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3 years ago

How do I solve (3x2y) (10x5y3)

Mathematics

1 answer:

dmitriy555 [2]3 years ago

4 0

Answer:

30x^7y^4

Step-by-step explanation:

Hope this helps!

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-11 is the answer

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2 years ago

5. Decide whether each equation is true for all, one, or no values of x.

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3 years ago

Sam was given a bag of 100 red and blue marbles. He believes

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Answer:

(D) There are probably more blue marbles than red marbles in the bag.

Step-by-step explanation:

There are a total of 100 marbles in the bag.

In the experiment of 50 trials, Sam had the following outcome:

Blue=35

Red=15

Relative Frequency of Blue Marbles =35/50=0.7

Relative Frequency of Red Marbles =15/50=0.3

Since the relative frequency of blue marbles is greater than the relative frequency of red marbles, <u>there are probably more blue marbles than red marbles in the bag.</u>

The correct option is D.

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3 years ago

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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3 years ago

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