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emmainna [20.7K]
3 years ago
11

Find the greatest common factor of

Mathematics
2 answers:
Gwar [14]3 years ago
6 0

For this case, we have by definition the greatest common factor or GFC is given by the greatest factor that divides two numbers. For example:

12 is divisible by 1,2,3,4,6,12

16 is divisible by 1,2,4,8,16

So the GFC of 12 and 16 is 4

Now, if we have:

x ^ 2y\ and\ xy ^ 2

The greatest common factor of both expressions is xy.

xy (x) = x ^ 2y\\xy (y) = xy ^ 2

Answer:

xy

Option D


labwork [276]3 years ago
5 0

Answer:

D. xy.

Step-by-step explanation:

The GCF of x and x^2 is x and GCF of y and y^2 is y.

Answer is xy.

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A cube is a special prism which has bases and faces that are all squares. A pyramid is a polyhedron with only one base. The shape of the base determines the specific name of each pyramid. The faces are always triangles that meet at a common vertex.

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Each year, all final year students take a mathematics exam. It is hypothesised that the population mean score for this test is 1
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Answer:

90% confidence interval for the population mean test score is [95.40 , 106.59]

Step-by-step explanation:

We are given that the population mean score for mathematics test is 115. It is known that the population standard deviation of test scores is 17.

Also, a random sample of 25 students take the exam. The mean score for this group is 101.

The, pivotal quantity for 90% confidence interval for the population mean test score is given by;

        P.Q. = \frac{Xbar-\mu}{\frac{\sigma}{\sqrt{n} } } ~ N(0,1)

where, X bar = sample mean = 101

              \sigma = population standard deviation

              n = sample size = 25

So, 90% confidence interval for the population mean test score, \mu is ;

P(-1.6449 < N(0,1) < 1.6449) = 0.90

P(-1.6449 < \frac{Xbar-\mu}{\frac{\sigma}{\sqrt{n} } } < 1.6449) = 0.90

P(-1.6449 * \frac{\sigma}{\sqrt{n} } < {Xbar-\mu} < 1.6449 * \frac{\sigma}{\sqrt{n} } ) = 0.90

P(X bar - 1.6449 * \frac{\sigma}{\sqrt{n} } < \mu < X bar + 1.6449 * \frac{\sigma}{\sqrt{n} } ) = 0.90

90% confidence interval for \mu = [ X bar - 1.6449 * \frac{\sigma}{\sqrt{n} } , X bar + 1.6449 * \frac{\sigma}{\sqrt{n} } ]

                                                  = [ 101 - 1.6449 * \frac{17}{\sqrt{25} } , 10 + 1.6449 * \frac{17}{\sqrt{25} } ]

                                                  = [95.40 , 106.59]

Therefore, 90% confidence interval for the population mean test score is [95.40 , 106.59] .

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