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Evgesh-ka [11]
2 years ago
8

When dealing with the number of occurrences of an event over a specified interval of time or space and when the occurrence or no

noccurrence in any interval is independent of the occurrence or nonoccurrence in any other interval, the appropriate probability distribution is a _____.
Mathematics
1 answer:
jeyben [28]2 years ago
7 0

Answer:

POISSON DISTRIBUTION

Step-by-step explanation:

When dealing with the number of occurrences of an event over a specified interval of time or space, the poisson distribution is often useful.

Poisson distribution is applicable if:

The probability of the occurrence of the event is the same for any two intervals of equal length.

The occurrence or nonoccurrence of the event in any interval is independent of the occurrence or nonoccurrence in any other interval.

The probability that two or more events will occur in an interval approaches zero as the interval becomes smaller.

Therefore, the appropriate probability distribution is POISSON PROBABILITY DISTRIBUTION.

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Which of the following is not a congruence theorem or postulate A. SSA B. SAS C. AAS D. SSS
bixtya [17]

Answer:ITS A

Step-by-step explanation:

SAS: side angle side

SSA: is not a congruence theorem

AAS:angle angle side

SSS:side side side

3 0
2 years ago
12x+7<−11 AND5x−8≥4012
Arte-miy333 [17]

Answer:

No solution

Step-by-step explanation:

Given: 12x+7 and  5x-8\geq 40

Solve both inequality separately.

12x+7              Subtraction property of inequality

12x

12x

x

x

and

5x-8\geq 40               Addition property of inequality

5x\geq 40+8

5x\geq 48

x\geq \dfrac{48}{5}

Hence, the solution of the compound inequality intersection of both solutions.

Please find attachment for number line solution.

No solution

8 0
2 years ago
Read 2 more answers
A veterinarian does research on the causes of enteroliths, stones that develop in the colon of horses. She suspects that feeding
skad [1K]

Answer: 8

Step-by-step explanation:

The degree of freedom for t-distribution of testing the differnce between the two population mean is given by :-

df= n_1+n_2-2

, where n_1 = Size of first sample.

n_1 = Size of second sample.

As per given ,

Veterinarian selects five horses that are known to have enteroliths and compares the number of flakes of alfalfa they have eaten over a month with the number of flakes eaten by five horses free of enteroliths.

i.e. n_1=5 and n_2=5

Now , If she calculates a two-sample confidence interval by hand for the difference in the mean number of flakes fed to horses with and without enteroliths the degrees of freedom she should use are

df= 5+5-2=8

Hence, the correct answer is 8.

8 0
1 year ago
G A professor has two light bulbs in his garage. When both are burned out, they are replaced, and the next day starts with two w
Aneli [31]

Answer:

2/7 or 0.2857

Step-by-step explanation:

The expected time before the first bulb burns out (two bulbs working) is given by the inverse of the probability that a bulb will go out each day:

E_1 = \frac{1}{0.02}=50\ days

The expected time before the second bulb burns out (one bulb working), after the first bulb goes out, is given by the inverse of the probability that the second bulb will go out each day:

E_2 = \frac{1}{0.05}=20\ days

Therefore, the long-run fraction of time that there is exactly one bulb working is:

t=\frac{E_2}{E_1+E_2}=\frac{20}{20+50}\\t=\frac{2}{7}=0.2857

There is exactly one bulb working 2/7 or 0.2857 of the time.

7 0
2 years ago
If f(x)=x2+10sinx, show that there is a number c such that f(c)=1000.
gayaneshka [121]
F(x) is continuous for all x.

Pick a point and show that f(x) is either negative or positive. Pick another point and show that f(x) is negative, if positive, or positive, if negative.

At x = 30, f(30) - 1000 = 900 + 10sin(30) - 1000 ≤ 0
Now, show at another point f(x) - 1000 is positive, and hence, there would be root between 30 and such point.

Let's pick 40.
At x = 40, f(40) - 1000 = 1600 + 10sin(40) - 1000 ≥ 0

Since f(x) - 1000 is continuous, there lies a root between 30 and 40, and hence, 30 ≤ c ≤ 40
7 0
2 years ago
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