Answer:
the equation would be y=1/2x-9/2
Step-by-step explanation:
next time try to put the answers like the question did because that's always helpful
Answer:
the width is 3.5 in.
Step-by-step explanation:
u get it by dividing 182 in and 52 in and u get 3.5 In
The answer is: The first triangle. The reasons are shown below:
1. All the triangles are rigth triangles, because they have an angle of 90°. So, let's calculate the others angles of the first one:
Tan(α)^-1= opposite leg/adjacent leg
Opposite leg=5
Adjacent leg=5√3
Tan(α)^-1= 5/5√3
Tan(α)^-1=30°
2. Let's calculate the other angle:
Tan(α)^-1= opposite leg/adjacent leg
Now, the opposite leg will be 5√3 and the adjacent leg will be 5. Then:
Tan(α)^-1= 5√3/5
Tan(α)^-1=60°
As you can see, the angles of first triangle are: 30°,60° and 90°.
Since there are more parakeets than canaries, it is not possible to have only 1 of each bird in each cage <u>and</u> have the same number of birds in each cage.
He could use 42 cages, putting a canary in with the parakeet in 18 of them. Then he would have 18 cages with 2 birds each, and 24 cages with 1 bird each.
The only way to have the same number of birds (1) in all cages is to have 60 cages, 42 of which have 1 parakeet, and 18 of which have 1 canary.
_____
If more than 1 of each kind of bird can be put in the cage, the collection of birds could be put into 6 cages, each of which would be home to 7 parakeets and 3 canaries.
Answer:
0.6,0.7,0.3 neither disjoint nor independent.
Step-by-step explanation:
Given that at a large university, 60% of the students have a Visa card and 40% of the students have a MasterCard.
A= visa card
B = Master card
P(A) = 0.60 and P(B) = 0.40
P(AUB)' = 0.30
i.e. P(AUB) = 0.70
Or P(A)+P(B)-P(AB) =0.70
P(AB)= 0.30
Randomly select a student from the university.
1) the probability that this student does not have a MasterCard.
2. the probability that this student has either a Visa card or a MasterCard.
=
3. Calculate the probability that this student has neither a Visa card nor a MasterCard.
=
4. Are the events A and B disjoint? Are the events A and B independent?
A and B have common prob 0.30 hence not disjoint.
P(AB) ≠P(A)P(B)
Hence not independent
