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2 years ago

Which of the following is a monomial?

Mathematics

1 answer:

Naya [18.7K]2 years ago

5 0

Mono (1) nomial == 1 term.

11x^2 is 1 term in combination so B is your answer.

D. is not, its a binomial along with C. and A.

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Question 11 Multiple Choice Worth 1 points)

Viefleur [7K]

Answer:

x = 35

Step-by-step explanation:

First, we need to add the angles on a straight line and subtract them by 180 (because angles on a straight line adds up to 180)

56 + 51 = 107

180 - 107 = 73

Now that we have found the angle of y, we can find the angle of x now. Because angles in a triangle adds up to 180, we have to add 72 and 73 now.

72 + 73 = 145

And now subtract it from 180

180 - 145 = 35

So x = 35

8 0

2 years ago

Is the confidence interval affected by the fact that the data appear to be from a population that is not normally distributed?

sp2606 [1]

Answer:

D. No, because the sample size is large enough.

Step-by-step explanation:

The central limit theorem states that "if we have a population with mean μ and standard deviation σ and take sufficiently large random samples from the population with replacement, then the distribution of the sample means will be approximately normally distributed. This will hold true regardless of whether the source population is normal or skewed, provided the sample size is sufficiently large".

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

If the sample size is higher than 30, on this case the answer would be:

D. No, because the sample size is large enough.

And the reason is given by The Central Limit Theorem since states if the individual distribution is normal then the sampling distribution for the sample mean is also normal.

From the central limit theorem we know that the distribution for the sample mean $\bar X$ is given by:

$\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})$

If the sample size it's not large enough n<30, on that case the distribution would be not normal.

7 0

3 years ago

1. Let f(x, y) be a differentiable function in the variables x and y. Let r and θ the polar coordinates,and set g(r, θ) = f(r co

Olenka [21]

Answer:

$g_{r}(\sqrt{2},\frac{\pi}{4})=\frac{\sqrt{2}}{2}\\$

Step-by-step explanation:

First, notice that:

$g(\sqrt{2},\frac{\pi}{4})=f(\sqrt{2}cos(\frac{\pi}{4}),\sqrt{2}sin(\frac{\pi}{4}))\\$

$g(\sqrt{2},\frac{\pi}{4})=f(\sqrt{2}(\frac{1}{\sqrt{2}}),\sqrt{2}(\frac{1}{\sqrt{2}}))\\$

$g(\sqrt{2},\frac{\pi}{4})=f(1,1)\\$

We proceed to use the chain rule to find $g_{r}(\sqrt{2},\frac{\pi}{4})$ using the fact that $X(r,\theta)=rcos(\theta)\ and\ Y(r,\theta)=rsin(\theta)$ to find their derivatives:

$g_{r}(r,\theta)=f_{r}(rcos(\theta),rsin(\theta))=f_{x}( rcos(\theta),rsin(\theta))\frac{\delta x}{\delta r}(r,\theta)+f_{y}(rcos(\theta),rsin(\theta))\frac{\delta y}{\delta r}(r,\theta)\\$