Answer:
Mean 24, standard error 0.8
Step-by-step explanation:
The Central Limit Theorem estabilishes that, for a random variable X, with mean
and standard deviation
, the sample means with size n of at least 30 can be approximated to a normal distribution with mean
and standard deviation, also called standard error ![s = \frac{\sigma}{\sqrt{n}}](https://tex.z-dn.net/?f=s%20%3D%20%5Cfrac%7B%5Csigma%7D%7B%5Csqrt%7Bn%7D%7D)
In this problem, we have that:
![\mu = 24, \sigma = 6.4, n = 64](https://tex.z-dn.net/?f=%5Cmu%20%3D%2024%2C%20%5Csigma%20%3D%206.4%2C%20n%20%3D%2064)
What are the mean and the standard error of the sample mean?
By the Central Limit Theorem, mean 24 and standard error ![s = \frac{6.4}{\sqrt{64}} = 0.8](https://tex.z-dn.net/?f=s%20%3D%20%5Cfrac%7B6.4%7D%7B%5Csqrt%7B64%7D%7D%20%3D%200.8)
Answer:
In the following image, AB is parallel to DC, and BC is a transversal intersecting both parallel lines.
and the angle marked q° are ________angles.
The angles marked p° and q° are ________
The angle marked p° and are _______angles.
The angles marked q° and n° are _______angles.
The angles marked n° and m° are _______angles.
DONT ANSWER UNLESS YOU KNOW IT
Answer:
0.01125 foot-candles
Step-by-step explanation:
According to the data, intensity of light measured in foot-candles varies inversely with the square of the distance from the light source
Therefore,
α 1/![d^{2}](https://tex.z-dn.net/?f=d%5E%7B2%7D)
= k/ ![d^{2}](https://tex.z-dn.net/?f=d%5E%7B2%7D)
where,
'k' is the constant
'd' is the distance from bulb
'
' is intensity of a light bulb
When,
d= 3meters
= 0.08foot-candles
k=
. ![d^{2}](https://tex.z-dn.net/?f=d%5E%7B2%7D)
k= 0.08 x 3² => 0.08 x 9
k= 0.72
Next is to determine the intensity level at 8 meters.
= k/ ![d^{2}](https://tex.z-dn.net/?f=d%5E%7B2%7D)
= 0.72/ 8²
= 0.01125 foot-candles
Therefore, the intensity level at 8 meters is 0.01125 foot-candles
First point- (-9,-9)
Second point- (-8,-6)
and so on going up 1 over 3 to the right and you should pass through (-5,3)