Answer:
The second graph near the bottom with a Y-intercept of -1
Answer:
The answer is <u>3.2</u>
Step-by-step explanation:
- <em>8 x 40</em>
- <em>________</em>
- <em>100</em>
- <em> = 3.2</em>
Total number of oz of honey = 12 oz.
Cost of 12 oz of honey = $5.40.
Cost per oz of honey = 5.40 divided 12 = $0.45 per oz.
Therefore,
a) Cost per ounce of honey = $0.45.
b) Number of ounces we can buy for one dollar = 12 divided by 5.40 = 2.22 oz approximatly.
c)
h = 2.22c, wherr h is the number of ounces of honey for c dollars.
c=0.45h, where c is the total cost of h number of ounces of honey.
d) Let us take c=0.45h equation for graph the situation.
x-axis would be for h number of ounces of honey and y-axis is the total cost c of h ounces of honey.
Let us find some values of h and c for the graph.
For equation: c=0.45h for h=1,we get c=$0.45.
If we plug h=2, we get c= 0.45*2 = $0.90.
If we plug h=3, we get c= 0.45*3 = $1.35.
Therefore, we would plot (1,0.45), (2,0.90) and (3,1.35) on the graph and draw a straight line.
Given that a display allows a customer to hook together any selection of components, one of each type. These are the types:
Receiver: Kenwood, Onkyo, Pioneer, Sony, Sherwood
CD player: Onkyo, Pioneer, Sony, Technics
Speakers: Boston, Infinity, Polk
Cassette: Onkyo, Sony, Teac, Technics:
Part (a):
In how many ways can one component of each type be selected?
The number of ways one type of receiver will be selected is given by 5C1 = 5
The number of ways one type of CD player will be selected is given by 4C1 = 4
The number of ways one type of speakers will be selected is given by 3C1 = 3
The number of ways one type of cassette will be selected is given by 4C1 = 4
Therefore, the number of ways one component of each type can be selected is given by 5 x 4 x 3 x 4 = 240 ways
Part (b):
In how many ways can components be selected if both the
receiver and the compact disc player are to be Sony?
The number of ways of selecting a Sony receiver is 1
The number of ways of selecting a Sony CD player is 1
The number of ways one type of speakers will be selected is given by 3C1 = 3
The number of ways one type of cassette will be selected is given by 4C1 = 4
Therefore, the number of ways components can be selected if both the
receiver and the compact disc player are to be Sony is given by 1 x 1 x 3 x 4 = 12
Part (c)
In how many ways can components be selected if none of them are Sony?
The number of ways one type of receiver that is not Sony will be selected is given by 4C1 = 4
The number of ways one type of CD player that is not Sony will be selected is given by 3C1 = 3
The number of ways one type of speakers that is not Sony will be selected is given by 3C1 = 3
The number of ways one type of cassette that is not Sony will be selected is given by 3C1 = 3
Therefore, the number of ways that components can be selected if none of them are Sony is given by 4 x 3 x 3 x 3 = 108
Part (d):
In how
many ways can a selection be made if at least one Sony component is
to be included?
The total number of ways of selecting one component of each type is 240
The number of ways that components can be selected if none of them are Sony is 108
Therefore, the number of ways of selecting at least one Sony component is given by 240 - 108 = 132
Part (e):
If someone flips switches on the selection in a
completely random fashion, what is the probability that the system
selected contains at least one Sony component?
The total number of ways of selecting one component of each type is 240
The number of ways of selecting at least one Sony component is 132
Therefore, the probability that a system
selected at random contains at least one Sony component is given by 132 / 240 = 0.55
Part (f):
If someone flips switches on the selection in a
completely random fashion, what is the probability that the system
selected contains exactly one Sony
component? (Round your answer to three decimal places.)
The number of ways of selecting only a Sony receiver is given by 1 x 3 x 3 x 3 = 27
The number of ways of selecting only a Sony CD player is given by 4 x 1 x 3 x 3 = 36
The number of ways of selecting only a Sony cassette is given by 4 x 3 x 3 x 1 = 36
Thus, the number of ways of selecting exactly one Sony component is given by 27 + 36 + 36 = 99
Therefore, the probability that a system
selected at random contains exactly one Sony
component is given by 99 / 240 = 0.413
