Two pound bags of candy are being used to fill containers that hold 1 1/2 pounds each. How many containers can be filled by each two pound bag of candy?
Answers
Answer:
4 Containers Can Be Filled By Each Two Pound Bag Of Candy .
Step-by-step explanation:
GIven Two Pound Bag Of Candy Used To Fill Container That hold Pound Each .
Now There are 4 Container of Capacity Pound Needed To fill Exactly 3 two Pound Bags .
Thus 4 container have Total capacity Of 6 Pound . Which Is Easily Filled By The 3 two Pound Bags . ( Without Any Loss )
Related Questions
If you had 9 pieces of licorice to share equally around 5 people, how much licorice would each person get? *
A triangle has one angle that measures 35°. which values could be the measures of the other two angles?
Helga needs to lay out her newly created pivot table. What are the two ways she can set up a pivot table? A. Helga can check the field or type in the data in the pivot table. B. Helga can slide cells to the four boxes or double-click properties to choose from a list of pivot table options. C. When Helga created the pivot table, Excel automatically put the fields to the appropriate cells of the pivot table. There aren't two ways to set up a pivot table. D. Helga can drag the fields to the four boxes in the pivot table field list or right-click a field name and choose its location from the shortcut menu.
Jorge lives 6/8 mile from school and 2/8 mile from a ballpark how much farther does Jorge live from school than from the balloark
B. 7 notebooks; 10 pencils
C. 10 notebooks; 7 pencils
D. 5 notebooks; 12 pencils
Answers
Answers
Distributhing - through parantheses
The answer is:
Answers
Answer:
neither A nor B will occur simultaneously, as they are mutually exclusive.
Step-by-step explanation:
To compute the probability that either event A occurs, or B occurs, or both occur, you can use the principle of the union of events. The probability of the union of two events A and B (denoted as A ∪ B) can be calculated as:
�
(
�
∪
�
)
=
�
(
�
)
+
�
(
�
)
−
�
(
�
∩
�
)
P(A∪B)=P(A)+P(B)−P(A∩B)
In this case, you're given:
�
(
�
∁
)
=
0.40
P(A
∁
)=0.40, which means the probability of the complement of A (i.e., the probability that A does not occur).
�
(
�
∣
�
)
=
0.80
P(B∣A)=0.80, which is the conditional probability of B occurring given that A has occurred.
Let's break it down:
�
(
�
∁
)
P(A
∁
) is the probability that event A does not occur, which is
1
−
�
(
�
)
1−P(A).
�
(
�
∣
�
)
P(B∣A) is the conditional probability that event B occurs given that A has occurred.
So, you can calculate
�
(
�
)
P(A) and
�
(
�
)
P(B) as follows:
�
(
�
)
=
1
−
�
(
�
∁
)
=
1
−
0.40
=
0.60
P(A)=1−P(A
∁
)=1−0.40=0.60
Now, you can use the formula for the union of events to calculate
�
(
�
∪
�
)
P(A∪B):
�
(
�
∪
�
)
=
�
(
�
)
+
�
(
�
)
−
�
(
�
∩
�
)
P(A∪B)=P(A)+P(B)−P(A∩B)
But before we calculate
�
(
�
∩
�
)
P(A∩B), note that events A and B are independent, so
�
(
�
∩
�
)
=
�
(
�
)
⋅
�
(
�
∣
�
)
P(A∩B)=P(A)⋅P(B∣A).
�
(
�
∩
�
)
=
�
(
�
)
⋅
�
(
�
∣
�
)
=
0.60
⋅
0.80
=
0.48
P(A∩B)=P(A)⋅P(B∣A)=0.60⋅0.80=0.48
Now, plug this value into the formula:
�
(
�
∪
�
)
=
0.60
+
�
(
�
)
−
0.48
P(A∪B)=0.60+P(B)−0.48
Solve for
�
(
�
)
P(B):
�
(
�
)
=
�
(
�
∪
�
)
+
0.48
−
0.60
P(B)=P(A∪B)+0.48−0.60
�
(
�
)
=
�
(
�
∪
�
)
−
0.12
P(B)=P(A∪B)−0.12
Now, you have the equation:
�
(
�
∪
�
)
=
0.60
+
�
(
�
∪
�
)
−
0.12
−
0.48
P(A∪B)=0.60+P(A∪B)−0.12−0.48
Simplify:
�
(
�
∪
�
)
=
0.60
−
0.12
−
0.48
P(A∪B)=0.60−0.12−0.48
�
(
�
∪
�
)
=
0.00
P(A∪B)=0.00
So, the probability that either event A occurs, or B occurs, or both occur is 0.00. This means that neither A nor B will occur simultaneously, as they are mutually exclusive.
Answers
f = gmn/d²
gmn = fd²
m = fd²/gn
Indus
Yangtze
Tigris and Euphrates
Answers