Answer:
The probability that the mean daily precipitation will be 0.11 inches or less for a random sample of November days
P(X≤ 0.11) = 0.4404
Step-by-step explanation:
<u><em>Step(i):-</em></u>
Given that the mean of the Population = 0.10 inches
Given that the standard deviation of the population = 0.07inches
Let 'X' be a random variable in a normal distribution

<u><em>Step(ii):-</em></u>
The probability that the mean daily precipitation will be 0.11 inches or less for a random sample of November days
P(X≤ 0.11) = P(Z≤0.1428)
= 1-P(Z≥0.1428)
= 1 - ( 0.5 +A(0.1428)
= 0.5 - A(0.1428)
= 0.5 -0.0596
= 0.4404
<u><em>Final answer:-</em></u>
The probability that the mean daily precipitation will be 0.11 inches or less for a random sample of November days
P(X≤ 0.11) = 0.4404
Answer:
105
Step-by-step explanation:
5 ÷ 35 = 7
140 ÷ 20 = 7
15 · 7 = 105
I hope this helps!
Central angle = 360 / 24 = 15 degrees
ARc length = r * theta, where theta is the angle in radians and r = radius
= 15 * 15 * pi/180
= 3.93 feet to nearest hundredth.
Area of sector = area of circle / 24 = pi*15^2 / 24 = 29.45 ft^2
First find the gradient of the line
Change in y/change in x
-3–3/-3-3
0/-6
=0 ( so the gradient m is equal to zero)
Y=0x+c
Input the coordinates of one point to find c
-3=(0*3)+c
-3=c
So the equation is
Y= -3