The mean salary if of 5 employees is $35900. the median is $37000. the mode is $382000. If the median payed employee gets a $310
1 answer:
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Answer:
a) 0.1423
b) 0.2977
c) 0.56
Step-by-step explanation:
For each driver stopped for speeding, there are only two possible outcomes. Either they have invalid licenses, or they do not. The probability of a driver having an invalid license is independent from other drivers. So we use the binomial probability distribution to solve this problem.
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.
In this problem we have that:
13 percent of the drivers stopped for speeding have invalid licenses.
This means that
14 drivers are stopped
This means that
(a) None will have an invalid license.
This is
(b) Exactly one will have an invalid license.
This is
(c) At least 2 will have invalid licenses.
Either less than 2 have invalid licenses, or at least 2 does. The sum of the probabilities of these events is decimal 1. Mathematically, this is
We want
So
In which
Answer:
x >= -7 ................(1a)
x >= 3 ...............(2a1)
Step-by-step explanation:
f(x) = .............(0)
find the domain.
To find the (real) domain, we need to ensure that each term remains a real number.
which means the following conditions must be met
x+7 >= 0 .....................(1)
AND
x^2+2x-15 >= 0 ..........(2)
To satisfy (1), x >= -7 .....................(1a) by transposition of (1)
To satisfy (2), we need first to find the roots of (2)
factor (2)
(x+5)(x-3) >= 0
This implis
(x+5) >= 0 AND (x-3) >= 0.....................(2a)
OR
(x+5) <= 0 AND (x-3) <= 0 ...................(2b)
(2a) is satisfied with x >= 3 ...............(2a1)
(2b) is satisfied with x <= -5 ................(2b1)
Combine the conditions (1a), (2a1), and (2b1),
x >= -7 ................(1a)
AND
(
x >= 3 ...............(2a1)
OR
x <= -5 ................(2b1)
)
which expands to
(1a) and (2a1) OR (1a) and (2b1)
( x >= -7 and x >= 3 ) OR ( x >= -7 and x <= -5 )
Simplifying, we have
x >= 3 OR ( -7 <= x <= -5 )
Referring to attached figure, the domain is indicated in dark (purple), the red-brown and white regions satisfiy only one of the two conditions.
