The length of a rectangle is 5t+6 and its height is t^6 , where t is time in seconds and the dimensions are in inches. Find the

Ber [7]

A, rotated then reflected

4 0

2 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

Use the diagram of point O. What is the length of OY to the nearest 10th of an Inch? XZ = 10 and OX= 10

Lady bird [3.3K]

From the diagram above,

XZ = 10 in and OX = 10 in

we are to find length of OY

XZ is a chord and line OY divides the chord into equal length

Hence, ZY=YX= 5 in

Now we solve the traingle OXY

To find OY we solve using pythagoras theorem

$(Hyp)^2=(Opp)^2+(Adj)^2$

applying values from the triangle above

$\begin{gathered} OX^2=XY^2+OY^2 \\ 10^2=5^2+OY^2 \\ 100=25+OY^2 \\ OY^2\text{ = 100 -25} \\ OY^2\text{ = 75} \\ OY\text{ = }\sqrt[]{75} \\ OY\text{ = }\sqrt[]{25\text{ }\times\text{ 3}} \\ OY\text{ = 5}\sqrt[]{3\text{ }}in \end{gathered}$

Therefore,

Length of OY =

$5\sqrt[]{3}$

8 0

9 months ago

A very large number of college men were asked what they thought their ideal weight was. A five number summary of the response (i

Lilit [14]

Answer:

Nearly 48.78% of the men gave a response that falls in the interval 155 to 195 pounds

Step-by-step explanation:

Given that a very large number of college men were asked what they thought their ideal weight was.

A five number summary of the response (in pounds) follows:

Median 175

Quartiles 155 196

Extremes 123 225

This implies that 25th percentile is 155 and 75th percentile is 196

This can be interpreted as approximately 50% of men lie between 155 and 196 pounds.

Since 195 is very near to 195, we can say approximately

i.e. 41 difference = 50%

40 difference = $\frac{40*50}{41} \\=48.78$

Nearly 48.78% of the men gave a response that falls in the interval 155 to 195 pounds

4 0

3 years ago

A manufacturer is interested in the output voltage of a power supply used in a PC. Output voltage is assumed to be normally dist

kompoz [17]

Answer: hello your question was poorly written but i was able to the get missing parts online which enabled me resolve your question

answer:

a) a = 0.1096

b) 1.89 watts

Step-by-step explanation:

Std of output voltage = 0.25 volt

H0 : μ = 5 volts

Ha : μ ≠ 5 volts

n = 16

a) Acceptance region = 4.9 ≤ X ≤ 5.1

Determine the value of a

value of a = 0.0548 + 0.0548

= 0.1096

<em>attached below is the reaming solution</em>

note : a is a type 1 error

b) power of test

True mean output voltage = 5.1 volts

P = - 1.89 watts

power cant be negative hence the power of the test = 1.89 watts

4 0

3 years ago