Determine if 0.515115111511115111115... is rational or irrational and give

When a certain basketball player takes his first shot in a game he succeeds with probability 1/2. If he misses his first shot, h

Gnom [1K]

Answer:

we are given

basketball player Chauncey Billups of the Detroit Pistons makes free throw shots 88% of the time

so, probability of making shot is

=88%

so, p=0.88

To find the probability of missing first shot and making the second shot

so, we can use formula

probability = p(1-p)

now, we can plug values

we get

So, the probability that he misses his first shot and makes the second is 0.1056........Answer

Step-by-step explanation:

3 0

2 years ago

Two number cubes are rolled to determine how a token moves on a game board. The sides of

ahrayia [7]

Answer:

D. The mathematical expectation of Option A is 1. The mathematical expectation of Option B is 1.5. Option B offers a greater likelihood of advancing to the finish line.

Step-by-step explanation:

The result of a product is odd only when the two numbers are odds.

There are 6*6 = 36 possible outcomes when two dice are rolled. Only 9 of them are a combination of two odd numbers: {1, 1} {1, 3} {1, 5} {3, 1} {3, 3} {3, 5} {5, 1} {5, 3} {5, 5}. Then 36 - 9 = 27 outcomes are even.

P(even) = 27/36 = 0.75

Option A) Mathematical expectation: 0.75*4 + 0.25*(-8) = 1

Option B) Mathematical expectation: 0.75*5 + 0.25*(-9) = 1.5

8 0

2 years ago

The lifespans of lions in a particular zoo are normally distributed. The average lion lives 12.512.512, point, 5 years; the stan

zaharov [31]

This question was not written properly

Complete Question

The lifespans of lions in a particular zoo are normally distributed. The average lion lives 12.5 years; the standard deviation is 2.4 years. Use the empirical rule (68-95-99.7\%)(68−95−99.7%) to estimate the probability of a lion living between 5.3 to 10. 1 years.

Answer:

Thehe probability of a lion living between 5.3 to 10. 1 years is 0.1585

Step-by-step explanation:

The empirical rule formula states that:

1) 68% of data falls within 1 standard deviation from the mean - that means between μ - σ and μ + σ.

2) 95% of data falls within 2 standard deviations from the mean - between μ – 2σ and μ + 2σ.

3) 99.7% of data falls within 3 standard deviations from the mean - between μ - 3σ and μ + 3σ.

Mean is given in the question as: 12.5

Standard deviation : 2.4 years

We start by applying the first rule

1) 68% of data falls within 1 standard deviation from the mean - that means between μ - σ and μ + σ.

μ - σ

12.5 -2.4

= 10.1

We apply the second rule

2) 95% of data falls within 2 standard deviations from the mean - between μ – 2σ and μ + 2σ.

μ – 2σ

12.5 - 2 × 2.4

12.5 - 4.8

= 7.7

We apply the third rule

3)99.7% of data falls within 3 standard deviations from the mean - between μ - 3σ and μ + 3σ.

μ - 3σ

= 12.5 - 3(2.4)

= 12.5 - 7.2

= 5.3

From the above calculation , we can see that

5.3 years corresponds to one side of 99.7%

Hence,

100 - 99.7%/2 = 0.3%/2

= 0.15%

And 10.1 years corresponds to one side of 68%

Hence

100 - 68%/2 = 32%/2 = 16%

So,the percentage of a lion living between 5.3 to 10. 1 years is calculated as 16% - 0.15%

= 15.85%

Therefore, the probability of a lion living between 5.3 to 10. 1 years

is converted to decimal =

= 15.85/ 100

= 0.1585

8 0

2 years ago

There are four activities on the critical path, and they have standard deviations of 1, 2, 4, and 2. what is the probability tha

Anit [1.1K]

Standard deviations of the four activities of the critical path are 1,2,4,2.

Standard deviation of this critical path = Sum of square root of variance of this corresponding critical path

Standard deviation of critical path $=\sqrt{1^2+2^2+4^2+2^2}$

$=\sqrt{1+4+16+4}$

$=\sqrt{25}$

$=5$

Now we need to find the probability that the project will completed in 38 weeks given that its expected completion time is 40 weeks.

That is, we need to find P(X<38) :

$z=\frac{38-40}{5}= \frac{-2}{5}=-0.4$

$P(X$

Probability $=0.5-0.16=0.34$

Thus the probability that the project will be completed in 38 weeks is 0.34.

8 0

2 years ago

Read 2 more answers

Imaginá que tenés 125 dados cúbicos del mismo tamaño ¿Cuantos dados de altura tiene el cubo de mayor tamaño que podés armar apil

kumpel [21]

Answer:

(i) Debemos apilar 5 dados para construir el cubo de mayor tamaño.

(ii) Se necesita 121 dados cuadrados para formar el cuadrado con la mayor cantidad de dados posibles, quedando 4 dados sobrantes.

Step-by-step explanation:

(i) Sabemos por la Geometría Euclídea del Espacio que un cubo es un sólido regular con 6 caras cuadradas y longitudes iguales. Cada dado tiene un volumen de 1 dado cúbico y 125 dados dan un volumen total de 125 dados cúbicos.

El volumen de un cubo está dado por la siguiente fórmula:

$V = L^{3}$

Donde:

$L$ - Longitud de la arista, medida en dados.

$V$ - Volumen del cubo, medido en dados cúbicos.

Ahora, necesitamos despejar la longitud de la arista para calcular la altura máxima posible:

$L = \sqrt[3]{V}$

Dado que $V = 125\,dados^{3}$ , encontramos que la altura del cubo de mayor tamaño sería:

$L =\sqrt[3]{125\,dados^{3}}$

$L = 5\,dados$

Debemos apilar 5 dados para construir el cubo de mayor tamaño.

(ii) El área cuadrada formada por cubos está determinada por la siguiente fórmula:

$A = L^{2}$

Donde:

$L$ - Longitud de arista, medida en dados.

$A$ - Área, medida en dados cuadrados.

Puesto que la longitud de arista se basa en un conjunto discreto, esto es, el número de dados disponibles, debemos encontrar el valor máximo de $L$ tal que no supere 125 y de un área entera. Es decir:

$L \leq 125\,dados$

Si cada cubo tiene un área de 1 dado cuadrado, entonces un cuadrado conformado por 125 dados tiene un área total de 125 dados cuadrados. Entonces:

$L^{2}< 125\,dados^{2}$

Esto nos lleva a decir que:

$L < 11.180\,dados$

Entonces, la longitud máxima del cuadrado con la mayor cantidad de cubos posible es de 11 dados. El número total requerido de cubos es el cuadrado de esa cifra, es decir:

$n = (11\,dados)^{2}$

$n = 121\,dados$

Se necesita 121 dados cuadrados para formar el cuadrado con la mayor cantidad de dados posibles, quedando 4 dados sobrantes.

4 0

2 years ago