What is the area of 2 yards

In the triple beam balance, what is the appropriate number of decimal places?

Lyrx [107]

Answer:

Then the correct answer would be to two decimal places. Credit to 25rx0162.

Step-by-step explanation

Hope this helps : )

4 0

3 years ago

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Solve for x: sinx-cosx=√2

kvv77 [185]

sinx - cosx =sqrt(2)
Taking square on both sides:
(sinx - cosx)^2 =sqrt(2)^2
sin^2(x) -2cos(x)sin(x) + cos^2(x) = 2
Rearranging the equation:
sin^2(x)+cos^2(x) -2cos(x)sin(x)=2
As,
sin^2(x)+cos^2(x) = 1
So,
1-2sinxcosx=2
1-1-2sinxcosx=2-1
-2sinxcosx = 1
Using Trignometric identities:
-2(0.5(sin(x+x)+sin(x-x))=1
-sin2x+sin0=1
As,
sin 0 = 0
So,
sin2x+0 = -1
sin2x = -1
2x=-90 degrees + t360
Dividing by 2 on both sides:
x=-45 degrees + t180
or 2x=270 degrees +t360
x= 135 degrees + t180 where t is integer

6 0

3 years ago

As part of the Pew Internet and American Life Project, researchers conducted two surveys in late 2009. The first survey asked a

REY [17]

Answer:

The 95% confidence interval for the difference between the proportion of all U.S. teens and adults who use social networking sites is (0.223, 0.297). This means that we are 95% sure that the true difference of the proportion is in this interval, between 0.223 and 0.297.

Step-by-step explanation:

Before building the confidence interval we need to understand the central limit theorem and the subtraction of normal variables.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean $\mu$ and standard deviation $\sigma$ , the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$ .

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

For a proportion p in a sample of size n, the sampling distribution of the sample proportion will be approximately normal with mean $\mu = p$ and standard deviation $s = \sqrt{\frac{p(1-p)}{n}}$

Subtraction between normal variables:

When two normal variables are subtracted, the mean is the difference of the means, while the standard deviation is the square root of the sum of the variances.

Sample of 800 teens. 73% said that they use social networking sites.

This means that:

$p_T = 0.73, s_T = \sqrt{\frac{0.73*0.27}{800}} = 0.0157$

Sample of 2253 adults. 47% said that they use social networking sites.

This means that:

$p_A = 0.47,s_A = \sqrt{\frac{0.47*0.53}{2253}} = 0.0105$

Distribution of the difference:

$p = p_T - p_A = 0.73 - 0.47 = 0.26$

$s = \sqrt{s_T^2+s_A^2} = \sqrt{0.0157^2+0.0105^2} = 0.019$

Confidence interval:

Is given by:

$p \pm zs$

In which

z is the zscore that has a pvalue of $1 - \frac{\alpha}{2}$ .

95% confidence level

So $\alpha = 0.05$ , z is the value of Z that has a pvalue of $1 - \frac{0.05}{2} = 0.975$ , so $Z = 1.96$ .

Lower bound:

$p - 1.96s = 0.26 - 1.96*0.019 = 0.223$

Upper bound:

$p + 1.96s = 0.26 + 1.96*0.019 = 0.297$

The 95% confidence interval for the difference between the proportion of all U.S. teens and adults who use social networking sites is (0.223, 0.297). This means that we are 95% sure that the true difference of the proportion is in this interval, between 0.223 and 0.297.

3 0

2 years ago

Can y’all help me simplify this

yKpoI14uk [10]

The simplified answer is 17.48

4 0

3 years ago

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For bone density scores that are normally distributed with a mean of 0 and a standard deviation of 1, find the percentage of sc

mr Goodwill [35]

In a normal distribution 68.27% of the values are within one standard deviation from the mean, 95.5% of the values are within two standard deviations from the mean, and 99.7 % of the values are within three standard deviations of the mean

With that you have the answer to the three questions:


a. significantly high (or at least 2 standard deviations above the mean).

99.5% of the values are within 2 standard deviations from the mean, half of 100% - 95.5% = 4.5% / 2 = 2.25% are above the mean, so the answer is 2.25%

b. significantly low (or at least 2 standard deviations below the mean).

The other half are below 2 standard deviations, so the answer is 2.25%

c. not significant (or less than 2 standard deviations away from the mean).

As said, 95.5% are within the band of two standard deviations from the mean, so the answer is 95.5%.

7 0

3 years ago

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