Whats the answerto this question -2y = 45 + 7y

A DEF has vertices D(-5,-1), E (3, 3), and F (1,-5). Find the equation of the altitude through F in

Tems11 [23]

Answer:

y = -2x -3

Step-by-step explanation:

- the altitude trough F is a perpendicular line to the line DE

- find slope of line DE

D ( x2 = -5, y2 = -1); E (x1 = 3, y1 = 3)

slope m = (y2-y1) / (x2-x1) = (-1-3) / (-5-3) = -4/ -8 = 1/2

-find equation of the altitude trough F

lines that are perpendicular have the slope negative reciprocal (negative reciprocal of 1/2 is -2)

y= -2x +b , for point F(1, -5)

-5 = -2*1 +b, add 2 to both sides

-5 +2 = b, combine like terms

-3 =b

equation of the altitude trough F is y = -2x -3

4 0

2 years ago

Can you please help me with this all I'm asking is just help for one question .

shutvik [7]

The value of “u” is 32°.

5 0

3 years ago

A manufacturer knows that their items have a normally distributed length, with a mean of 15.4 inches, and standard deviation of

Masteriza [31]

Answer:

0.9452 = 94.52% probability that their mean length is less than 16.8 inches.

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal Probability Distribution:

Problems of normal distributions can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the z-score of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem establishes 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 $s = \frac{\sigma}{\sqrt{n}}$ .