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1 year ago

Brainly to whoever is first and right:) - In quadrilateral ABCD, AB = BC = 1, m∠B=100°, m∠D=130°. Find BD

Mathematics

1 answer:

n200080 [17]1 year ago

7 0

Answer: 1

B is the center of a circle of radius 1. A and C are on the circle 100 degrees apart. So arc AC is 100 degrees. Arc AC the long way around is 360-100=260 degrees. That means any point D on arc AC will subtend a 130 degree angle. BD is a radius for all of those, and the radius is 1.

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-6y + 3(12y) = 20(y - 1) + 15 is it true?

xxMikexx [17]

Answer:

y= - 1/2 (negative half) = -0.5

Step-by-step explanation:

−6y+3(12y)=20(y−1)+15

Multiply 3 and 12 to get 36.

−6y+36y=20(y−1)+15

Combine −6y and 36y to get 30y.

30y=20(y−1)+15

Use the distributive property to multiply 20 by y−1.

30y=20y−20+15

Add −20 and 15 to get −5.

30y=20y−5

Subtract 20y from both sides.

30y−20y=−5

Combine 30y and −20y to get 10y.

10y=−5

Divide both sides by 10

y= -5/10

Reduce the fraction -5/10 = -0.5 to lowest terms by extracting and cancelling out 5 .

5 0

1 year ago

It is believed that as many as 23% of adults over 50 never graduated from high school. We wish to see if this percentage is the

JulijaS [17]

Answer:

1) n=48

2) n=298

3) n=426

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

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".

$p$ represent the real population proportion of interest

$\hat p$ represent the estimated proportion for the sample

n is the sample size required (variable of interest)

$z$ represent the critical value for the margin of error

The population proportion have the following distribution

$p \sim N(p,\sqrt{\frac{\hat p(1-\hat p)}{n}})$

Part 1

In order to find the critical value we need to take in count that we are finding the interval for a proportion, so on this case we need to use the z distribution. Since our interval is at 90% of confidence, our significance level would be given by $\alpha=1-0.90=0.10$ and $\alpha/2 =0.05$ . And the critical value would be given by:

$z_{\alpha/2}=-1.64, z_{1-\alpha/2}=1.64$

The margin of error for the proportion interval is given by this formula:

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

And on this case we have that $ME =\pm 0.1$ and we are interested in order to find the value of n, if we solve n from equation (a) we got:

$n=\frac{\hat p (1-\hat p)}{(\frac{ME}{z})^2}$ (b)

We can assume that the estimated proportion is 0.23 for the 25 to 30 group. And replacing into equation (b) the values from part a we got:

$n=\frac{0.23(1-0.23)}{(\frac{0.1}{1.64})^2}=47.63$

And rounded up we have that n=48

Part 2

The margin of error on this case changes to 0.04 so if we use the same formula but changing the value for ME we got:

$n=\frac{0.23(1-0.23)}{(\frac{0.04}{1.64})^2}=297.7$

And rounded up we have that n=298

Part 3

In order to find the critical value we need to take in count that we are finding the interval for a proportion, so on this case we need to use the z distribution. Since our interval is at 95% of confidence, our significance level would be given by $\alpha=1-0.95=0.05$ and $\alpha/2 =0.025$ . And the critical value would be given by:

$z_{\alpha/2}=-1.96, z_{1-\alpha/2}=1.96$

The margin of error for the proportion interval is given by this formula:

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

And on this case we have that $ME =\pm 0.04$ and we are interested in order to find the value of n, if we solve n from equation (a) we got:

$n=\frac{\hat p (1-\hat p)}{(\frac{ME}{z})^2}$ (b)

We can assume that the estimated proportion is 0.23 for the 25 to 30 group. And replacing into equation (b) the values from part a we got:

$n=\frac{0.23(1-0.23)}{(\frac{0.04}{1.96})^2}=425.22$

And rounded up we have that n=426

3 0

2 years ago

If the heights of 300 students are normally distributed with mean 68.0 inches and standard deviation 3.0 inches, how many studen

Lunna [17]

Given:
μ = 68 in, population mean
σ = 3 in, population standard deviation

Calculate z-scores for the following random variable and determine their probabilities from standard tables.

x = 72 in:
z = (x-μ)/σ = (72-68)/3 = 1.333
P(x) = 0.9088

x = 64 in:
z = (64 -38)/3 = -1.333
P(x) = 0.0912

x = 65 in
z = (65 - 68)/3 = -1
P(x) = 0.1587

x = 71:
z = (71-68)/3 = 1
P(x) = 0.8413

Part (a)
For x > 72 in, obtain
300 - 300*0.9088 = 27.36

Answer: 27

Part (b)
For x ≤ 64 in, obtain
300*0.0912 = 27.36

Answer: 27

Part (c)
For 65 ≤ x ≤ 71, obtain
300*(0.8413 - 0.1587) = 204.78

Answer: 204

Part (d)
For x = 68 in, obtain
z = 0
P(x) = 0.5
The number of students is
300*0.5 = 150

Answer: 150

3 0

2 years ago

Complete the sentence below using one of the words. 'Power, Base, Exponent'

uysha [10]

The number is 10 to the -6 power

6 0

2 years ago

If X is a normal random variable with parameters µ = 10 and σ 2 = 36, compute

alex41 [277]

Answer:

Step-by-step explanation:

Given that X is a normal random variable with parameters µ = 10 and σ 2 = 36,