Write an equation of the line with a slope of and y-intercept of -8.

Dafna11 [192]

It's a binomial times a binomial, so you FOIL
First multiply the First numbers
3z*z = 3z(^2)
Then multiply Outers:
3z* -4 = -12z
Nest, Inners:
9*z = 9z
Finally, the Lasts:
-4*9 = -36
Then, throw it all together in descending order, starting with the variable of highest degree to lowest:
3z^2 - 12z + 9z - 36
Last of all, simplify coefficients with like terms and you're good to go:
3z^2 - 3z - 36
Hope this helps!

3 0

3 years ago

The average value of all the pennies, nickels, dimes, and quarters in Paula's purse is 20 cents. If she had one more quarter, th

sveta [45]

Answer:

Step-by-step explanation:

The average value of all the pennies, nickels, dimes, and quarters in Paula's purse is 20 cents. If she had one more quarter, the average value would be 21 cents. How many dimes does she have in her purse

7 0

1 year ago

What is the equation, in standard form, of a parabola that models the values in the table?

love history [14]

Answer:

$Y=4x^2+3x-6$

Step-by-step explanation:

For the standard form equation to model the values in the table, each value of x in the table should give the matching the y value when substituted into the equation. We will test each equation:

$Y=3x^2+4x-6$ for (-2,4)

$Y=3(-2)^2+4(-2)-6=3(4)+-8-6=12+-8-6=-2\\$

This does not give 4 as the answer and is not a solution.

$Y=4x^2+3x-6$ for (-2,4)

$Y=4(-2)^2+3(-2)-6=4(4)+-6-6=16+-6-6=-4\\$

This does give 4 as the answer and is a possible solution.

$Y=4x^2-3x-6$ for (-2,4)

$Y=4(-2)^2-3(-2)-6=4(4)+6-6=16+6-6=16\\$

This does not give 4 as the answer and is not a solution.

$Y=-4x^2-3x-6$ for (-2,4)

$Y=-4(-2)^2-3(-2)-6=-4(4)+6-6=-16+6-6=-16\\$

This does not give 4 as the answer and is not a solution.

The only possible solution is $Y=4x^2+3x-6$

3 0

3 years ago

Select the most appropriate unit for the measure of time that the graph represents.

kolezko [41]

i think its A. hours

3 0

3 years ago

Read 2 more answers

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