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hoa [83]
3 years ago
12

What is the greatest common factor of 9x^2y and 5x^4y^2

Mathematics
1 answer:
ELEN [110]3 years ago
6 0
Xy(9x)(5x^3y) so the greatest common factor is xy 
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What is the rate of change of the function?
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-2

Step-by-step explanation:

Rise over run

Pick a point and check how much it goes vertically, then divide it by how much it goes horizontally.

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The cost of 5 books is 71$ . What is the cost of each book
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$14.20 is the cost of each book
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There are 32 times students can work in the computer lab during one week. If 24 students can work in the computer lab at one tim
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Sonali invests 15% of her monthly salary in insurance policies.
strojnjashka [21]

Answer:

₹42,500

Step-by-step explanation:

Let Sonali's monthly income be ₹x

<h2>According to the question </h2>

15\% \:ofx + 55\% \: of \: x + 12750 = x

=  >  \frac{15}{100} x +  \frac{55}{100} x + 12750 = x

=  >  \frac{70}{100} x + 12750 = x

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Police estimate that​ 25% of drivers drive without their seat belts. If they stop 6 drivers at​ random, find the probability tha
Furkat [3]

Answer:

17.80% probability that all of them are wearing their seat belts.

Step-by-step explanation:

For each driver stopped, there are only two possible outcomes. Either they are wearing their seatbelts, or they are not. The drivers are chosen at random, which mean that the probability of a driver wearing their seatbelts is independent from other drivers. So we use the normal probability distribution to solve this problem.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

In which C_{n,x} is the number of different combinations of x objects from a set of n elements, given by the following formula.

C_{n,x} = \frac{n!}{x!(n-x)!}

And p is the probability of X happening.

Police estimate that​ 25% of drivers drive without their seat belts.

This means that 75% wear their seatbelts, so p = 0.75

If they stop 6 drivers at​ random, find the probability that all of them are wearing their seat belts.

This is P(X = 6).

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

P(X = 6) = C_{6,6}.(0.75)^{6}.(0.25)^{0} = 0.1780

17.80% probability that all of them are wearing their seat belts.

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