78 x 42 in area model and standard algorithm

If a diameter intersects a chord of a circle at a perpendicular, what conclusion can be made?

Agata [3.3K]

If a diameter intersects a chord of a circle at a perpendicular then the chord is bisected.

<h3>What is the relation between the chord and diameter of a circle?</h3>

The segments of a circle whose endpoints are on its perimeter are called chords.
A circle's diameter is the chord that passes across its center.
The circle's longest chord equals its diameter.
Any line that is perpendicular to a chord on the circle cuts it in half and goes through the center.

The solution to the problem:

A circle's chord is intersected perpendicularly by its diameter, which also passes through the center of the circle. The chord and diameter are perpendicular. Any line that is perpendicular to a chord on the circle and passes through its center divides it. The chord is divided equally by the diameter.

Therefore, the chord is bisected.

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7 0

2 years ago

Please help fast mathswatch!! I

netineya [11]

Answer:

Step-by-step explanation:

find the volume of one and multiply by six

6 * 2 * 2 * 12 =

288 cm³

4 0

2 years ago

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(b) Rationalise the denominator and simplify fully <br> 10 +3√2 / √2

kotegsom [21]

Answer:

I hope it helped you☺️☺️☺️☺️☺️

5 0

3 years ago

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A random sample of n measurements was selected from a population with unknown mean mu and standard deviation sigmaequals50 for e

Andre45 [30]

Answer:

a) (26.50;57.50)

b) (117.34;128.66)

c) (12.13;27.87)

d) (-4.73;11.01)

e) No. Since the sample sizes are large (n ≥ 30), the central limit theorem guarantees that $\bar x$ is approximately normal, so the confidence intervals are valid

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".

The confidence interval is given by this formula:

$\bar X \pm z_{\alpha/2} \frac{\sigma}{\sqrt{n}}$ (1)

And for a 95% of confidence the significance is given by $\alpha=1-0.95=0.05$ , and $\frac{\alpha}{2}=0.025$ . Since we know the population standard deviation we can calculate the critical value $z_{0.025}= \pm 1.96$

Part a

$n=40,\bar X=42,\sigma=50$

If we use the formula (1) and we replace the values we got:

$42 - 1.96 \frac{50}{\sqrt{40}}=26.50$

$42 + 1.96 \frac{50}{\sqrt{40}}=57.50$

The 95% confidence interval is given by (26.50;57.50)

Part b

$n=300,\bar X=123,\sigma=50$

If we use the formula (1) and we replace the values we got:

$123 - 1.96 \frac{50}{\sqrt{300}}=117.34$

$123 + 1.96 \frac{50}{\sqrt{300}}=128.66$

The 95% confidence interval is given by (117.34;128.66)

Part c

$n=155,\bar X=20,\sigma=50$

If we use the formula (1) and we replace the values we got:

$20 - 1.96 \frac{50}{\sqrt{155}}=12.13$

$20 + 1.96 \frac{50}{\sqrt{155}}=27.87$

The 95% confidence interval is given by (12.13;27.87)

Part d

$n=155,\bar X=3.14,\sigma=50$

If we use the formula (1) and we replace the values we got:

$3.14 - 1.96 \frac{50}{\sqrt{155}}=-4.73$

$3.14 + 1.96 \frac{50}{\sqrt{155}}=11.01$

The 95% confidence interval is given by (-4.73;11.01)

Part e

No. Since the sample sizes are large (n ≥ 30), the central limit theorem guarantees that $\bar x$ is approximately normal, so the confidence intervals are valid

8 0

3 years ago

CONSTRUCTION You are asked to copy a segment CD, construct a segment bisector by paper folding, and label the

lilavasa [31]

The steps on the construction of a segment bisector by paper folding, and label the midpoint M is given below.

<h3>What are the steps of this construction?</h3>

1. First, one need to open a Compass so that it is said to be more than half the length of the said segment.

2. Without altering it, with the aid of the compass, do draw an art above and also below the said line segment from one of the segment endpoints.

3. Also without altering it and with use the compass, do draw another pair of arts from the other and points. One arc will be seen above the segment and the other or the second arc will be seen below.

4. Then do draw the point of intersection that is said to exist between the pair of arts below the line segment and also in-between the pair of arts as seen below the line segment

5. Lastly, do make use of a straight edge to link the intersection points between the both pair of arts.

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5 0

2 years ago