Which of the following are among the five basic postulates of Euclidean geometry?

Mathematics

1 answer:

lana66690 [7]1 year ago

6 0

Answer:

Choices B, C, and D are correct.

Step-by-step explanation:

Euclid's Five postulates are as follows:

1. A straight line segment may be drawn from any given point to any other.

2. A straight line may be extended to any finite length.

3. A circle can be drawn with any center and radius.

4. All right angles are congruent/equal to each other.

5. Through a point not on a given line, there is exactly one line parallel to the given line.

Now looking at these postulates and the choices we are given we see that only choice B, C, and D match with above.

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Milton is floating in an inner tube in a wave pool. He is 1.5 m from the bottom of the pool when he is at the trough of a wave.

posledela

This is the concept of sinusoidal, to solve the question we proceed as follows;
Using the formula;
g(t)=offset+A*sin[(2πt)/T+Delay]
From sinusoidal theory, the time from trough to crest is normally half the period of the wave form. Such that T=2.5
The pick magnitude is given by:
Trough-Crest=
2.1-1.5=0.6 m
amplitude=1/2(Trough-Crest)
=1/2*0.6
=0.3
The offset to the center of the circle is 0.3+1.5=1.8
Since the delay is at -π/2 the wave will start at the trough at [time,t=0]
substituting the above in our formula we get:
g(t)=1.8+(0.3)sin[(2*π*t)/2.5]-π/2]
g(t)=1.8+0.3sin[(0.8πt)/T-π/2]

3 0

1 year ago

A bin of 5 transistors is known to contain 2 that are defective. The transistors are to be tested, one at a time, until the defe

xxMikexx [17]

Answer:

$P(N_1 = a , N_2 = b)= \frac{1}{5-a C 1} * \frac{5-a C 1}{5C2} = \frac{1}{5C2}=\frac{1}{10}$

Step-by-step explanation:

For the random variable $N_1$ we define the possible values for this variable on this case $[1,2,3,4,5]$ . We know that we have 2 defective transistors so then we have 5C2 (where C means combinatory) ways to select or permute the transistors in order to detect the first defective:

$5C2 = \frac{5!}{2! (5-2)!}= \frac{5*4*3!}{2! 3!}= \frac{5*4}{2*1}=10$

We want the first detective transistor on the ath place, so then the first a-1 places are non defective transistors, so then we can define the probability for the random variable $N_1$ like this:

$P(N_1 = a) = \frac{5-a C 1}{5C2}$

For the distribution of $N_2$ we need to take in count that we are finding a conditional distribution. $N_2$ given $N_1 =a$ , for this case we see that $N_2 \in [1,2,...,5-a]$ , so then exist $5-a C 1$ ways to reorder the remaining transistors. And if we want b additional steps to obtain a second defective transistor we have the following probability defined: