1) The outcomes for rolling two dice, the sample space, is as follows:
(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6)
(2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6)
(3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6)
(4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6)
(5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6)
(6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)
There are 36 outcomes in the sample space.
2) The ways to roll an odd sum when rolling two dice are:
(1, 2), (1, 4), (1, 6), (2, 1), (2, 3), (2, 5), (3, 2), (3, 4), (3, 6), (4, 1), (4, 3), (4, 5), (5, 2), (5, 4), (5, 6), (6, 1), (6, 3), (6, 5). There are 18 outcomes in this event.
3) The probability of rolling an odd sum is 18/36 = 1/2 = 0.5
Answer:
Using the ratio table the dogs weight is:
30 pounds = 13.5 kilograms
Step-by-step explanation:
For this case we have the following conversion:
20 pounds = 9 kilograms
To use the table what we must do is find another relationship that allows us to find the weight in kilograms for 30 pounds.
For example, half the weight in pounds is half the weight in kilograms.
Therefore, the given conversion is:
10 pounds = 4.5 kilograms
So, for 30 pounds, we multiply this last ratio obtained by three on both sides:
30 pounds = 13.5 kilograms
Then, the table is:
Pounds 10 20 30
This question is not complete
Complete Question
A boat sails 4km on a bearing of 038 degree and then 5km on a bearing of 067 degree.(a)how far is the boat from its starting point.(b) calculate the bearing of the boat from its starting point
Answer:
a)8.717km
b) 54.146°
Step-by-step explanation:
(a)how far is the boat from its starting point.
We solve this question using resultant vectors
= (Rcos θ, Rsinθ + Rcos θ, Rsinθ)
Where
Rcos θ = x
Rsinθ = y
= (4cos38,4sin38) + (5cos67,5sin67)
= (3.152, 2.4626) + (1.9536, 4.6025)
= (5.1056, 7.065)
x = 5.1056
y = 7.065
Distance = √x² + y²
= √(5.1056²+ 7.065²)
= √75.98137636
= √8.7167296826
Approximately = 8.717 km
Therefore, the boat is 8.717km its starting point.
(b)calculate the bearing of the boat from its starting point.
The bearing of the boat is calculated using
tan θ = y/x
tan θ = 7.065/5.1056
θ = arc tan (7.065/5.1056)
= 54.145828196°
θ ≈ 54.146°