The standard form of the equation of a parabola is x-y^2+10y+22.what is the vertex form of the equation?

72/1000 as a decimal

polet [3.4K]

Answer: 0.072

Explanation: There are 3 zeros in 1000, so there must be 3 decimal digits.

5 0

3 years ago

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The bawden family is shopping for a new television.They are making their decision based on brand and sizes.The family has 20 pos

nalin [4]

I think it’s 10that’s what I think

5 0

3 years ago

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Help me with the first one please

natulia [17]

D- 2x-1
plug in a number for x and you'll see

6 0

3 years ago

Aaron wants to mulch his garden. His garden is x^2+18x+81 ft^2 One bag of mulch covers x^2-81 ft^2 . Divide the expressions and

blsea [12.9K]

Answer:

Step-by-step explanation:

Given

Garden: $x^2+18x+81$

One Bag: $x^2 - 81$

Requires

Determine the number of bags to cover the whole garden

This is calculated as thus;

$Bags = \frac{x^2+18x+81}{x^2 - 81}$

Expand the numerator

$Bags = \frac{x^2+9x+9x+81}{x^2 - 81}$

$Bags = \frac{x(x+9)+9(x+9)}{x^2 - 81}$

$Bags = \frac{(x+9)(x+9)}{x^2 - 81}$

Express 81 as 9²

$Bags = \frac{(x+9)(x+9)}{x^2 - 9\²}$

Evaluate as difference of two squares

$Bags = \frac{(x+9)(x+9)}{(x - 9)(x+9)}$

$Bags = \frac{(x+9)}{(x - 9)}$

Hence, the number of bags is $Bags = \frac{(x+9)}{(x - 9)}$

3 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