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3 years ago

Explain why there must be at least two lines on any given plane,

Mathematics

1 answer:

Alex_Xolod [135]3 years ago

4 0

Answer:

there must be at least two lines on any plane because a plane is defined by 3 non-collinear points. Explanation: ... These lines may or may not intersect. If two of the 3 points are collinear, then we have a line through those 2 points as well as a line through the 3rd point.

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Line E passes through the points (2, 3) and (4, -3). What is the slope of a line perpendicular to line E?

lilavasa [31]

$\bf \begin{array}{ccccccccc}
&&x_1&&y_1&&x_2&&y_2\\
% (a,b)
&&(~ 2 &,& 3~) 
% (c,d)
&&(~ 4 &,& -3~)
\end{array}
\\\\\\
% slope = m
slope = m\implies 
\cfrac{\stackrel{rise}{ y_2- y_1}}{\stackrel{run}{ x_2- x_1}}\implies \cfrac{-3-3}{4-2}\implies \cfrac{-6}{2}\implies \cfrac{-3}{1}$

now, a line perpendicular to that one, will have a "negative reciprocal" slope, thus

$\bf \textit{perpendicular, negative-reciprocal slope for}\quad \cfrac{-3}{1}\\\\
negative\implies +\cfrac{3}{ 1}\qquad reciprocal\implies +\cfrac{ 1}{3}\implies \cfrac{1}{3}$

7 0

3 years ago

I need a models or visuals for that problem

Leya [2.2K]

Answer:I will be a mode

Step-by-step explanation:

4 0

4 years ago

Which statement is true about whether A and B are independent events? A and B are independent events because P(A∣B) = P(A) = 0.1

love history [14]

The events A and B are independent if the probability that event A occurs does not affect the probability that event B occurs.
A and B are independent if the equation P(A∩B) = P(A) P(B) holds true.
P(A∩B) is the probability that both event A and B occur.
Conditional probability is the probability of an event given that some other event first occurs.
P(B|A)=P(A∩B)/P(A)
In the case where events A and B are independent the conditional probability of event B given event A is simply the probability of event B, that is P(B).
Statement 1:A and B are independent events because P(A∣B) = P(A) = 0.12. This is true.
Statement 2:A and B are independent events because P(A∣B) = P(A) = 0.25.
This is true.
Statement 3:A and B are not independent events because P(A∣B) = 0.12 and P(A) = 0.25.
This is true.
Statement 4:A and B are not independent events because P(A∣B) = 0.375 and P(A) = 0.25
This is true.

5 0

3 years ago

Read 2 more answers

34% of college students say they use credit cards because of the rewards program. You randomly select 10 college students and a

finlep [7]

Answer:

a) There is a 18.73% probability that exactly two students use credit cards because of the rewards program.

b) There is a 71.62% probability that more than two students use credit cards because of the rewards program.

c) There is a 82% probability that between two and five students, inclusive, use credit cards because of the rewards program.

Step-by-step explanation:

There are only two possible outcomes. Either the student use credit cards because of the rewards program, or they use for other reason. So, we can solve this problem by the binomial distribution.

Binomial probability

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

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

In which $C_{n,x}$ is the number of different combinatios of x objects from a set of n elements, given by the following formula.

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

And $\pi$ is the probability of X happening.

In this problem, we have that:

10 student are sampled, so $n = 10$

34% of college students say they use credit cards because of the rewards program, so $\pi = 0.34$

(a) exactly two

This is P(X = 2).

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

$P(X = 2) = C_{10,2}.(0.34)^{2}.(0.66)^{8} = 0.1873$

There is a 18.73% probability that exactly two students use credit cards because of the rewards program.

(b) more than two

This is $P(X > 2)$ .

Either a value is larger than two, or it is smaller of equal. The sum of the decimal probabilities must be 1. So:

$P(X \leq 2) + P(X > 2) = 1$

$P(X > 2) = 1 - P(X \leq 2)$

In which

$P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2)$

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

$P(X = 0) = C_{10,0}.(0.34)^{0}.(0.66)^{10} = 0.0157$

$P(X = 1) = C_{10,1}.(0.34)^{1}.(0.66)^{9} = 0.0808$

$P(X = 2) = C_{10,2}.(0.34)^{2}.(0.66)^{8} = 0.1873$

$P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2) = 0.0157 + 0.0808 + 0.1873 = 0.2838$

$P(X > 2) = 1 - P(X \leq 2) = 1 - 0.2838 = 0.7162$

There is a 71.62% probability that more than two students use credit cards because of the rewards program.

(c) between two and five inclusive

This is:

$P = P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5)$

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

$P(X = 2) = C_{10,2}.(0.34)^{2}.(0.66)^{8} = 0.1873$

$P(X = 3) = C_{10,3}.(0.34)^{3}.(0.66)^{7} = 0.2573$

$P(X = 4) = C_{10,4}.(0.34)^{4}.(0.66)^{6} = 0.2320$

$P(X = 5) = C_{10,5}.(0.34)^{5}.(0.66)^{5} = 0.1434$

$P = P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) = 0.1873 + 0.2573 + 0.2320 + 0.1434 = 0.82$

There is a 82% probability that between two and five students, inclusive, use credit cards because of the rewards program.

6 0

3 years ago

Please Help URGENT!!

AlekseyPX

Y + 79 + 37 = 180
y + 116 = 180
y = 64

3 0

3 years ago

Read 2 more answers