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3 years ago

Inscribed circles on the arc; improvement answer is required in full.

Mathematics

1 answer:

e-lub [12.9K]3 years ago

8 0

Answer:

The measure of an inscribed angle is half the measure of the intercepted arc

∠EFG creates Arc EDG

if ∠EFG is 1/2 arc EDG then 115° x 2 = 230°

A circle is 360°.

The remaining arc EFG is 360° - 230° = 130°

∠EDG is 1/2 of arc EFG so, 130° ÷ 2 = 65°

∠EDG = 65°

You might be interested in

Find an exact value.

leonid [27]

Answer:

(√2 - √6) / 4

C. square root of two minus square root of six divided by four.

Step-by-step explanation:

sine of negative eleven pi divided by twelve.

We have :

sin(-11π/12)

sin((4 - 15)π / 12) = sin(4π/12 - 15π/12)

sin(4π/12 - 15π/12) = sin(π/3 - 5π/4)

Recall:

Angle difference formula:

sin(A - B) = sinAcosB - sinBcosA

Hence,

sin(π/3 - 5π/4) = sin(π/3) cos(5π/4) − sin(5π/4) cos(π/3)

From trigonometry:

sinπ/3 = √3/2

cos5π/4 = -√2/2

sin5π/4 = -√2/2

cos π/3 = 1/2

(√3/2) (-√2/2) − (-√2/2) (1/2)

-√6/4 - -√2/4

-√6/4 + √2/4

√2/4 - √6/4

(√2 - √6) / 4

7 0

3 years ago

Wendy throws a dart at this square shaped target: Part A: is the probability of hitting the black circle inside the target close

KonstantinChe [14]

Since the area of the circle is only 3.14. that means 3.14 out of one hundred units can be hit. so you have a 3.14% chance of hitting the circle in the middle. SO the probability is very close to 0.

100 - 3.14 = 96.86, so we have a 97 percent chance (approximately) of landing in the white space, This is very close to one.

3 0

3 years ago

Fill in the blanks

olga nikolaevna [1]

The answer is
5mn(KUASA DUA) -6gh(KUASA DUA) +2

4 0

3 years ago

Standard Error from a Formula and a Bootstrap Distribution Sample A has a count of 30 successes with and Sample B has a count of

tia_tia [17]

Answer:

Using a formula, the standard error is: 0.052

Using bootstrap, the standard error is: 0.050

Comparison:

The calculated standard error using the formula is greater than the standard error using bootstrap

Step-by-step explanation:

Given

Sample A Sample B

$x_A = 30$ $x_B = 50$

$n_A = 100$ $n_B =250$

Solving (a): Standard error using formula

First, calculate the proportion of A

$p_A = \frac{x_A}{n_A}$

$p_A = \frac{30}{100}$

$p_A = 0.30$

The proportion of B

$p_B = \frac{x_B}{n_B}$

$p_B = \frac{50}{250}$

$p_B = 0.20$

The standard error is:

$SE_{p_A-p_B} = \sqrt{\frac{p_A * (1 - p_A)}{n_A} + \frac{p_A * (1 - p_B)}{n_B}}$

$SE_{p_A-p_B} = \sqrt{\frac{0.30 * (1 - 0.30)}{100} + \frac{0.20* (1 - 0.20)}{250}}$

$SE_{p_A-p_B} = \sqrt{\frac{0.30 * 0.70}{100} + \frac{0.20* 0.80}{250}}$

$SE_{p_A-p_B} = \sqrt{\frac{0.21}{100} + \frac{0.16}{250}}$

$SE_{p_A-p_B} = \sqrt{0.0021+ 0.00064}$

$SE_{p_A-p_B} = \sqrt{0.00274}$

$SE_{p_A-p_B} = 0.052$

Solving (a): Standard error using bootstrapping.

Following the below steps.

Open Statkey
Under Randomization Hypothesis Tests, select Test for Difference in Proportions
Click on Edit data, enter the appropriate data
Click on ok to generate samples
Click on Generate 1000 samples ---- <em>see attachment for the generated data</em>

From the randomization sample, we have:

Sample A Sample B

$x_A = 23$ $x_B = 57$