What is a linear equation that is perpendicular to y=5

Please answer. Geometry problem. Unit: Degrees

Zinaida [17]

Answer:

17

Step-by-step explanation:

∠C is common in both triangles

equate them

∠C = 180- (x + 100) = 180 - ( 4x + 49)

x + 100 = 4x + 49

100 - 49 = 3x

51 = 3x

x = 51/3 = 17

3 0

3 years ago

(2+x)dy/dx=3y need help

den301095 [7]

$(2+x) \frac{dy}{dx} =3y$
$\frac{1}{3y} dy= \frac{1}{2+x} dx$
$\int { \frac{1}{3y} } \, dy = \int { \frac{1}{2+x} } \, dx$
$\frac{1}{3}\ln(3y) = \ln(2+x) + C$
(raise both to the power e): $e^{ \frac{1}{3}\ln(3y)}=e^{\ln(k(2+x))}$
$(3y)^ \frac{1}{3} =k(2+x)$
$3y=k(2+x)^3$
$y= k(2+x)^3$

5 0

3 years ago

Helppp plsssss someone D:

zimovet [89]

〽Hola User_______________

⭐Here is Your Answer..!!!

______________________

↪Cartesian Coordinates of the system....!!

↪X = rcos⊙ and Y = rsin⊙

↪since here Radius (r) = 10 units and Theta (⊙) = 225°

↪there fore substituting we get as ..

↪X = 10 ( cos 225° ) = 10 * -1/ root 2 = -7.071

↪Y = 10 ( sin 225° ) = 10 * -1/root2 = = -7.071

↪the cordinates are as follows

↪(x,y) = (-7.071, -7.071 )

4 0

3 years ago

John has two open packs of paper. One pack has 31 pieces of paper. The second pack has 45 pieces of paper. How many pieces of pa

Tems11 [23]

D 76 pieces of paper. Just add

8 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