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3 years ago

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intro urna sunt 24 de bile albe si 16 bile negre. se extrage la intamplare o bila . probabilitatea ca bila extrasa sa fie alba e

ste..

1 answer:

garri49 [273]3 years ago

4 0

Answer:

$\frac{3}{5}$

Step-by-step explanation:

Given parameters:

Number of white balls = 24

Number of black balls = 16

Unknown:

The probability that white ball is drawn at random = ?

Solution:

The probability of an event is the likelihood of such an event to occur. That an event will occur has a probability of 1, it will not occur have a probability of zero.

In this problem, the total number of outcomes of drawing any ball has sample space of (24 + 16)outcomes = 40outcomes.

Probability of an event = $\frac{number of successful outcome}{sample space}$

Pr(white balls) = $\frac{24}{40}$ = $\frac{3}{5}$

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PolarNik [594]

Answer:

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3 years ago

What is an algebraic expression

Annette [7]

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3 years ago

3.<br>June has 248 oranges. She sells of these oranges. How many oranges did June sell?<br>Answer.

uranmaximum [27]

Answer:

93. Divide 248 by 8. One eighth of 248 = 31.

June sold 3/8 of the 248 oranges.

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Step-by-step explanation:

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2 years ago

Read 2 more answers

Let M be the set of all nxn matrices. Define a relation on won M by A B there exists an invertible matrix P such that A = P BP S

Sophie [7]

Answer:

Recall that a relation is an <em>equivalence relation</em> if and only if is symmetric, reflexive and transitive. In order to simplify the notation we will use A↔B when A is in relation with B.

<em>Reflexive: </em>We need to prove that A↔A. Let us write J for the identity matrix and recall that J is invertible. Notice that $A=J^{-1}AJ$ . Thus, A↔A.

<em>Symmetric</em>: We need to prove that A↔B implies B↔A. As A↔B there exists an invertible matrix P such that $A=P^{-1}BP$ . In this equality we can perform a right multiplication by $P^{-1}$ and obtain $AP^{-1} =P^{-1}B$ . Then, in the obtained equality we perform a left multiplication by P and get $PAP^{-1} =B$ . If we write $Q=P^{-1}$ and $Q^{-1} = P$ we have $B = Q^{-1}AQ$ . Thus, B↔A.

<em>Transitive</em>: We need to prove that A↔B and B↔C implies A↔C. From the fact A↔B we have $A=P^{-1}BP$ and from B↔C we have $B=Q^{-1}CQ$ . Now, if we substitute the last equality into the first one we get

$A=P^{-1}Q^{-1}CQP = (P^{-1}Q^{-1})C(QP)$ .

Recall that if P and Q are invertible, then QP is invertible and $(QP)^{-1}=P^{-1}Q^{-1}$ . So, if we denote R=QP we obtained that

$A=R^{-1}CR$ . Hence, A↔C.

Therefore, the relation is an <em>equivalence relation</em>.

4 0

3 years ago

If a cone shaped vase can hold 45 cubic centimeter of water and has a radius of 2 centimeters what is the height of the vase use

xz_007 [3.2K]

Answer:

The height of the vase is $h=10.75\ cm$

Step-by-step explanation:

we know that

The volume of a cone is equal to

$V=\frac{1}{3}\pi r^{2} h$

we have

$V=45\ cm^{3}$

$r=2\ cm$

$\pi =3.14$

substitute and solve for h

$45=\frac{1}{3}(3.14)(2)^{2} h$

$135=(3.14)(4)h$

$135=12.56h$

$h=135/12.56$

$h=10.75\ cm$

3 0

3 years ago