Will mark brainliest pls help
1 answer:
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Answer:
Step-by-step explanation:
Universal set
U = 500
The number that likes snickers
n(S) = 42
The number that like Twix.
n(T) = 110
The number that like Reeses
n(R) = 125
n(S n T) = 33
n(T n R) = 62
n(S n R) = 26
n( S n R n T) = 22
Then,
n(S n T) only = n(S n T) - n(S n R n T)
n(S n T) only =33 - 22 = 11
n(T n R) only = n(T n R) - n(S n R n T)
n(T n R) only =62 - 22 = 40
n(S n R) only = n(S n R) - n(S n R n T)
n(S n R) only =26 - 22 = 4.
Also,
n(S) only = n(S) - n(S n R) - n(S n T) only
n(S) only = 42 - 26 - 11 = 5
n(T) only = n(T) - n(T n R) - n(T n S) only
n(T) only = 110 - 62 - 11 = 37
n(R) only = n(R) - n(S n R) - n(R n T) only
n(R) only = 124 - 26 - 40 = 58
Then, to know if some student don't like any of the of chocolate,
Let know the number of students that like the chocolate candy
n(S) + n(T)only + n(T n R)only + n(R) only
42 + 37 + 40 + 58 = 177 students.
Therefore, the total students that like chocolate candy is 177, so those that does not like any of them are
n(S U R U T)' = U – n(S U R U T)
n(S U R U T) = 500 - 177 = 323.
So, the question is how many student likes at most 2 kinds of these chocolates, this means that they can like exactly 2 or less or even none.
So, this category are
n(2 most) = n(S n T)only + n(R n T)only + n(S n R)only + n(s)only + n(T)only + n(R)only + n(S U R U T)'
n(2 most) = 11 + 40 + 4 + 5 + 37 + 58 + 323
n(2 most) = 478
The correct answer is E.
Check attachment for Venn diagram
The number of months until the insect population reaches 40 thousand is 14.29 months and the limiting factor on the insect population as time progresses is 250 thousands.
Given that population P(t) (in thousands) of insects in t months after being transplanted is P(t)=(50(1+0.05t))/(2+0.01t).
(a) Firstly, we will find the number of months until the insect population reaches 40 thousand by equating the given population expression with 40, we get
P(t)=40
(50(1+0.05t))/(2+0.01t)=40
Cross multiply both sides, we get
50(1+0.05t)=40(2+0.01t)
Apply the distributive property a(b+c)=ab+ac, we get
50+2.5t=80+0.4t
Subtract 0.4t and 50 from both sides, we get
50+2.5t-0.4t-50=80+0.4t-0.4t-50
2.1t=30
Divide both sides with 2.1, we get
t=14.29 months
(b) Now, we will find the limiting factor on the insect population as time progresses by taking limit on both sides with t→∞, we get
(c) Further, we will sketch the graph of the function using the window 0≤t≤700 and 0≤p(t)≤700 as shown in the figure.
Hence, when the population P(t) (in thousands) of insects in t months after being transplanted by P(t)=(50(1+0.05t))/(2+0.01t) then the number of months until the insect population reaches 40 thousand 14.29 months and the limiting factor on the insect population is 250 thousand and the graph is shown in the figure.
Learn more about limiting factor from here brainly.com/question/18415071.
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Answer:
For this case the reliability of the sytem would be given by:
R1= 0.95, R2 =0.95, R3= 0.5, R4 = 0.79 , R5 = 0.6
And replacing we got:
Step-by-step explanation:
We can assume that the system work in series
If we have in general n units the reliability of the system is the probability that unit 1 succeeds and unit 2 succeeds and all of the other units in the system succeed. fror n units must succeed for the system to succeed. The reliability of the system is then given by:
For this case the reliability of the sytem would be given by:
R1= 0.95, R2 =0.95, R3= 0.5, R4 = 0.79 , R5 = 0.6
And replacing we got: