Dado una población N(321, 54), para una muestra aleatoria de tamaño n = 28, Ejercicio 0.5. La distribución de la media muestral X¯, es: (a) Una normal con media 321 y varianza 54 (b) Una normal con media 321 y varianza 1.3887 (c) Una normal con media 321 y varianza 1.9286 (d) Una normal con media 321 y varianza 1 Ejercicio 0.6. Hallar la probabilidad que la media muestral barX asuma valores en el intervalo [289, 3573]. (a) 0.9633 (b) 1.0000 (c) 0.6912 (d) 0.9543 Ejercicio 0.7. La probabilidad que la media muestral barX difiera de la media poblacional µ en más de 5.67 es: (a) 0.9016 (b) 0.8961 (c) 0.9942 (d) 0.9966

m∠cdf = 43° , m∠gdf = 114°, m∠gde = 23°, m∠edh = 23° , m∠cdh = 157°

step-by-step explanation:

since, it is given that de bisects the angle gdh ⇒ de divides the angle gdh into two equal angles.

⇒ m∠gde = m∠edh

⇒ 8·x   - 1 = 6·x + 5

⇒ x = 3

so, m∠gde = 8·x - 1 = 23°

and m∠edh = 23°

now, it is given ∠cde is an straight angle so, m∠cde = 180°

⇒ m∠cdf + m∠fdg + m∠gde = 180°

⇒ 43° + m∠fdg + 23° = 180°

⇒ m∠fdg = 114°

again, m∠cde = m∠cdh + m∠edh = 180°

m∠cdh + 23° = 180°

⇒ m∠cdh = 157°

the solution of the equation is 2.23 ⇒ answer b

step-by-step explanation:

* lets revise the meaning of exponential function

- the form of the exponential function is y = ab^x, where a ≠ 0, b > 0 ,  

  b ≠ 1, and x is any real number

- it has a constant base b

- it has a variable exponent x

- to solve an exponential equation, take the log or ln of both sides,  

  and solve for the variable

* lets solve the problem

∵

- the base is 7 and the exponent is x

- insert ㏑ in both sides

∴

- use the rule

∴ x ㏑(7) = ㏑(77)

- to find x divide both sides by ㏑(7)

∴

* the solution of the equation is 2.23 to the nearest 2 decimal places