Answer:
hope it's help you ok have a good day
Answer:
The integer -37 represents the direction and the distance covered by Jose from Gainesville to Ocala.
Step-by-step explanation:
The direction is represented by the sign that accompanies the distance, since Jose is returning from Gainesville, then the direction must represented by a minus sign (-), since he is travelling southwards. The distance is the magnitude of the length covered by Jose during his return. Hence, distance is represented by the natural number 37.
Finally, the integer -37 represents the direction and the distance covered by Jose from Gainesville to Ocala.
Step-by-step explanation:
Hello there!
Just divide 1000 by 100, because 100 cm = 1 m.
1000/100= 10 m.
:)
Using the slope concept, it is found that the thundercloud is 6.43 miles above the ground.
<h3>What is a slope?</h3>
The slope is given by the <u>vertical change divided by the horizontal change</u>, and it's also the tangent of the angle of depression.
In this problem, we have that:
- The vertical distance is the height h.
- The horizontal distance is of 3 miles, as the sound takes 5 seconds to travel 1 mile, and it took 15 seconds.
Hence:
tan(65º) = h/3.
h = 3 x tan(65º)
h = 6.43 miles.
The thundercloud is 6.43 miles above the ground.
More can be learned about the slope concept at brainly.com/question/18090623
#SPJ1
Answer:
Mean=685
Variance=36.7
Step-by-step explanation:
The mean of uniform discrete distribution can be expressed as the average of the boundaries
mean=( b+a)/2
The variance of uniform discrete distribution can be expressed as the difference of the boundaries decreased by 1 and squared, decreased by 1 and divided by 12.
σ²=[(b-a+1)^2 - 1]/12
We were given the wavelength from from 675 to 695 nm which means
a= 675, b= 695
We can now calculate the mean by using the expresion below
mean=( b+a)/2
Mean=( 675 + 695)/2
=685
The variance can be calculated by using the expression below
σ²=[(b-a+1)^2 - 1]/12
σ²=[(695-675+1)^2 -1]/12
σ²=440/12
σ²=36.7
Therefore, the the mean and variance, of the wavelength distribution for this radiation are 685 and 36.7 respectively
