ABC and QRS are complementary angles. ABC=52, what is QRS?

The equation of a rational function does not have to contain a rational expression

jok3333 [9.3K]

Your answer is false

7 0

3 years ago

Eighty percent of the light aircraft that disappear while in flight in a certain country are subsequently discovered. Of the air

gladu [14]

Answer:

a) P(B'|A) = 0.042

b) P(B|A') = 0.625

Step-by-step explanation:

Given that:

80% of the light aircraft that disappear while in flight in a certain country are subsequently discovered

Of the aircraft that are discovered, 63% have an emergency locator,

whereas 89% of the aircraft not discovered do not have such a locator.

From the given information; it is suitable we define the events in order to calculate the probabilities.

So, Let :

A = Locator

B = Discovered

A' = No Locator

B' = No Discovered

So; P(B) = 0.8

P(B') = 1 - P(B)

P(B') = 1- 0.8

P(B') = 0.2

P(A|B) = 0.63

P(A'|B) = 1 - P(A|B)

P(A'|B) = 1- 0.63

P(A'|B) = 0.37

P(A'|B') = 0.89

P(A|B') = 1 - P(A'|B')

P(A|B') = 1 - 0.89

P(A|B') = 0.11

Also;

P(B ∩ A) = P(A|B) P(B)

P(B ∩ A) = 0.63 × 0.8

P(B ∩ A) = 0.504

P(B ∩ A') = P(A'|B) P(B)

P(B ∩ A') = 0.37 × 0.8

P(B ∩ A') = 0.296

P(B' ∩ A) = P(A|B') P(B')

P(B' ∩ A) = 0.11 × 0.2

P(B' ∩ A) = 0.022

P(B' ∩ A') = P(A'|B') P(B')

P(B' ∩ A') = 0.89 × 0.2

P(B' ∩ A') = 0.178

Similarly:

P(A) = P(B ∩ A ) + P(B' ∩ A)

P(A) = 0.504 + 0.022

P(A) = 0.526

P(A') = 1 - P(A)

P(A') = 1 - 0.526

P(A') = 0.474

The probability that it will not be discovered given that it has an emergency locator is,

P(B'|A) = P(B' ∩ A)/P(A)

P(B'|A) = 0.022/0.526

P(B'|A) = 0.042

(b) If it does not have an emergency locator, what is the probability that it will be discovered?

The probability that it will be discovered given that it does not have an emergency locator is:

P(B|A') = P(B ∩ A')/P(A')

P(B|A') = 0.296/0.474

P(B|A') = 0.625

4 0

3 years ago

An 8-oz bottle of ketchup costs $1.39. If the unit rate stays the same, how much should a 14-oz bottle cost? $0.79 $2.43 $2.78 $

Thepotemich [5.8K]

To find the unit rate you do

$1.39/8 oz.=.17

Then, you multiply the unit rate by the ounces

.17 * 14 oz. = $2.43

6 0

3 years ago

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

Can someone help me please.

slega [8]

6^log(216) is the answer

5 0

3 years ago