Whether dividing constant terms or polynomials, we always have definitive terms when it comes to division. Suppose we say, 10x divided by 2. The dividend is the 10x and the divisor is the 2. In other words, the dividend is the number to be divided by the divisor, to obtain the answer called the quotient.

When dividing polynomials, your main goal is to be able to divide the dividend evenly into the divisor. For example, we divide x²+2x+1 by x+1. The first thing you're going to focus is, what term will completely divide the first term of the polynomial? That would be x. Why? Because when you multiply x with x+1, the product is x²+x. When you subtract this from the polynomial, the x² will cancel out. All you have to do is subtract x from 2x, yielding x. Then, you carry down the last term of the equation: +1. You do the steps again. The term that will completely divide x+1 by x+1 is 1. When you subtract the two, you will come up with zero. That means there is no remainder. The polynomial is divisible by the divisor.
x + 1
------------------------------------
x+1| x²+2x+1
- x²+x
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x +1
- x +
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2 years ago

Repeated student samples. Of all freshman at a large college, 16% made the dean’s list in the current year. As part of a class p

Dima020 [189]

Answer:

a) p-hat (sampling distribution of sample proportions)

b) Symmetric

c) σ=0.058

d) Standard error

e) If we increase the sample size from 40 to 90 students, the standard error becomes two thirds of the previous standard error (se=0.667).

Step-by-step explanation:

a) This distribution is called the sampling distribution of sample proportions (p-hat).

b) The shape of this distribution is expected to somewhat normal, symmetrical and centered around 16%.

This happens because the expected sample proportion is 0.16. Some samples will have a proportion over 0.16 and others below, but the most of them will be around the population mean. In other words, the sample proportions is a non-biased estimator of the population proportion.

c) The variability of this distribution, represented by the standard error, is:

$\sigma=\sqrt{p(1-p)/n}=\sqrt{0.16*0.84/40}=0.058$

d) The formal name is Standard error.

e) If we divided the variability of the distribution with sample size n=90 to the variability of the distribution with sample size n=40, we have:

$\frac{\sigma_{90}}{\sigma_{40}}=\frac{\sqrt{p(1-p)/n_{90}} }{\sqrt{p(1-p)/n_{40}}}}= \sqrt{\frac{1/n_{90}}{1/n_{40}}}=\sqrt{\frac{1/90}{1/40}}=\sqrt{0.444}= 0.667$

If we increase the sample size from 40 to 90 students, the standard error becomes two thirds of the previous standard error (se=0.667).

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2 years ago