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2 years ago

Ethan rolls a 6-sided number cube. What is the probability that he gets a number greater than 2?

2 answers:

6 0

D.
This is because the probability of getting a number greater than 2 is 4/6.
Therefore, simplifying the expression gives us the fraction 2/3.

posledela2 years ago

6 0

Answer: D. $\frac{2}{3}$ .

Step-by-step explanation:

When we rolls a 6-sided number cube., the total outcomes would be 6.

The number of cube greater than 2 = (3,4,5,6)=4

Therefore, the probability that he gets a number greater than 2 is given by :-

$P(\text{Greater than 2})=\frac{\text{Number greater than 2}}{\text{Total number}}\\\\=\frac{4}{6}=\frac{2}{3}$

Hence, the probability that he gets a number greater than 2 is $\frac{2}{3}$ .

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A helicopter flies 8 km due north from A to B. It then flies 5 km from B to C and returns to A as shown in the figure. Angle ABC

In-s [12.5K]

Please consider the attached graph.

We have been given that a helicopter flies 8 km due north from A to B. It then flies 5 km from B to C and returns to A as shown in the figure. The measure of angle ABC is 150°. We are asked to find the area of triangle ABC.

We will use trigonometric area formula to solve our given problem.

$A=\frac{1}{2}a\cdot c\cdot\text{sin}(b)$ , where angle b is angle between sides a and c.

For our given triangle $a=5,c=8$ and measure of angle b is 150 degrees.

$A=\frac{1}{2}(5)\cdot (8)\cdot\text{sin}(150^{\circ})$

$A=5\cdot 4\cdot(0.5)$

$A=10$

Therefore, the area of the triangle ABC is 10 square kilo-meters and option 'c' is the correct choice.

6 0

2 years ago

Ken normally leaves work at 5:00 pm, but he is leaving 20 minutes late today. He decides to make up time by taking the toll road

NemiM [27]

Answer:

y =1/3x+20

Step-by-step explanation:

He starts 20 minutes late so we need to add 20 minutes to the time

He travels 3 times faster so he takes 1/3 of the normal time x

y =1/3x+20

4 0

2 years ago

The results of a mathematics placement exam at two different campuses of Mercy College follow: Campus Sample Size Sample Mean Po

Leona [35]

Answer:

$z=\frac{(33-31)-0}{\sqrt{\frac{8^2}{330}+\frac{7^2}{310}}}}=3.37$

$p_v =P(Z>3.37)=1-P(Z$

Comparing the p value with a significance level for example $\alpha=0.05$ we see that $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, and we can say that the mean for the Campus 1 is significantly higher than the mean for the group 2.

Step-by-step explanation:

Data given

Campus Sample size Mean Population deviation

1 330 33 8

2 310 31 7

$\bar X_{1}=33$ represent the mean for sample 1

$\bar X_{2}=31$ represent the mean for sample 2

$\sigma_{1}=8$ represent the population standard deviation for 1

$\sigma_{2}=7$ represent the population standard deviation for 2

$n_{1}=330$ sample size for the group 1

$n_{2}=310$ sample size for the group 2

$\alpha$ Significance level provided

z would represent the statistic (variable of interest)

Concepts and formulas to use

We need to conduct a hypothesis in order to check if the mean for Campus 1 is higher than the mean for Campus 2, the system of hypothesis would be:

Null hypothesis: $\mu_{1}-\mu_{2}\leq 0$

Alternative hypothesis: $\mu_{1} - \mu_{2}> 0$

We have the population standard deviation's, and the sample sizes are large enough we can apply a z test to compare means, and the statistic is given by:

$z=\frac{(\bar X_{1}-\bar X_{2})-\Delta}{\sqrt{\frac{\sigma^2_{1}}{n_{1}}+\frac{\sigma^2_{2}}{n_{2}}}}$ (1)

z-test: Is used to compare group means. Is one of the most common tests and is used to determine whether the means of two groups are equal to each other.

With the info given we can replace in formula (1) like this:

$z=\frac{(33-31)-0}{\sqrt{\frac{8^2}{330}+\frac{7^2}{310}}}}=3.37$

P value

Since is a one right tailed test the p value would be:

$p_v =P(Z>3.37)=1-P(Z$

5 0

2 years ago

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

How much heat is required to raise the temperature of

Elina [12.6K]

Answer:

The amount of heat required to raise the temperature of liquid water is 9605 kilo joule .

Step-by-step explanation:

Given as :

The mass of liquid water = 50 g

The initial temperature = $T_1$ = 15°c

The final temperature = $T_2$ = 100°c

The latent heat of vaporization of water = 2260.0 J/g

Let The amount of heat required to raise temperature = Q Joule

Now, From method

Heat = mass × latent heat × change in temperature

Or, Q = m × s × ΔT

or, Q = m × s × ( $T_2$ - $T_1$ )

So, Q = 50 g × 2260.0 J/g × ( 100°c - 15°c )

Or, Q = 50 g × 2260.0 J/g × 85°c

∴ Q = 9,605,000 joule

Or, Q = 9,605 × 10³ joule

Or, Q = 9605 kilo joule

Hence The amount of heat required to raise the temperature of liquid water is 9605 kilo joule . Answer

4 0

1 year ago