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3 years ago

Given the parabola whose equation is , find the focus and determine if the parabola opens up or down.

1 answer:

4 0

The equation of a parabola is:
$y = ax {}^{2} + bx + c$
If a>0, then the parabola opens up;
If a<0, then the parabola opens down.
The coordinates of focus are:
$( - \frac{b}{2a} \: \: \frac{1 - b {}^{2} + 4ac}{4a} )$

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At Cindy's birthday each guest will get one-third of a medium sized pizza. There will be 12 people at the party, including Cindy

tiny-mole [99]

Answer: 4 pizzas

Step-by-step explanation:

There are to be 12 people at the party.

Each person is expected to take 1/3 of a pizza.

The total number of pizza needed is therefore:

= No. of people * quantity of pizza

= 12 * 1/3

= 4 pizzas

Note: Answer would be 6 if each guest wanted 1/2 of a pizza.

3 0

3 years ago

Find the common difference of the arithmetic sequence –9, -7, -5, ...

MrMuchimi

It’s add positive 2 every time I believe

if I’m wrong I’m so sorry and please tell me I’m wrong for other that have to use this.
If I’m right ya!! XD
Hope this helps you!
-Pam Pam

3 0

3 years ago

Which line has a slope of 1 and a y-intercept of 2?

Vedmedyk [2.9K]

Answer:

bottom left :)

Step-by-step explanation:

3 0

3 years ago

Order these number (smallest to largest) 35.6 0.356 35.63 3.65 356.06 356.6

Alenkinab [10]

Answer: 0.356, 3.65, 35.6, 35.63, 356.06, 356.6

Step-by-step explanation:

8 0

3 years ago

A candidate for a US Representative seat from Indiana hires a polling firm to gauge her percentage of support among voters in he

mojhsa [17]

Answer:

(A) The minimum sample size required achieve the margin of error of 0.04 is 601.

(B) The minimum sample size required achieve a margin of error of 0.02 is 2401.

Step-by-step explanation:

Let us assume that the percentage of support for the candidate, among voters in her district, is 50%.

(A)

The margin of error, MOE = 0.04.

The formula for margin of error is:

$MOE=z_{\alpha /2}\sqrt{\frac{p(1-p)}{n}}$

The critical value of z for 95% confidence interval is: $z_{\alpha/2}=1.96$

Compute the minimum sample size required as follows:

$MOE=z_{\alpha /2}\sqrt{\frac{p(1-p)}{n}}\\0.04=1.96\times \sqrt{\frac{0.50(1-0.50)}{n}}\\(\frac{0.04}{1.96})^{2} =\frac{0.50(1-0.50)}{n}\\n=600.25\approx 601$

Thus, the minimum sample size required achieve the margin of error of 0.04 is 601.

(B)

The margin of error, MOE = 0.02.

The formula for margin of error is:

$MOE=z_{\alpha /2}\sqrt{\frac{p(1-p)}{n}}$

The critical value of z for 95% confidence interval is: $z_{\alpha/2}=1.96$

Compute the minimum sample size required as follows:

$MOE=z_{\alpha /2}\sqrt{\frac{p(1-p)}{n}}\\0.02=1.96\times \sqrt{\frac{0.50(1-0.50)}{n}}\\(\frac{0.02}{1.96})^{2} =\frac{0.50(1-0.50)}{n}\\n=2401.00\approx 2401$

Thus, the minimum sample size required achieve a margin of error of 0.02 is 2401.

8 0

3 years ago