What is 18 7/8 -14 7/9

What is 6 plus 12 times 8-3

9966 [12]

Answer:

99

Step-by-step explanation: Use PEMDAS

12+8 = 96

96+6 = 102

102-3 = 96

Hope this helped!

4 0

3 years ago

Divide.<br>(3x4 - 3x2 - 6x - 10) = (x - 2)

alekssr [168]

Answer:

3x^3 + 6x^2 + 15x + 36 + 62 / 3x^4 +3x^2 +6x + 10

Step-by-step explanation:

Use Syenthetic Division

4 0

2 years ago

Need awnsers please help

sleet_krkn [62]

The line of reflection is what the graph flips over. You can find the line with two points, and a point on the reflection line is the midpoint of a point and the corresponding point in the after-image.
The first one reflects over the y-axis, or x=0. One point is (-2, 1) and its corresponding point is (2, -1). The midpoint is found by the average of the two coordinates, which is (0,0). Pick another pair of points and find the midpoint which you should get (x,0).
You have two points (0,0) and (x,0) and they form a line, which is the y-axis, or x=0.
The line of reflection for the 1st one is x=0 (y-axis).

4 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

A chocolate company makes dark chocolate bars. The quality control department of the company has found that the weights of the c

devlian [24]

D and e i’m pretty sure :)

3 0

3 years ago

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