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3 years ago

-1/64, -1/8, -1, -8, ...

Mathematics

1 answer:

Taya2010 [7]3 years ago

6 0

Answer:

I think it's B: an = 8 ⋅ an–1

Step-by-step explanation:

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Which of the following are the coordinates of the vertex of y = −x2 + 2x + 1? (2, 1) (1, –2) (1, 2) (–2, 1)

IceJOKER [234]

Answer:

Step-by-step explanation:

We want to determine the vertex of the quadratic equation:

$\displaystyle y = -x^2 + 2x + 1$

Recall that the vertex is given by the formulas:

$\displaystyle \text{Vertex} = \left(-\frac{b}{2a}, f\left(-\frac{b}{2a}\right)\right)$

In this case, a = -1, b = 2, and c = 1.

Determine the x-coordinate of the vertex:

$\displaystyle x = -\frac{(2)}{2(-1)} = 1$

To determine the y-coordinate, evaluate the function at x = 1:

$\displaystyle \begin{aligned} y(1)&= -(1)^2 + 2(1) + 1 \\ &= 2\end{aligned}$

In conclusion, the vertex of the quadratic equation is (1, 2).

Hence, our answer is C.

3 0

3 years ago

Determine if the relation is a function.

bixtya [17]

The first one I think

8 0

3 years ago

Because of staffing decisions, managers of the Gibson-Marion Hotel are interested in the variability in the number of rooms occu

olchik [2.2K]

Answer:

a) $s^2 =30^2 =900$

b) $\frac{(19)(30)^2}{30.144} \leq \sigma^2 \leq \frac{(19)(30)^2}{10.117}$

$567.28 \leq \sigma^2 \leq 1690.224$

c) $23.818 \leq \sigma \leq 41.112$

Step-by-step explanation:

Assuming the following question: Because of staffing decisions, managers of the Gibson-Marimont Hotel are interested in the variability in the number of rooms occupied per day during a particular season of the year. A sample of 20 days of operation shows a sample mean of 290 rooms occupied per day and a sample standard deviation of 30 rooms

Part a

For this case the best point of estimate for the population variance would be:

$s^2 =30^2 =900$

Part b

The confidence interval for the population variance is given by the following formula:

$\frac{(n-1)s^2}{\chi^2_{\alpha/2}} \leq \sigma^2 \leq \frac{(n-1)s^2}{\chi^2_{1-\alpha/2}}$

The degrees of freedom are given by:

$df=n-1=20-1=19$

Since the Confidence is 0.90 or 90%, the significance $\alpha=0.1$ and $\alpha/2 =0.05$ , the critical values for this case are:

$\chi^2_{\alpha/2}=30.144$

$\chi^2_{1- \alpha/2}=10.117$

And replacing into the formula for the interval we got:

$\frac{(19)(30)^2}{30.144} \leq \sigma^2 \leq \frac{(19)(30)^2}{10.117}$

$567.28 \leq \sigma^2 \leq 1690.224$

Part c

Now we just take square root on both sides of the interval and we got:

$23.818 \leq \sigma \leq 41.112$

5 0

4 years ago

Sweatshirt = $32 t-shirts = $14 ,

rodikova [14]

Cost of a sweatshirt = $32
Then
Cost of 14 sweatshirts = (32 * 14) dollars
= 448 dollars
Cost of a t-shirt = $14
Cost of 32 t-shirts = (32 * 14) dollars
= 448 dollars
So it can seen from the deduction that the cost of 14 sweatshirts and 32 t-shirts comes out to be same and equal to $448. I hope the answer and the procedure of doing this problem is clear to you. It is important to look at all the details given in the question and then start solving the problem.

4 0

3 years ago

Read 2 more answers

HELP ME PLEASE!!!!

QveST [7]

$54=9\cdot6\\24=4\cdot6\\\\3\sqrt{54}-4\sqrt{24}=3\sqrt{9\cdot6}-4\sqrt{4\cdot6}=3\cdot\sqrt9\cdot\sqrt6-4\cdot\sqrt4\cdot\sqrt6\\\\=3\cdot3\cdot\sqrt6-4\cdot2\cdot\sqrt6=9\sqrt6-8\sqrt6=1\sqrt6=\sqrt6\\\\\text{Answer:}\ \sqrt6\\\\\text{Used:}\ \sqrt{a\cdot b}=\sqrt{a}\cdot\sqrt{b}$

4 0

3 years ago