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2 years ago

Mila graphs the point A (5, 0) on a coordinate plane.

Mathematics

1 answer:

alexdok [17]2 years ago

8 0

Answer:

B,E

Step-by-step explanation:

(5,0) (5=x, 0=y)

A is false because y = 0

B is true because x=5

c is false because y = 0

D is false because again y = 0

E is true because 5,0 is on the x axis

- Gage Millar, Algebra 2 tutor

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The table displays the number of students in each grade at Richmond High School who will or will not be attending the amusement

Pani-rosa [81]

Answer:

$P( eleventh\: \: |\: \: FT)= 0.2453 \\\\P( eleventh\: \: |\: \: FT)= 24.53\% \\\\$

Step-by-step explanation:

We are given a joint probability table.

There are four different graders in a school

1. Grade Ninth

2. Grade Tenth

3. Grade Eleventh

4. Grade Twelfth

Field trip refers to the students who will attending the amusement park field trip.

No field trip refers to the students who will not be attending the amusement park field trip.

We want to find out the probability that the selected student is an eleventh grader given that the student is going on a field trip.

$P( eleventh\: \: |\: \: FT)= \frac{P( eleventh\: \: and \: \: FT)}{P(FT)}$

Where P(eleventh and FT) is the probability of students who are in eleventh grade and will be going to field trip

$P(eleventh\: and \: FT) = \frac{13}{92} \\\\P(eleventh \: and \: FT) = 0.1413$

Where P(FT) is the probability of students who will be going to field trip

$P(FT) = \frac{12}{92} + \frac{9}{92} + \frac{13}{92} + \frac{19}{92}\\\\P(FT) = 0.1304 + 0.0978 + 0.1413 + 0.2065\\\\P(FT) = 0.576 \\\\$

So the required probability is

$P( eleventh\: \: |\: \: FT)= \frac{P( eleventh\: \: and \: \: FT)}{P(FT)} \\\\P( eleventh\: \: |\: \: FT)= \frac{0.1413}{0.576} \\\\P( eleventh\: \: |\: \: FT)= 0.2453 \\\\P( eleventh\: \: |\: \: FT)= 24.53\% \\\\$

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kirill115 [55]

Answer:

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Step-by-step explanation:

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