What is the slope of a line with points (3, 12)(7, 16)?

A line passes through (1,2) and is perpendicular to y=3.

dybincka [34]

Make a graph and put a point at exactly (1,2) and then make a like at y=3 make a line through the point and you'll see that this is an undefined equation x cant equal all those numbers

6 0

3 years ago

Read 2 more answers

PLSSS HELP IF YOU TURLY KNOW THISS

Crank

Answer:

7/10

Step-by-step explanation:

hope this helps

6 0

3 years ago

B is the midpoint of AC, what in the measure of AB if AC 82cm?

Bess [88]

Answer:

AB = 41 cm

Step-by-step explanation:

If B is the midpoint of AC, then B splits AC into 2 equal halves: AB & BC.

Therefore, AB = ½ of AC.

Given, AC = 82 cm.

So,

AB = ½ × AC

AB = 82/2

AB = 41 cm

_____

Hope it helps ⚜

8 0

2 years ago

A ______________ consists of an a sample space and a list of events with their probabilities.

motikmotik

Sample space َََ َ َ َ َ َ َ َ َ َ

5 0

3 years ago

Birds arrive at a birdfeeder according to a Poisson process at a rate of six per hour.

m_a_m_a [10]

Answer:

a) time= $10 \frac{1}{6}=\frac{10}{6}=1.67 hours$

b) $P(T\geq 0.25h)=e^{-(6)0.25}=0.22313$

c) $P(T\leq 0.0833)=1-e^{-(6)0.0833}=0.39347$

Step-by-step explanation:

Definitions and concepts

The Poisson process is useful when we want to analyze the probability of ocurrence of an event in a time specified. The probability distribution for a random variable X following the Poisson distribution is given by:

$P(X=x) =\lambda^x \frac{e^{-\lambda}}{x!}$

And the parameter $\lambda$ represent the average ocurrence rate per unit of time.

The exponential distribution is useful when we want to describ the waiting time between Poisson occurrences. If we assume that the random variable T represent the waiting time btween two consecutive event, we can define the probability that 0 events occurs between the start and a time t, like this:

$P(T>t)= e^{-\lambda t}$

a. What is the expected time you would have to wait to see ten birds arrive?

The original rate for the Poisson process is given by the problem "rate of six per hour" and on this case since we want the expected waiting time for 10 birds we have this:

time= $10 \frac{1}{6}=\frac{10}{6}=1.67 hours$

b. What is the probability that the elapsed time between the second and third birds exceeds fifteen minutes?

Assuming that the time between the arrival of two birds consecutive follows th exponential distribution and we need that this time exceeds fifteen minutes. If we convert the 15 minutes to hours we have 15(1/60)=0.25 hours. And we want to find this probability:

$P(T\geq 0.25h)$

And we can use the result obtained from the definitions and we have this:

$P(T\geq 0.25h)=e^{-(6)0.25}=0.22313$

c. If you have already waited five minutes for the first bird to arrive, what is the probability that the bird will arrive within the next five minutes?

First we need to convert the 5 minutes to hours and we got 5(1/60)=0.0833h. And on this case we want a conditional probability. And for this case is good to remember the "Markovian property of the Exponential distribution", given by :

$P(T \leq a +t |T>t)=P(T\leq a)$

Since we have a waiting time for the first bird of 5 min = 0.0833h and we want that the next bird will arrive within 5 minutes=0.0833h, we can express on this way the probability of interest:

$P(T\leq 0.0833+0.0833| T>0.0833)$

$P(T\leq 0.1667| T>0.0833)$

And using the Markovian property we have this:

$P(T\leq 0.0833)=1-e^{-(6)0.0833}=0.39347$

3 0

3 years ago