Given:
Two coins are tossed.
To find:
1. P(H on first coin)?
2. P(H on second coin)?
3. List the paired outcomes for tossing two coins.
4. How many ways are there for two coins to land?
5. What is P(HH)?
Solution:
If a a coin is tossed, then we have to possible outcomes, i.e., heads (H) and tails (T).
It is given that two coins are tossed.
1. The probability of getting a heads on first coin is:
2. The probability of getting a heads on second coin is:
3. If two coins are tossed, then the total possible outcomes are:
4. The number of ways for two coins to land is 4.
5. The probability of the heads on both tosses is:
Therefore, the required solution are:
1.
2.
3. List of possible outcomes is .
4. Number of possible outcomes is 4.
5.
The answer is 40 square m
Answer:
HH TT HT TH
Step-by-step explanation:
My extremely bad makeshift tree diagram:
the first column is our first flip, second column is our second
H
H-
T
possible outcomes -
H
T -
T
It would equal 0.45
Step-by-step explanation:
9 × 5 = 45 put a decimal in front of the 45
Answer:
No. 2/3 of the space to Grano and 1/3 of the space to Wheatie.
Step-by-step explanation:
Allocating about 57% to Wheatie and 43% to Grano, means to allocate 60(.57)=34.2 ft2 of Wheatie and 60(.43)=25.8 ft2 of Grano. In that case there would be 34.2/.4=85.5≈85 boxes of Wheatie and 25.8/.2=129 boxes of Grano. The total profit would be 129(1)+85(1.35)=$243.75
Best option:
200 Granos boxes and 50 Wheaties boxes on the shelf.
200(.2)=40ft^2 will allocate Granos boxes
50(.4)=20ft^2 will allocate Wheaties boxes.
This means that 40/60=2/3=66.6% of the space will allocate Granos boxes and 20/60=1/3=33.3% of the space will allocate Wheaties boxes.
The total profit would be 200(1)+50(1.35)=$267.5
EXTRA:
This is a optimization problem.
Let X1 be the number of Granos boxes
Let X2 be the number of Wheaties boxes
Objective:
Max Z=1(X1)+1.35(X2)
Subjecto to
0.2(X1)+0.4(X2)<=60,
X1<=200,
X2<=120,
X1,X2>=0.
You can solve it using the simplex method. Check the image for more details.
