Answer:
Hay 160 maneras
Step-by-step explanation:
Para calcular de cuántas maneras se puede seleccionar x elementos de un grupo de n elementos podemos usar la siguiente fórmula:
Si vas a elegir 3 personas de entre 6 matrimonios y dos miembros de la misma pareja no pueden ser elegidos, entonces podemos decir que se está eligiendo 3 matrimonios y en cada matrimonio se está eligiendo un representante.
Entonces, podemos calcular de cuántas maneras se puede escoger 3 matrimonios de los 6, así:
Adicionalmente, para cada uno de los 3 matrimonios hay 2 opciones para ser representantes. Esto significa que hay 
maneras de escoger representantes en cada una de las 20 formas calculadas anteriormente.
Por lo tanto se puede formar el grupo de 3 personas de 160 maneras diferentes:
All you need to do is multiply the different types of each choice.
For salad, there are three different items. For entree, there are four different items. For side dish, there are three different items. for dessert, there are three different items.
Multiply the four numbers:
<u>108</u> dinner specials are possible.
Answer:
1
Step-by-step explanation:
There is one whole 3 inside of 4 1/2 or 4.5.
Multiples of 3 are 3x1=3 3x2=6....
Only a single whole 3 can be inside of 4.5.
Answer:
a) F
b) B, E, D
Step-by-step explanation:
a) The segment with the greatest gradient has the largest change in y-values per unit change in x-values
From the given option, the rate of change of the <em>y </em>to the<em> </em>x-values of B = the gradient = (4 units)/(2 units) = 2
The gradient of F = (-3units)/(1 unit) = -3
The gradient of A = 4/4 = 1
The gradient of C = -2/5
The gradient of D = 2/6 = 1/3
The gradient of E = 3/4
The segment with the greatest gradient is F
b) The steepest segment has the higher gradient
From their calculated we have;
The gradient of segment B = 2 therefore, B is steeper than E that has a gradient of 3/4, and E is steeper than D, as the gradient of D = 1/3
Therefore, we have;
B, E, D.
What page is it because I have the same math book?Unless you're not in 8th grade then I can't help.
