Answer:
There is a 2.17% probability that a randomly selected person aged 40 years or older is male and jogs.
It would be unusual to randomly select a person aged 40 years or older who is male and jogs.
Step-by-step explanation:
We have these following probabilities.
A 13.9% probability that a randomly selected person aged 40 years or older is a jogger, so 
.
In addition, there is a 15.6% probability that a randomly selected person aged 40 years or older is male comma given that he or she jogs. I am going to say that P(B) is the probability that is a male. 
is the probability that the person is a male, given that he/she jogs. So 
The Bayes theorem states that:
In which 
is the probability that the person does both thigs, so, in this problem, the probability that a randomly selected person aged 40 years or older is male and jogs.
So
There is a 2.17% probability that a randomly selected person aged 40 years or older is male and jogs.
A probability is unusual when it is smaller than 5%.
So it would be unusual to randomly select a person aged 40 years or older who is male and jogs.
Answer:
<em>680</em>
Step-by-step explanation:
Given that:
All the pupils at a primary school come from one of three villages: Elmswell, Haughley or Woolpit.
of the pupils come from Elmswell
of the pupils come from Haughley
102 pupils come from Woolpit.
To find:
Total number of pupils in the school.
Solution:
Let total number of pupils = 
So, number of pupils that come from Elmswell = of 100x
OR
And, number of pupils that come from Haughley= of 100x
OR
Number of pupils that come from Woolpit = 
As per given statement:
Number of pupils from Elmswell = 25x = 25 
6.8 = 170
Number of pupils from Haughley = 60x = 60 6.8 = 408
Total number of pupils at the school = 
Answer:
Both ends go up.
Step-by-step explanation:
(x + 3) (x + 2) = 0
To solve it, the most appropriate technique is:
1.) zero product property
The solutions are:
(x + 3) = 0
x = -3
(x + 2) = 0
x = -2
x² + 6 = 31
To solve it, the most appropriate technique is:
2.) square root property
x² = 31-6
x² = 25
x = +/- root (25)
x = +/- 5
The solutions are:
x = 5
x = -5
$=\sqrt{4^2}\times\sqrt{2^2\cdot3}$
$=4\times2\sqrt3=8\sqrt3$
