Hey,
We can see here 2 of the angles are equal and the third one is a right angle right?
That means it is an isosceles triangle since they have 2 equel angles and a right anlgle.

Another feature of the isosceles triangle is that 2 sides are equal.
We are given one side is 5.5 right?
So the side BC has to be 5.5 or the side AB has to be 5.5
In isosceles triangles the side opposite to the 90 degree angle is the hypotenuse.This means it is the longest angle.
So it has to be AB which is equal to 5.5 since BC is the hypotenuse.

Hope this helps,Happy New Year

8 0

3 years ago

What is the mean of 87 104 and 99

sveticcg [70]

It is 96.6 repeating. to get this answer, you add up all the numbers then divide the answer by hiw many numbers there are

6 0

3 years ago

Read 2 more answers

A certain Bookstore that sells A Million books wants to hire you, and you get the job if you can answer these problems correctly

Maslowich

Answer:

a) P ( X = 2 ) = 0.23028

b) P ( X < 4 ) = 0.95689

c) P ( X ≥ 3 | X ≥ 2 ) = 0.38292

Step-by-step explanation:

Given:-

- The parameter for the poisson distribution is given, λ = 1.3.

- Declare a random variable (X) which is the number of books sold in the next minute:

X ~ Po (1.3)

Find:-

a) What's the probability that the store will sell 2 books in the next minute? b) What's the probability that the store will sell less than 4 books in the next minute? c) What's the probability that the store will sell at least 3 books in the next minute given that it sells at least 2 books in the next minute?

Solution:-

a) The required probability P ( X = 2 ). Can be computed by using the pmf for the poisson distribution:

$P(X = x ) =\frac{ (lambda)^k e^(^-^l^a^m^b^d^a^)}{k!}\\\\P(X = x ) =\frac{ (1.3)^k e^(^-^1^.^3^)}{k!}$

Where, "k" is the number of books sold in next minute.

- Now compute P ( X = 2 ) :

$P(X = 2 ) =\frac{ (1.3)^2 e^(^-^1^.^3^)}{2!}\\\\P(X = 2 ) = 0.23028$

b) The required probability P ( X < 4 ). Can be computed by using the pmf for the poisson distribution and summing individual terms from 0 - 3:

$P(X < 4 ) = P ( X = 0) + P ( X = 1 ) + P ( X = 2 ) + P ( X = 3 )\\\\P(X < 4 ) = \frac{ (1.3)^0 e^(^-^1^.^3^)}{0!}+ \frac{ (1.3)^1 e^(^-^1^.^3^)}{1!}+ \frac{ (1.3)^2 e^(^-^1^.^3^)}{2!} + \frac{ (1.3)^3 e^(^-^1^.^3^)}{3!}\\\\P(X < 4 ) = 0.27253 + 0.35429 + 0.23028 + 0.09979 = 0.95689$

c) The required probability P ( X ≥ 3 | X ≥ 2 ). We have to consider the conditional probability as follows:

P ( X ≥ 3 | X ≥ 2 ) = P ( X ≥ 3 & X ≥ 2 ) / P (X ≥ 2 )

= P ( X ≥ 3 ) / P (X ≥ 2 )

= P ( X > 2 ) / P ( X > 1 )

= [ 1 - P ( X ≤ 2 ) ] / [ 1 - P ( X ≤ 1 ) ]

- Compute P ( X ≤ 2 ) & P ( X ≤ 1 ) using pmf:

P ( X ≤ 2 ) = 0.27253 + 0.35429 + 0.23028

= 0.8571

P ( X ≤ 1 ) = 0.27253 + 0.35429

= 0.62682

- Use the expression developed for conditional probability to evaluate the required probability:

P ( X ≥ 3 | X ≥ 2 ) = [ 1 - P ( X ≤ 2 ) ] / [ 1 - P ( X ≤ 1 ) ]

= [ 1 - 0.8571 ] / [ 1 - 0.62682 ]

= 0.38292

8 0

3 years ago