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Gemiola [76]
3 years ago
15

Consider the following statements:

Mathematics
1 answer:
Black_prince [1.1K]3 years ago
3 0

Answer:

b. I and II are both false.

Step-by-step explanation:

I. For a significance level, the two tailed hypothesis is not always accurate than the one tailed hypothesis test. The hypothesis testing is carried to find out the correctness of a claim of a population parameter. The two tail hypothesis test which used both positive and negative tails of the distribution is not always more accurate than one tailed test.

II. The process of the point estimation involves the utilization of the values of a statistic which is obtained from the sample data to obtain the best estimate of a corresponding unknown parameter in the given population.

Hence, both the statements are false.

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Katen [24]
You can find a common demominator and then change 8 1/4 into a mixed number and just subtract
7 0
3 years ago
Felipe and Makayla each tried to solve the same
amid [387]
Felipe subtract 3x from both side to get rid of x on one side. Ik this isn’t part of the answer but the correct way to do this would be to subtract 2x from each side after adding 8 to each side of the equation
7 0
3 years ago
solo el 20% de los empleados de la población civil que está en una base militar restringida porta su identificación personal. Si
prisoha [69]

Usando la distribución binomial, hay una probabilidad de 0.8926 = 89.26% de que el guardia de seguridad encuentre al menos uno en la base militar restringida.

<h3>¿Qué es la distribución binomial?</h3>

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

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

Los parámetros son:

  • n es el número de ensayos.
  • p es la probabilidad de éxito en un ensayo
  • x es el número de éxitos

En este problema, hay que:

  • 20% de los empleados de la población civil que está en una base militar restringida porta su identificación personal, o sea p = 0.2.
  • Llegan 10 empleados, o sea, n = 10.

La probabilidad de que el guardia de seguridad encuentre al menos uno en la base militar restringida es dada por:

P(X \geq 1) = 1 - P(X = 0)

En que:

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

P(X = 0) = C_{10,0}.(0.2)^{0}.(0.8)^{10} = 0.1074

Por eso:

P(X \geq 1) = 1 - P(X = 0) = 1 - 0.1074 = 0.8926

Hay una probabilidad de 0.8926 = 89.26% de que el guardia de seguridad encuentre al menos uno en la base militar restringida.

Puede-se aprender más a cerca de la distribución binomial en brainly.com/question/25132113

3 0
2 years ago
a city doubles its size every 90 years if the population is currently 90,000, what will the population be in 270 years DUE IN FI
kati45 [8]

Answer:

the world would be filled with people

we would all be dead

Step-by-step explanation:

8 0
3 years ago
Read 2 more answers
Suppose two dice are tossed and the numbers on the upper faces are observed. Let S denote the set of all possible pairs that can
Thepotemich [5.8K]

Answer:

▪A = {(1,2) (1,4) (1,6) (2,2) (2,4) (2,6) (3,2) (3,4) (3,6) (4,2) (4,4) (4,6) (5,2) (5,4) (5,6) (6,2) (6,4)(6,6)}

▪C bar = {(2,2) (2,4) (2,6) (4,2) (4,4) (4,6) (6,2) (6,4) (6,6)}

▪A∩B = {(2,2) (2,4) (2,6) (4,2) (4,4) (4,6) (6,2) (6,4) (6,6)}

▪A∩B bar = {(1,2) (1,4) (1,6) (3,2) (3,4) (3,6) (5,2) (5,4) (5,6)}

▪A bar∪B = {(1,1) (1,3) (1,5) (2,1) (2,2) (2,3) (2,4) (2,5) (2,6) (3,1) (3,3) (3,5) (4,1) (4,2) (4,3) (4,4) (4,5) (4,6) (5,1) (5,3) (5,5) (6,1) (6,2) (6,3) (6,4) (6,5) (6,6)}

▪A bar∩C = {(1,1) (1,3) (1,5) (2,1) (2,3) (2,5) (3,1) (3,3) (3,5) (4,1) (4,3) (4,5) (5,1) (5,3) (5,5) (6,1) (6,3) (6,5)}

Step-by-step explanation:

S = {(1,1) (1,2) (1,3) (1,4) (1,5) (1,6) (2,1) (2,2) (2,3) (2,4) (2,5) (2,6) (3,1) (3,2) (3,3) (3,4) (3,5) (3,6) (4,1) (4,2) (4,3) (4,4) (4,5) (4,6) (5,1) (5,2) (5,3) (5,4) (5,5) (5,6) (6,1) (6,2) (6,3) (6,4) (6,5) (6,6)}

A = {(1,2) (1,4) (1,6) (2,2) (2,4) (2,6) (3,2) (3,4) (3,6) (4,2) (4,4) (4,6) (5,2) (5,4) (5,6) (6,2) (6,4)(6,6)} (second die is even)

B = {(1,1) (1,3) (1,5) (2,2) (2,4) (2,6) (3,1) (3,3) (3,5) (4,2) (4,4) (4,6) (5,1) (5,3) (5,5) (6,2) (6,4) (6,6)} (sum of the two numbers is even)

C = {(1,1) (1,2) (1,3) (1,4) (1,5) (1,6) (2,1) (2,3) (2,5) (3,1) (3,2) (3,3) (3,4) (3,5) (3,6) (4,1) (4,3) (4,5) (5,1) (5,2) (5,3) (5,4) (5,5) (5,6) (6,1) (6,3) (6,5)} (at least one in the pair is odd i.e one of the pair is odd or both are odd)

A bar = {(1,1) (1,3) (1,5) (2,1) (2,3) (2,5) (3,1) (3,3) (3,5) (4,1) (4,3) (4,5) (5,1) (5,3) (5,5) (6,1) (6,3) (6,5)} (the pairs that are not in A)

B bar = {(1,2) (1,4) (1,6) (2,1) (2,3) (2,5) (3,2) (3,4) (3,6) (4,1) (4,3) (4,5) (5,2) (5,4) (5,6) (6,1) (6,3) (6,5)} (the pairs that are not in B)

C bar = {(2,2) (2,4) (2,6) (4,2) (4,4) (4,6) (6,2) (6,4) (6,6)} (the pairs that are not in C)

▪A = {(1,2) (1,4) (1,6) (2,2) (2,4) (2,6) (3,2) (3,4) (3,6) (4,2) (4,4) (4,6) (5,2) (5,4) (5,6) (6,2) (6,4)(6,6)}

▪C bar = {(2,2) (2,4) (2,6) (4,2) (4,4) (4,6) (6,2) (6,4) (6,6)}

▪A∩B = {(2,2) (2,4) (2,6) (4,2) (4,4) (4,6) (6,2) (6,4) (6,6)} (intersection: the pairs that are common to both A and B)

▪A∩B bar = {(1,2) (1,4) (1,6) (3,2) (3,4) (3,6) (5,2) (5,4) (5,6)} (intersection: the pairs that are common to both A and B bar)

▪A bar∪B = {(1,1) (1,3) (1,5) (2,1) (2,2) (2,3) (2,4) (2,5) (2,6) (3,1) (3,3) (3,5) (4,1) (4,2) (4,3) (4,4) (4,5) (4,6) (5,1) (5,3) (5,5) (6,1) (6,2) (6,3) (6,4) (6,5) (6,6)} (union: all the pairs in A bar and B )

▪A bar∩C = {(1,1) (1,3) (1,5) (2,1) (2,3) (2,5) (3,1) (3,3) (3,5) (4,1) (4,3) (4,5) (5,1) (5,3) (5,5) (6,1) (6,3) (6,5)} (intersection: the pairs that are common to both A bar and C)

6 0
3 years ago
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