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Andrew [12]
3 years ago
8

Which statement about numbers on the horizontal number line is true?

Mathematics
2 answers:
kari74 [83]3 years ago
8 0
<span>C.) On the horizontal number line, -0.75 is located to the left of -0.4

[ It is because, When you go from left to right,number decreased upto zero, and then increases with positive sign,so more negative number would be in left and more positive in the right]

Hope this helps!</span>
sattari [20]3 years ago
6 0
On a horizontal number line, -ve is on left, 0 in middle n +ve on right

so ans is <span>C.) On the horizontal number line, -0.75 is located to the left of -0.4.</span>
You might be interested in
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
WILL GIVE BRAINLIEST
kifflom [539]
Speed boat departed 1.75 hrs after steam boat1.8 hrs after speed boat departure, steam boat has sailed for a total: 1.8+1.75 = 3.55 hrssteam boat's speed is 36.8 km/hr, so it has traveled: 3.55*36.8 = 130.64 km
speed boat is 86.9 km behind steam boat. so it is 130.64-86.9 = 43.74 km from portspeed boat's speed = 43.74/1.8 = 24.3 km/hr

8 0
3 years ago
Read 2 more answers
[figured out on own]
Yuri [45]
What happend ohhhhh Ok Good
3 0
3 years ago
Last week, a store sold laptops worth a total of $10,885. Each laptop cost $1,555. Hoy many laptops did the store sell last week
aliina [53]
1,555 x 7 = 10,885 so 7 is the answer
4 0
3 years ago
Read 2 more answers
A standardized​ exam's scores are normally distributed. In a recent​ year, the mean test score was 14901490 and the standard dev
kicyunya [14]
<h2>Answer with explanation:</h2>

Given : A standardized​ exam's scores are normally distributed.

Mean test score : \mu=1490

Standard deviation : \sigma=320

Let x be the random variable that represents the scores of students .

z-score : z=\dfrac{x-\mu}{\sigma}

We know that generally , z-scores lower than -1.96 or higher than 1.96 are considered unusual .

For x= 1900

z=\dfrac{1900-1490}{320}\approx1.28

Since it lies between -1.96 and 1.96 , thus it is not unusual.

For x= 1240

z=\dfrac{1240-1490}{320}\approx-0.78

Since it lies between -1.96 and 1.96 , thus it is not unusual.

For x= 2190

z=\dfrac{2190-1490}{320}\approx2.19

Since it is greater than 1.96 , thus it is unusual.

For x= 1240

z=\dfrac{1370-1490}{320}\approx-0.38

Since it lies between -1.96 and 1.96 , thus it is not unusual.

5 0
3 years ago
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