A single die is rolled twice. the set of 36 equally likely outcomes is {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (2, 1),
Maksim231197 [3]
<span>We need to find the rolls whose sum is greater than 10. By looking at the outcomes, we see that (5,6), (6,5), and (6,6) all have a sum greater than 10. Therefore, there are 3 chances to get a sum greater than 10. Since there are 36 chances overall, the probability of rolling greater than 10 are 3/36 = 1/12.</span>
Answer:
V = 792 m3
I am kinda guessing on this since you didn't provide a unit or label what the height, width, and length are. Also you should specify what prism...
Finding the midpoint coordinates of any segment really boils down to finding the midpoints of each individual coordinate.
The x-coordinates of the two points are -12 and -8 - the number halfway between those two is -10, so that'll be the midpoint's x-coordinate. The y-coordinates are -7 and -4 - -5.5 is halfway between these two, so the y-coordinate will be 5.5.
Putting the two together, the midpoint of the segment WT has the coordinates (-10, -5.5).
Answer:
The answer is -2, -17
Step-by-step explanation:
You write a system of equations
Then you solve:
-2, -17
Answer:
7.30167%
Step-by-step explanation:
Usando la fórmula de puntuación z
z = (x-μ) / σ, donde x es la puntuación bruta, μ es la media de la población y σ es la desviación estándar de la población
Para x <0.20 pulgadas
z = 0.20 - 0.25 / 0.02
z = -2.5
Valor de probabilidad de Z-Table:
P (x <0.20) = 0.0062097
Para x> 0.28 pulgadas
z = 0.28 - 0.20 / 0.02
z = 1.5
Valor de probabilidad de Z-Table:
P (x <0.28) = 0.93319
P (x> 0.28) = 1 - P (x <0.28) = 0.066807
La probabilidad de que se produzcan tornillos defectuosos cuando el tornillo se considera defectuoso si su diámetro es inferior a 0.20 pulgadas o superior a 0.28 pulgadas es
P (x <0.20) + P (x> 0.28)
= 0.0062097 + 0.066807
= 0.0730167
Conversión a porcentaje
= 0.0730167 × 100
= 7.30167%
El porcentaje de tornillos defectuosos producidos es
7.30167%
