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3 years ago

Clueless!!!! Simplify the expression 42(uv^2)^3/(7u)^3v^4

Mathematics

2 answers:

Mrac [35]3 years ago

7 0

Answer:

6v^2 is the answer

Daniel [21]3 years ago

3 0

(42u^3 * 42v^5) / (343u^3)

I think that's the answer I'm not sure

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PLZ HELP IM IN DESPERATE NEED OF AN ANSWER!!!

Irina18 [472]

I believe 4(x - 3)=32 could help I haven’t done an equation from the unit in months so hopefully that looks familiar to you.

5 0

3 years ago

In the equation y=-2x + 3, if x = 7, what does y equal

Ahat [919]

Answer:

-11

Step-by-step explanation:

-14+3=-11

7 0

3 years ago

solo el 20% de los empleados de la población civil que está en una base militar restringida porta su identificación personal. Si

prisoha [69]

Usando la distribución binomial, hay una probabilidad de 0.8926 = 89.26% de que el guardia de seguridad encuentre al menos uno en la base militar restringida.

<h3>¿Qué es la distribución binomial?</h3>

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$C_{n,x} = \frac{n!}{x!(n-x)!}$

Los parámetros son:

n es el número de ensayos.

p es la probabilidad de éxito en un ensayo

x es el número de éxitos

En este problema, hay que:

20% de los empleados de la población civil que está en una base militar restringida porta su identificación personal, o sea p = 0.2.

Llegan 10 empleados, o sea, n = 10.

La probabilidad de que el guardia de seguridad encuentre al menos uno en la base militar restringida es dada por:

$P(X \geq 1) = 1 - P(X = 0)$

En que:

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{10,0}.(0.2)^{0}.(0.8)^{10} = 0.1074$

Por eso:

$P(X \geq 1) = 1 - P(X = 0) = 1 - 0.1074 = 0.8926$

Hay una probabilidad de 0.8926 = 89.26% de que el guardia de seguridad encuentre al menos uno en la base militar restringida.

Puede-se aprender más a cerca de la distribución binomial en brainly.com/question/25132113

3 0

2 years ago

Determine whether the set of vectors is a basis for ℛ3. Given the set of vectors , decide which of the following statements is t

schepotkina [342]

Answer:

(A) Set A is linearly independent and spans $R^3$ . Set is a basis for $R^3$ .

Step-by-Step Explanation

<u>Definition (Linear Independence)</u>

A set of vectors is said to be linearly independent if at least one of the vectors can be written as a linear combination of the others. The identity matrix is linearly independent.

<u>Definition (Span of a Set of Vectors)</u>

The Span of a set of vectors is the set of all linear combinations of the vectors.

<u>Definition (A Basis of a Subspace).</u>

A subset B of a vector space V is called a basis if: (1)B is linearly independent, and; (2) B is a spanning set of V.

Given the set of vectors $A= \left(\begin{array}{[c][c][c][c]}1 & 0 & 0 & 0\\ 0 & 1 & 0 & 1\\ 0 & 0 & 1 & 1\end{array} \right)$ , we are to decide which of the given statements is true:

In Matrix $A= \left(\begin{array}{[c][c][c][c]}(1) & 0 & 0 & 0\\ 0 & (1) & 0 & 1\\ 0 & 0 & (1) & 1\end{array} \right)$ , the circled numbers are the pivots. There are 3 pivots in this case. By the theorem that The Row Rank=Column Rank of a Matrix, the column rank of A is 3. Thus there are 3 linearly independent columns of A and one linearly dependent column. $R^3$ has a dimension of 3, thus any 3 linearly independent vectors will span it. We conclude thus that the columns of A spans $R^3$ .

Therefore Set A is linearly independent and spans $R^3$ . Thus it is basis for $R^3$ .

8 0

3 years ago

What is 22 3/4÷2 as a fraction,?<br>

nordsb [41]

Answer:

11375 / 1000

Hope This Helps! Have A Nice Day!!

4 0

3 years ago