Answer:
Area=36.66 cm^2
Step-by-step explanation:
First, you need the height.
Use the Pythagorean theorem:
hypotenuse^2=height^2+(1/2 base)^2
It's one-half base because it's an equilateral triangle.
1/2 base =4.6
(9.2)^2-4.6^2=height^2
84.64-21.16=height^2
height^2=63.48
height=7.97
Area=(1/2)(base)(height)
Area=(1/2)(9.2)(7.97)=36.66 cm^2
We define the following variables:
x = r * cos (θ)
y = r * sine (θ)
Substituting the variables we have:
x = -y ^ 2
r * cos (θ) = - (r * sin (θ)) ^ 2
Rewriting:
r * cos (θ) = - (r ^ 2 * sin ^ 2 (θ))
We cleared r:
r = - ((cos (θ)) / (sin ^ 2 (θ)))
We rewrite:
r = - ((cos (θ)) / (sin (θ))) * (1 / sin (θ))
r = - cot (θ) * csc (θ)
Answer:
a polar equation of the form r = f (θ) for the curve represented by the cartesian equation x = -y2 is:
r = - cot (θ) * csc (θ)
Answer:
a.) 0.7063
b.) 23
Step-by-step explanation:
a.)
Let X be an event in which at least 2 students have same birthday
Y be an event in which no student have same birthday.
Now,
P(X) + P(Y) = 1
⇒P(X) = 1 - P(Y)
as we know that,
Probability of no one has birthday on same day = P(Y)
⇒P(Y) =
where there are n people in a group
As given,
n = 30
⇒P(Y) =
= 0.2937
∴ we get
P(X) = 1 - 0.2937 = 0.7063
So,
The probability that at least two of them have their birthdays on the same day = 0.7063
b.)
Given, P(X) > 0.5
As
P(X) + P(Y) = 1
⇒P(Y) ≤ 0.5
As
P(Y) =
We use hit and trial method
If n = 1 , then
P(Y) =
= 1
0.5
If n = 5 , then
P(Y) =
= 0.97
0.5
If n = 10 , then
P(Y) =
= 0.88
0.5
If n = 15 , then
P(Y) =
= 0.75
0.5
If n = 20 , then
P(Y) =
= 0.588
0.5
If n = 22 , then
P(Y) =
= 0.52
0.5
If n = 23 , then
P(Y) =
= 0.49
0.5
∴ we get
Number of students should be in class in order to have this probability above 0.5 = 23
The answer she most likely be d