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NikAS [45]
3 years ago
12

Plz help its due tonight!!!!

Mathematics
1 answer:
Natalka [10]3 years ago
8 0

Answer:

Domain is (infinity, infinity) so it would hit every single number of x. The range which is y would be greater than or equal to -3. Range starts at -3 and hits every number above that.

You might be interested in
10 black balls and 5 white balls are placed in an urn. Two balls are then drawn in succession. What is the probability that the
Alex787 [66]

Answer:

The probability that the second ball drawn is a white ball if the second ball is drawn without replacing the first ball is \frac{1}{3} or 0.3333

Step-by-step explanation:

Probability is the greater or lesser possibility that a certain event will occur. In other words, the probability establishes a relationship between the number of favorable events and the total number of possible events. Then, the probability of any event A is defined as the ratio between the number of favorable cases (number of cases in which event A may or may not occur) and the total number of possible cases. This is called Laplace's Law:

P(A)\frac{number of favorable cases of A}{total number of possible cases}

Each of the results obtained when conducting an experiment is called an elementary event. The set of all elementary events obtained is called the sample space, so that every subset of the sample space is an event.

The total number of possible cases is 15 (10 black balls added to the 5 white balls).

As each extraction is without replacement, the events are dependent. For that, the dependent probabilities are defined first

Two events are dependent on each other when the fact that one of them is verified influences the probability of the other being verified.  In other words, the probability of A happening is affected because B has happened or not.

The probability of two events A and B of two successive simple experiments in a dependent compound experiment is:

A and B are dependent ⇔ P (A ∩ B) = P (A) · P (B / A)

                                              P (A ∩ B) = P (B) · P (B / A)

As the color of the first ball that is extracted is unknown, there are two cases: that the ball is black or that the ball is white.

<em> It will be assumed first that the first ball drawn is black</em>. Then the probability of this happening is \frac{10}{15} since the number of black balls in the urn is 10 and the total number of cases is 15. It is now known that the second ball extracted will be white. Then the number of favorable cases will be 5 (number of white balls inside the ballot urn), but now the number of total cases is 14, because a ball was previously removed that was not replaced.  So the probability of this happening is \frac{5}{14}

So the probability that the first ball is black and the second white is:

<em>\frac{10}{15} *\frac{5}{14} =\frac{5}{21}</em>

<em>It will now be assumed first that the first ball that is drawn is white.</em> Then the probability of this happening is \frac{5}{15} since the number of white balls in the urn is 5 and the total number of cases is 15. And it is known that the second ball drawn will be white. Then, the number of favorable cases will be 4 (number of white balls inside the urn, because when removing a white ball and not replacing it, its quantity will decrease), and the total number of cases is 14, same as in the previous case  So, the probability of this happening is  \frac{4}{14}

So the probability that the first ball is white and the second white is:

<em>\frac{5}{15} *\frac{4}{14} =\frac{2}{21}</em>

If A and B are two incompatible events, that is, they cannot occur at the same time, the probability of occurrence A or of occurrence B will be the sum of the probabilities of each event occurring separately.

These are events are incompatible, since I cannot, in a first extraction, extract a black and white ball at the same time. So:

<em>\frac{5}{21} +\frac{2}{21} =\frac{7}{21}=\frac{1}{3} =0.3333</em>

Finally, <u><em>the probability that the second ball drawn is a white ball if the second ball is drawn without replacing the first ball is \frac{1}{3} or 0.3333</em></u>

3 0
3 years ago
I’m lost please help
vagabundo [1.1K]

Answer:

See proof below

Step-by-step explanation:

Two triangles are said to be congruent if one of the 4 following rules is valid

  1. The three sides are equal
  2. The three angles are equal
  3. Two angles are the same and a corresponding side is the same
  4. Two sides are equal and the angle between the two sides is equal

Let's consider the two triangles ΔABC and ΔAED.

ΔABC sides are AB, BC and AC

ΔAED sides are AD, AE and ED

We have AE = AC and EB = CD

So AE + EB = AC + CD

But AE + EB = AB and AC+CD = AD

We have

AB of ΔABC  = AD of ΔAED

AC of ΔABC =  AE of ΔAED

Thus two sides the these two triangles. In order to prove that the triangles are congruent by rule 4, we have to prove that the angle between the sides is also equal. We see that the common angle is ∡BAC = ∡EAC

So  triangles ΔABC and ΔAED are congruent

That means all 3 sides of these triangles are equal as well as all the angles

Since BC is the third side of ΔABC and ED the third side of ΔAED, it follows that

BC = ED  Proved

5 0
1 year ago
Does any body know how to prove a parallelogram with statements and reasons
murzikaleks [220]
Because the lines are parallel,

The width of that rhombus, or parallelogram.

If you conutined that line, they will never ever touch.

That is the meaning of parallel, and why it is a parallelogram.

7 0
3 years ago
Can someone help me please
STatiana [176]
A is the correct answer
7 0
2 years ago
7. The new York Volleyball Association
Yuri [45]

Using a geometric sequence, it is found that after 3 rounds, 8 teams are left.

In a geometric series, the quotient between consecutive terms is always the same, and it is called common ratio q.

The general equation of a geometric series is given by:

a_n = a_1q^{n-1}

In which a_1 is the first term.

In this problem:

  • 64 teams were invited, thus a_1 = 64.
  • After each round, half the teams are eliminated, thus q = \frac{1}{2}.

The <u>number of teams after 3 rounds</u> is the 4th term of sequence, as the first is the initial number(0 rounds), thus:

a_n = a_1q^{n-1}

a_4 = 64\left(\frac{1}{2}\right)^{4-1}

a_4 = 8

After 3 rounds, 8 teams are left.

A similar problem is given at brainly.com/question/25317689

4 0
2 years ago
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