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3 years ago

Find the measure of one interior angle for the following regular polygon

Mathematics

1 answer:

Tanya [424]3 years ago

3 0

Answer:

108 degrees

Step-by-step explanation:

The formula for the total sum of the angles in any REGULAR shape is

$(n-2) * 180$ , where n represents the number of sides.

The shape in the picture shown is a pentagon (5 sides).

$(5-2) * 180\\$

$3 * 180$

= $540$

So the angle sum is 540 degrees.

Again, we are told that it is a regular polygon, meaning all sides and angles in the shape are equal.

So to find one angle, just divide it by the number of sides.

$540/5 = 108$

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Which of the following parabolas opens down?

Veseljchak [2.6K]

ANSWER

The correct answer is C.

EXPLANATION

The directrix of the parabolas are parallel to the x-axis.
This means that, the orientation of the parabola is parallel to the y-axis.
We compare the focus and the directrix to determine whether the parabola opens downwards or upwards.
If the directrix is above the focus, then the parabola opens downwards.
On the other hand if the directrix is below the focus , then the parabola must open upwards.
If you look at option C, y=-2 isthe directrix. The focus is (3,-5). Since the the y-value of the focus , -5 is less than y=-2 which is the directrix, this parabola must open down.

The correct answer is C.

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3 years ago

In a large lecture class, the professor announced that the scores on a recent exam were normally distributed with a range from 5

Mars2501 [29]

Answer:

The standard deviation of the scores on a recent exam is 6.

The sample size required is 25.

Step-by-step explanation:

Let X = scores of students on a recent exam.

It is provided that the random variable X is normally distributed.

According to the Empirical rule, 99.7% of the normal distribution is contained in the range, μ ± 3σ.

That is, P (μ - 3σ < X < μ + 3σ) = 0.997.

It is provided that the scores on a recent exam were normally distributed with a range from 51 to 87.

This implies that:

P (51 < X < 87) = 0.997

So,

μ - 3σ = 51...(i)

μ + 3σ = 87...(ii)

Subtract (i) and (ii) to compute the value of σ as follows:

μ - 3σ = 51

(-)μ + (-)3σ = (-)87

______________

-6σ = -36

σ = 6

Thus, the standard deviation of the scores on a recent exam is 6.

The (1 - α)% confidence interval for population mean is given by:

$CI=\bar x\pm z_{\alpha/2}\frac{\sigma}{\sqrt{n}}$

The margin of error of this interval is:

$MOE = z_{\alpha/2}\frac{\sigma}{\sqrt{n}}$

Given:

MOE = 2

σ = 6

Confidence level = 90%

Compute the z-score for 90% confidence level as follows:

$z_{\alpha/2}=z_{0.10/2}=z_{0.05}=1.645$

*Use a z-table.

Compute the sample required as follows:

$MOE = z_{\alpha/2}\frac{\sigma}{\sqrt{n}}\\2=1.645\times \frac{6}{\sqrt{n}}\\n=(\frac{1.645\times 6}{2})^{2}\\n=24.354225\\n\approx 25$

Thus, the sample size required is 25.

4 0

3 years ago

What does coeffient stand for

azamat

A numerical or constant quantity place before and multiplying the variable in an algebraic expression

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3 years ago

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"is it possible to have more than one set of seven numbers having a mean of 50 and a standard deviation of 0?"

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The answer is 1.6. How??

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3 years ago

If the area of the region bounded by the curve y^2 =4ax and the line x= 4a is 256/3 Sq units, using integration find the value o

almond37 [142]

If the area of the region bounded by the curve $y^2 =4ax$ and the line $x= 4a$ is $\frac{256}{3}$ Sq units, then the value of $a$ will be $2$ .

<h3>What is area of the region bounded by the curve ?</h3>

An area bounded by two curves is the area under the smaller curve subtracted from the area under the larger curve. This will get you the difference, or the area between the two curves.

Area bounded by the curve $=\int\limits^a_b {x} \, dx$

We have,

$y^2 =4ax$

⇒ $y=\sqrt{4ax}$

$x= 4a$ ,

Area of the region $=\frac{256}{3}$ Sq units

Now comparing both given equation to get the intersection between points;

$y^2=16a^2$

$y=4a$

So,

Area bounded by the curve $= \[ \int_{0}^{4a} y \,dx \]$

$\frac{256}{3} =\[ \int_{0}^{4a} \sqrt{4ax} \,dx \]$

$\frac{256}{3}= \[\sqrt{4a} \int_{0}^{4a} \sqrt{x} \,dx \]$

$\frac{256}{3}= \[2\sqrt{a} \int_{0}^{4a} x^{\frac{1}{2} } \,dx \]$

$\frac{256}{3}= 2\sqrt{a} \left[\begin{array}{ccc}\frac{(x)^{\frac{1}{2}+1 } }{\frac{1}{2}+1 }\end{array}\right] _{0}^{4a}$

$\frac{256}{3}= 2\sqrt{a} \left[\begin{array}{ccc}\frac{(x)^{\frac{3}{2} } }{\frac{3}{2} }\end{array}\right] _{0}^{4a}$

$\frac{256}{3}= 2\sqrt{a} *\frac{2}{3} \left[\begin{array}{ccc}(x)^{\frac{3}{2}\end{array}\right] _{0}^{4a}$

On applying the limits we get;

$\frac{256}{3}= \frac{4}{3} \sqrt{a} \left[\begin{array}{ccc}(4a)^{\frac{3}{2} \end{array}\right]$

$\frac{256}{3}= \frac{4}{3} \sqrt{a} *\sqrt{(4a)^{3} }$

$\frac{256}{3}= \frac{4}{3} \sqrt{a} * 8 *a^{2} \sqrt{a}$

$\frac{256}{3}= \frac{4}{3} * 8 *a^{3}$

⇒ $a^{3} =8$

$a=2$

Hence, we can say that if the area of the region bounded by the curve $y^2 =4ax$ and the line $x= 4a$ is $\frac{256}{3}$ Sq units, then the value of $a$ will be $2$ .

To know more about Area bounded by the curve click here

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2 years ago

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