Answer: It's 72.
Step-by-step explanation: They deleted my answer <:(
We have that
A(-2,-4) B(8,1) <span>
let
M-------> </span><span>the coordinate that divides the directed line segment from A to B in the ratio of 2 to 3
we know that
A--------------M----------------------B
2 3
distance AM is equal to (2/5) AB
</span>distance MB is equal to (3/5) AB
<span>so
step 1
find the x coordinate of point M
Mx=Ax+(2/5)*dABx
where
Mx is the x coordinate of point M
Ax is the x coordinate of point A
dABx is the distance AB in the x coordinate
Ax=-2
dABx=(8+2)=10
</span>Mx=-2+(2/5)*10-----> Mx=2
step 2
find the y coordinate of point M
My=Ay+(2/5)*dABy
where
My is the y coordinate of point M
Ay is the y coordinate of point A
dABy is the distance AB in the y coordinate
Ay=-4
dABy=(1+4)=5
Mx=-4+(2/5)*5-----> My=-2
the coordinates of point M is (2,-2)
see the attached figure
Answer: 0.0035
Step-by-step explanation:
Given : The readings on thermometers are normally distributed with a mean of 0 degrees C and a standard deviation of 1.00 degrees C.
i.e.
and
Let x denotes the readings on thermometers.
Then, the probability that a randomly selected thermometer reads greater than 2.17 will be :_
![P(X>2.7)=1-P(\xleq2.7)\\\\=1-P(\dfrac{x-\mu}{\sigma}\leq\dfrac{2.7-0}{1})\\\\=1-P(z\leq2.7)\ \ [\because\ z=\dfrac{x-\mu}{\sigma}]\\\\=1-0.9965\ \ [\text{By z-table}]\ \\\\=0.0035](https://tex.z-dn.net/?f=P%28X%3E2.7%29%3D1-P%28%5Cxleq2.7%29%5C%5C%5C%5C%3D1-P%28%5Cdfrac%7Bx-%5Cmu%7D%7B%5Csigma%7D%5Cleq%5Cdfrac%7B2.7-0%7D%7B1%7D%29%5C%5C%5C%5C%3D1-P%28z%5Cleq2.7%29%5C%20%5C%20%5B%5Cbecause%5C%20z%3D%5Cdfrac%7Bx-%5Cmu%7D%7B%5Csigma%7D%5D%5C%5C%5C%5C%3D1-0.9965%5C%20%5C%20%5B%5Ctext%7BBy%20z-table%7D%5D%5C%20%5C%5C%5C%5C%3D0.0035)
Hence, the probability that a randomly selected thermometer reads greater than 2.17 = 0.0035
The required region is attached below .
ATQ


Now




From all equations




Now
Adding 3 and 4


One side won't be covered hence
Answer:

Step-by-step explanation:
It is given that,
The average distance between the Earth and the Moon is 384 400 km.
We need to express in in standard form.
1 km = 1000 m
It means,
384400 km = 384400000 km
or
= 
Hence, the average distance between the Earth and the Moon is
.