Answer:
m∠QPM=43°
Step-by-step explanation:
see the attached figure to better understand the problem
we know that
m∠NPQ=m∠MPN+m∠MPQ
we have
m∠NPQ=(9x-25)°
m∠MPN=(4x+12)°
m∠MPQ=(3x-5)°
substitute the given values and solve for x
(9x-25)°=(4x+12)°+(3x-5)°
(9x-25)°=(7x+7)°
9x-7x=25+7
2x=32
x=16
Find the measure of angle QPM
Remember that
m∠QPM=m∠MPQ
m∠MPQ=(3x-5)°
substitute the value of x
m∠MPQ=(3(16)-5)=43°
therefore
m∠QPM=43°
Answer:
Step-by-step explanation:
we know that
The equation of the line into point slope form is equal to
In this problem we have
substitute the given values

therefore
y minus StartFraction one-third EndFraction equals StartFraction 3 Over 4 EndFraction left-parenthesis x minus 4 right-parenthesis.(x – 4)
Please provide table or else i can't calculate!
Answer: is ) $0.60 cents.
Answer:
3.54% probability of observing at most two defective homes out of a random sample of 20
Step-by-step explanation:
For each house that this developer constructs, there are only two possible outcomes. Either there are some major defect that will require substantial repairs, or there is not. The probability of a house having some major defect that will require substantial repairs is independent of other houses. So we use the binomial probability distribution to solve this question.
Binomial probability distribution
The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

In which
is the number of different combinations of x objects from a set of n elements, given by the following formula.

And p is the probability of X happening.
30% of the houses this developer constructs have some major defect that will require substantial repairs.
This means that 
If the allegation is correct, what is the probability of observing at most two defective homes out of a random sample of 20
This is
when n = 20. So






3.54% probability of observing at most two defective homes out of a random sample of 20
Answer:
110.7 cm³ to the nearest tenth
Step-by-step explanation:
The volume V of a cone whose circular base has a radius r and a height h is given as
V = 1/3 πr²h
where π = 22/7
Given the circumference of the circular base as 18.5cm, we must find the radius before we can computed the volume. The relationship between the radius and the circumference is such that the circumference C
= 2πr
hence,
r = C/2π
= 18.5/2π
since the height of the cone is 12.2 cm, the volume V
= 1/3 * π * (18.5/2π)² * 12.2
= 110.71 cm³