Answer:
For tingle #1
We can find angle C using the triangle sum theorem: the three interior angles of any triangle add up to 180 degrees. Since we know the measures of angles A and B, we can find C.



We cannot find any of the sides. Since there is noting to show us size, there is simply just not enough information; we need at least one side to use the rule of sines and find the other ones. Also, since there is nothing showing us size, each side can have more than one value.
For triangle #2
In this one, we can find everything and there is one one value for each.
- We can find side c
Since we have a right triangle, we can find side c using the Pythagorean theorem






- We can find angle C using the cosine trig identity




- Now we can find angle A using the triangle sum theorem



For triangle #3
Again, we can find everything and there is one one value for each.
- We can find angle A using the triangle sum theorem



- We can find side a using the tangent trig identity




- Now we can find side b using the Pythagorean theorem




Answer: a) 30, b) 0.2138, c) 0.0820
Step-by-step explanation:
Since we have given that
X be the number of calls in 1 hour = 60 minutes
Rate of one call every two minutes.

(a) What is the expected number of 911 calls in one hour?

(b) What is the probability of three 911 calls in five minutes? If required, round your answer to four decimal places.

Number of calls = 3
So, P(X=3) is given by

(c) What is the probability of no 911 calls during a five-minute period? If required, round your answer to four decimal places.

Hence, a) 30, b) 0.2138, c) 0.0820
Answer:


Step-by-step explanation:
Mixed fraction
can be converted into like or unlike fractions.
Here, 
So, convert all the given fractions as follows:





Here, No Two fractions on addition gives a sum greater than 5.
But an approximate sum of 5 is given by adding 2.4 + 2.6 = 5
⇒
or ⇒
Answer:
w = 10 cm
Step-by-step explanation:
The area (A) of the triangle is calculated as
A =
base × height
here base = 6 and height = 8, thus
A = 0.5 × 6 × 8 = 3 × 8 = 24 cm²
The area of the rectangle = 5 × 24 = 120 cm²
area of rectangle = lw ( l is length, w is width )
Here l = 12, thus
120 = 12w ( divide both sides by 12 )
w = 10 cm