Briefly describe this graph of a Ferris wheel.

Please answer! I’m desperate.

lara31 [8.8K]

Answer:

Hi! This is the answer for part C, or finding the factors.

Step-by-step explanation:

Checked by my teacher. hope this part helps. Will post the rest soon.

8 0

1 year ago

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Which hyperbola has one focus point in common with the hyperbola (y+11)^2/(15^2)-(x-7)^2/(8^2)=1

irina [24]

$\bf \textit{hyperbolas, vertical traverse axis }\\\\
\cfrac{(y-{{ k}})^2}{{{ a}}^2}-\cfrac{(x-{{ h}})^2}{{{ b}}^2}=1
\qquad 
\begin{cases}
center\ ({{ h}},{{ k}})\\
vertices\ ({{ h}}, {{ k}}\pm a)
\end{cases}\\\\
-------------------------------\\\\
\cfrac{(y+11)^2}{15^2}-\cfrac{(x-7)^2}{8^2}=1\implies \cfrac{(y-(-11))^2}{15^2}-\cfrac{(x-7)^2}{8^2}=1\\\\
-------------------------------\\\\
c=\textit{distance from the center to either foci}\\\\
c=\sqrt{a^2+b^2}\implies c=\sqrt{15^2+8^2}\implies \boxed{c=17}$

now, if you notice, the positive fraction, is the one with the "y" variable, and that simply means, the hyperbola traverse axis is over the y-axis, so it more or less looks like the picture below.

now, from the hyperbola form, we can see the center is at (7, -11), and that the "c" distance is 17.

so, from -11 over the y-axis, we move Up and Down 17 units to get the foci, which will put them at (7, -11-17) or (7, -28) and (7 , -11+17) or (7, 6)

5 0

2 years ago

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In Juneau, Alaska, the 30-year annual snowfall average is 86.7 inches with a standard deviation of 40.4 inches. The last four ye

marusya05 [52]

The expected value of the amount of average snowfall for over 30 years is 86.7 inches with a standard deviation of 40.4 inches. To verify if this particular trend continues, we must check the significance value of the amount snowfall for the past four years. Given that the snowfall for past years are as follows: 115.7 inches, 62.9 inches, 168.5 inches, and 135.7 inches. Thus the mean of the sample would be: (115.7 + 62.9 + 168.5 + 135.7)/4 = 120.7 inches. To compute for the z-score, we have z-score = (x – μ) / (σ / √n) where x is the computed/measured value, μ is the expected mean, σ is the standard deviation, and n is the number of samples. Using the information we have, z-score (z) = (120.7 - 86.7) / (40.4/ √4) = 1.68 In order to reject the null hyptohesis our probability value must be less than the significance level of 5%. For our case, since z = 1.68, P-value = 0.093 > 0.05. Therefore, the answer is B.

5 0

2 years ago

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A school principal has to choose randomly among the six best students in each grade to be the school captain every month. For th

Viefleur [7K]

Since there are 6 students out of which one needs to be selected, the principal chose two die on which there are six numbers each numbered from 1 , 2, 3, 4, 5, 6.

Since there are two dice, the total possible outcome is 36.

Hence, the probability of getting one number each is 1/36

Hence, the principal used a fair method because each result is an equally likely possible outcome.

Option B is correct.

5 0

2 years ago

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Peter's salary is twice Anne's salary and half of David's salary. Then the average salary of Anne and David is _ of Peter's sala

satela [25.4K]

<h3><u>Answer:</u></h3>

Average of salary of Anne and David is 1.25 of Peter's Salary

<h3><u>Explanation:</u></h3>

In the question, relations between salaries of Peter, Anne, and David is given. From the given relations, we need to make few equations and calculate the average of Anne and David in terms of Peter's Salary.

Let us assume Peter's Salary as P.

Then we can calculate Anne's salary A = $\frac{P}{2}$ .