What is the range of the function y= 5x + 4 with the domain of x = 0
1 answer:
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Answer:
2.87%
Step-by-step explanation:
We have the following information:
mean (m) = 200
standard deviation (sd) = 50
sample size = n = 40
the probability that their mean is above 21.5 is determined as follows:
P (x> 21.5) = P [(x - m) / (sd / n ^ (1/2))> (21.5 - 200) / (50/40 ^ (1/2))]
P (x> 21.5) = P (z> -22.57)
this value is very strange, therefore I suggest that it is not 21.5 but 215, therefore it would be:
P (x> 215) = P [(x - m) / (sd / n ^ (1/2))> (215 - 200) / (50/40 ^ (1/2))]
P (x> 215) = P (z> 1.897)
P (x> 215) = 1 - P (z <1.897)
We look for this value in the attached table of z and we have to:
P (x> 215) = 1 - 0.9713 (attached table)
P (x> 215) =.0287
Therefore the probability is approximately 2.87%
Answer:
see explanation
Step-by-step explanation:
(a)
A recursive formula allows any term in the sequence to be found by adding the common difference d to the previous term.
Here d = - 4 , then recursive formula is
=
- 4 with a₁ = 2
(b)
The explicit formula for an arithmetic sequence is
= a₁ + (n - 1)d
where a₁ is the first term and d the common difference
Here a₁ = 2 and d = - 4, thus
= 2 - 4(n - 1) = 2 - 4n + 4 = 6 - 4n ← explicit formula
(c)
Using the recursive formula
a₁ = 2
a₂ = 2 - 4 = - 2
a₃ = - 2 - 4 = - 6
Using the explicit formula
a₅ = 6 - 4(5) = 6 - 20 = - 14
a₁₀ = 6 - 4(10) = 6 - 40 = - 34
a₁₀₀ = 6 - 4(100) = 6 - 400 = - 394