Answer:
a) P(X=2)= 0.29
b) P(X<2)= 0.59
c) P(X≤2)= 0.88
d) P(X>2)= 0.12
e) P(X=1 or X=4)= 0.24
f) P(1≤X≤4)= 0.59
Step-by-step explanation:
a) To find P(X=2), we calculate: P(X=2)= 1 - P(X=0) - P(X=1) - P(X=3) - P(X=4) which equals 1 - 0.41 - 0.18 - 0.06 - 0.06, resulting in 0.29
b) For P(X<2), we sum P(X=0) and P(X=1): 0.41 + 0.18 yields 0.59
c) To obtain P(X≤2), we add P(X=0), P(X=1), and P(X=2): 0.41 + 0.18 + 0.29 equals 0.88
d) To calculate P(X>2), we find P(X=3) + P(X=4): 0.06 + 0.06 gives us 0.12
e) For P(X=1 or X=4), we use the union of probabilities: P(X=1) + P(X=4) which is 0.18 + 0.06, resulting in 0.24
f) P(1≤X≤4) is found by adding P(X=1), P(X=2), P(X=3), and P(X=4): 0.18 + 0.29 + 0.06 + 0.06 results in 0.59
Response:
The probability that a student has a pet, given they do not have any siblings is:
Option: D ( 60%)
Step-by-step breakdown:
Let A represent the situation where a student lacks a sibling.
Let B signify the occurrence that a student has a pet.
Consequently, A∩B refers to the event in which a student is without siblings but possesses a pet.
Let P denote the chance of an event happening.
We need to determine:
P(B|A)
From our knowledge:
From the data provided:
P(A)=0.25
and P(A∩B)=0.15
Thus,
which expressed as a percentage is:
Therefore, the probability is:
60%
(5r - 4)(r² - 6r + 4)
uses the distributive property for multiplication.
This expands to 5r(r² - 6r + 4) - 4(r² - 6r + 4)
which results in 5r³ - 30r² + 20r - 4r² + 24r - 16
as you combine like terms and simplify.
The outcome is 5r³ - 30r² - 4r² + 20r + 24r - 16
leading to a final expression of 5r³ - 34r² + 44r - 16. Choice A.
<span><span>Response
26 1/2 does not correspond to an integer but can be rounded to 27, which is an integer.
Clarification
An integer comprises whole numbers only. It does not include fractions.
253/2=25+3/2=25+1 1/2=26 1/2
</span><span>26 1/2 does not correspond to an integer but can be rounded to 27, which is an integer.
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