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leva [86]
3 years ago
12

How does the y-intercept appear in these three representations: table,equation,and graph?

Mathematics
2 answers:
devlian [24]3 years ago
6 0
In a table it's  EXAMPLE: Cars/Drivers the cars is x and the drivers is y(y-intercept). In an equation, EXAMPLE using y=mx+b the b is the y-int., and in a graph it is (x,y) the y being the y-int.
Vikentia [17]3 years ago
6 0

Answer:

Table: The y coordinate value corresponding to x = 0

Equation: c is the y-intercept.

Graph: curve passes the y- axis

Step-by-step explanation:

We have to find how does the y-intercept appear in these three representations: table,equation,and graph.

Y-intercept:

  • The y-intercept of this line is the value of y at the point where the line crosses the y axis.
  • Thus, y-intercept is the value of y when x is 0.

In table:

The table shows the x coordinate and the corresponding y coordinates. The y coordinate value corresponding to x = 0 in the table is the y intercept.

In equation:

The slope intercept form of straight line is given by:

y = mx + c

where m is the slope of the line and c is the y-intercept.

In graph:

The point on the graph where the curve passes the y- axis that is the x value is zero.

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According to a​ survey, 60 ​% of the residents of a city oppose a downtown casino. Of these 60 ​% about 8 out of 10 strongly opp
katrin [286]

Answer:

(a) 0.48

(b) 0.20

(c) it is not unusual for a radomly selected resident to oppose the casino and strongly oppose the​ casino.

Step-by-step explanation:

​(a) Find the probability that a randomly selected resident opposes the casino and strongly opposes the casino. ​

The probability that a radomly selected resident opposes the casino and strongly opposes the cassino is the product of the two probabilities, that a resident opposes the casino and that it strongly opposes the casino (once it is in the first group) as it is shown below.

Use this notation:

  • Probability that a radomly selected resident opposes the casino: P(A)

  • Probability that a resident who opposes the casino strongly opposes it: P(B/A), because it is the probability of event B given the event A

i) Determine the <em>probability that a radomly selected resident opposes the casino</em>, P(A)

Probability = number of favorable outcomes / number of possible outcomes

  • P(A) is <em>given as 60%</em>, which in decimal form is 0.60

ii) Next, determine,the <em>probability that a resident who opposes the casino strongly opposes it</em>, P(B/A):

  • It is given as 8 out of 10 ⇒ P(B/A) = 8/10

iii) You want the probability of both events, which is the joint probability or  intersection: P(A∩B).

So, you can use the definition of conditional probability:

  • P(B/A) = P(A∩B) / P(A)

iv) From which you can solve for P(A∩B)

  • P(A∩B) = P(B/A)×P(A) =  (8/10)×(0.60) = 0.48

(b) Find the probability that a randomly selected resident who opposes the casino does not strongly oppose the casino.

In this case, you just want the complement of the probability that <em>a radomly selected resident who opposes the casino does strongly oppose the casino</em>, which is 1 - P(B/A) = 1 - 8/10 = 1 - 0.8 = 0.2.

​(c) Would it be unusual for a randomly selected resident to oppose the casino and strongly oppose the​ casino?

You are being asked about the joint probability (PA∩B), which you found in the part (a) and it is 0.48.

That is almost 0.50 or half of the population, so you conclude it is not unusual for a radomly selected resident to oppose the casino and strongly oppose the​ casino.

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