simplest form

You might be interested in

How can you quickly calculate 20 percent of any bill amount?

il63 [147K]

The answer is C an example is 20% of $20 is 4, so you would lose the zero and double the 2 to get 4

8 0

2 years ago

Read 2 more answers

2x ^2 −5=3x

expeople1 [14]

Answer:a

Step-by-step explanation:

3 0

3 years ago

The water level at Lake Penny changed −1 1/10 in. this year. Next year, it is predicted to change another −1 2/5

frutty [35]

- 1 1/10 + (-1 2/5)

- 1 1/10 + (-1 4/5)

-2 5/10

-2 1/2

7 0

3 years ago

A single die is rolled. How many ways can you roll a number that is prime, followed by a 6?

Grace [21]

Answer:

3 ways

Step-by-step explanation:

Since 6 will remain constant throughout the testing, we just need to find all prime numbers 1-6.

1 - is not prime nor composite

2 - is prime

3 - is prime

4 - 2x2=4, so composite

5 - is prime

6 - 2x3=6, so composite

Therefore, 2, 3, and 5 are prime numbers, and there are 3 of them.

5 0

2 years ago

Read 2 more answers

Let X and Y be discrete random variables. Let E[X] and var[X] be the expected value and variance, respectively, of a random vari

Ulleksa [173]

Answer:

(a) $E[X+Y]=E[X]+E[Y]$

(b) $Var(X+Y)=Var(X)+Var(Y)$

Step-by-step explanation:

Let X and Y be discrete random variables and E(X) and Var(X) are the Expected Values and Variance of X respectively.

(a)We want to show that E[X + Y ] = E[X] + E[Y ].

When we have two random variables instead of one, we consider their joint distribution function.

For a function f(X,Y) of discrete variables X and Y, we can define

$E[f(X,Y)]=\sum_{x,y}f(x,y)\cdot P(X=x, Y=y).$

Since f(X,Y)=X+Y

$E[X+Y]=\sum_{x,y}(x+y)P(X=x,Y=y)\\=\sum_{x,y}xP(X=x,Y=y)+\sum_{x,y}yP(X=x,Y=y).$

Let us look at the first of these sums.

$\sum_{x,y}xP(X=x,Y=y)\\=\sum_{x}x\sum_{y}P(X=x,Y=y)\\\text{Taking Marginal distribution of x}\\=\sum_{x}xP(X=x)=E[X].$

Similarly,

$\sum_{x,y}yP(X=x,Y=y)\\=\sum_{y}y\sum_{x}P(X=x,Y=y)\\\text{Taking Marginal distribution of y}\\=\sum_{y}yP(Y=y)=E[Y].$

Combining these two gives the formula:

$\sum_{x,y}xP(X=x,Y=y)+\sum_{x,y}yP(X=x,Y=y) =E(X)+E(Y)$

Therefore:

$E[X+Y]=E[X]+E[Y] \text{ as required.}$

(b)We want to show that if X and Y are independent random variables, then:

$Var(X+Y)=Var(X)+Var(Y)$

By definition of Variance, we have that:

$Var(X+Y)=E(X+Y-E[X+Y]^2)$

$=E[(X-\mu_X +Y- \mu_Y)^2]\\=E[(X-\mu_X)^2 +(Y- \mu_Y)^2+2(X-\mu_X)(Y- \mu_Y)]\\$Since we have shown that expectation is linear$\\=E(X-\mu_X)^2 +E(Y- \mu_Y)^2+2E(X-\mu_X)(Y- \mu_Y)]\\=E[(X-E(X)]^2 +E[Y- E(Y)]^2+2Cov (X,Y)$

Since X and Y are independent, Cov(X,Y)=0

$=Var(X)+Var(Y)$

Therefore as required:

$Var(X+Y)=Var(X)+Var(Y)$

7 0

2 years ago