Answer:
it is 358
Step-by-step explanation:
Hope this helped have an amazing day!
We need to 'standardise' the value of X = 14.4 by first calculating the z-score then look up on the z-table for the p-value (which is the probability)
The formula for z-score:
z = (X-μ) ÷ σ
Where
X = 14.4
μ = the average mean = 18
σ = the standard deviation = 1.2
Substitute these value into the formula
z-score = (14.4 - 18) ÷ 1..2 = -3
We are looking to find P(Z < -3)
The table attached conveniently gives us the value of P(Z < -3) but if you only have the table that read p-value to the left of positive z, then the trick is to do:
1 - P(Z<3)
From the table
P(Z < -3) = 0.0013
The probability of the runners have times less than 14.4 secs is 0.0013 = 0.13%
Answer:
Step-by-step explanation:
Part A
x-intercepts of the graph → x = 0, 6
Maximum value of the graph → f(x) = 120
Part B
Increasing in the interval → 0 ≤ x ≤ 3
Decreasing in the interval → 3 < x ≤ 6
As the price of goods increase in the interval [0, 3], profit increases.
But in the price interval of (3, 6] profit of the company decreases.
Part C
Average rate of change of a function 'f' in the interval of x = a and x = b is given by,
Average rate of change =
Therefore, average rate of change of the function in the interval x = 1 and x = 3 will be,
Average rate of change =
=
= 30
Answer:
[0,11]
Step-by-step explanation:
Required
Represent Set of numbers from 0 to 11
<em>In interval notation; the square bracket [ ... ] implies that the number in question is inclusive of the set.</em>
So; Given that
Beginning = 0
End = 11
The interval notation can be represented as [0,11] because both numbers are inclusive of the set.
Answer:
Difference between the approximation and the actual value of f′(0.5) is 0.4330
Step-by-step explanation:
As complete question is not give, so considering the most relevant question to given statement in Fig below.
from table
x 0 1
f(x) 1 2
f'(0.5)=?
from numerical differentiation
Given function is
Differentiating w.r.to x
Difference between the approximation and the actual value of f′(0.5) is 0.4330