Can Someone Help Me?

Mathematics

1 answer:

Hunter-Best [27]3 years ago

3 0

Answer:

D. Skewed Right

Step-by-step explanation:

Now some might think that this would be symmetrical, and that's even what my mind jumped to at first, but truly its skewed right. The definition of symmetrical is things or data that mirror itself perfectly on both sides of a midline, or things that can be folded in half to line up perfectly. In this case, the graph just has a positive correlation and is going up towards the right. It is not symmetrical, it just has a pattern.

Hope this helped ^-^

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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 probability that a radomly selected resident opposes the casino, P(A)

Probability = number of favorable outcomes / number of possible outcomes

P(A) is given as 60%, which in decimal form is 0.60

ii) Next, determine,the probability that a resident who opposes the casino strongly opposes it, 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 a radomly selected resident who opposes the casino does strongly oppose the casino, 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.

8 0

3 years ago

What is 7/24 written as a decimal ?

mash [69]

Pretty sure its 0.2917, not sure tho,

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2 years ago

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Find the value of y. Round your answers to the nearest tenth

ANTONII [103]

Answer:

y ≈ 4.3