−2x − 10y = 10 D: {−10, −5, 0, 5} A. x=-10 B. x=-5 C. x=0 D. x=5

The man is pushing with a force of 75 N. The rock is not moving. This is an example of what kind of force?

Oksi-84 [34.3K]

Kinetic energy
Pushing is energy in motion and energy in motion is classified as (KE) kinetic energy

3 0

3 years ago

- <img src="https://tex.z-dn.net/?f=%20-%203%5Cpi%20%2B%20w%20%3D%202%5Cpi" id="TexFormula1" title=" - 3\pi + w = 2\pi" alt="

pantera1 [17]

Isolate the w. Note the equal sign. What you do to one side, you do to the other.

Add 3π to both sides

- 3π (+3π) + w = 2π (+3π)

w = 2π + (3π)

w = 5π

w = 5π is your answer

hope this helps

4 0

2 years ago

The taxi and takeoff time for commercial jets is a random variable x with a mean of 8.9 minutes and a standard deviation of 2.9

Eva8 [605]

Answer:

a) 0.2981 = 29.81% probability that for 37 jets on a given runway, total taxi and takeoff time will be less than 320 minutes.

b) 0.999 = 99.9% probability that for 37 jets on a given runway, total taxi and takeoff time will be more than 275 minutes

c) 0.2971 = 29.71% probability that for 37 jets on a given runway, total taxi and takeoff time will be between 275 and 320 minutes

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 $s = \frac{\sigma}{\sqrt{n}}$ .