What is the slope of the line

Simple interest 34,100 at 4% for 3 years

Kaylis [27]

Answer:

Step-by-step explanation:

.....

4 0

3 years ago

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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

Write an inequality for the situation

Firdavs [7]

A) Shane’s pets >5
B) Money Earned > 200
3) Bre’s Score > 75
4) Sarah’s Dance practice ≤ 2 hours
5) Shannon’s Age < 8 years old

4 0

2 years ago

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Question 16

den301095 [7]

Answer:

Statement B and D are correct.

Step-by-step explanation:

The number of minutes Gabriel spends grading essays can be presented as a function: f(x) = 4x, where x is the number of graded essays and 4 is the number of minutes Gabriel spends on grading each essay.

By definition, domain of a function f(x) is the set of all values for which the function is defined, and the range of the function is the set of all values that f takes.

So in this case, domain of f(x) is the set of all values of x which is an integer going from 0 to 105. Statement D is accurate.

Range of f(x) is the set of all values that f takes and can be calculated by multiplying 4 with (0,105), equal (0,420). Statement B is accurate.

4 0

3 years ago

MATH HELP PLEASE!!!!

AleksandrR [38]

Monomial is a polynomial with one term.
A.)
C.)

8 0

3 years ago