Answer:
x= 5 , for c to be same on each eqn
Answer:
P ( 5 < X < 10 ) = 1
Step-by-step explanation:
Given:-
- Sample size n = 49
- The sample mean u = 8.0 mins
- The sample standard deviation s = 1.3 mins
Find:-
Find the probability that the average time waiting in line for these customers is between 5 and 10 minutes.
Solution:-
- We will assume that the random variable follows a normal distribution with, then its given that the sample also exhibits normality. The population distribution can be expressed as:
X ~ N ( u , s /√n )
Where
s /√n = 1.3 / √49 = 0.2143
- The required probability is P ( 5 < X < 10 ) minutes. The standardized values are:
P ( 5 < X < 10 ) = P ( (5 - 8) / 0.2143 < Z < (10-8) / 0.2143 )
= P ( -14.93 < Z < 8.4 )
- Using standard Z-table we have:
P ( 5 < X < 10 ) = P ( -14.93 < Z < 8.4 ) = 1
Look at the graph below carefully
Observe the results of shifting ={2}^{x}f(x)=2x
vertically:
The domain, (−∞,∞) remains unchanged.
When the function is shifted up 3 units to ={2}^{x}+3g(x)=2x +3:
The y-intercept shifts up 3 units to (0,4).
The asymptote shifts up 3 units to y=3y=3.
The range becomes (3,∞).
When the function is shifted down 3 units to ={2}^{x}-3h(x)=2 x −3:
The y-intercept shifts down 3 units to (0,−2).
The asymptote also shifts down 3 units to y=-3y=−3.
The range becomes (−3,∞).
Using translation concepts, it is found that the equation that represents the graph of g(x) is:
.
<h3>What is a translation?</h3>
A translation is represented by a change in the function graph, according to operations such as multiplication or sum/subtraction in it's definition.
In this problem, we have that g(x) is a shift left of 2 units of f(x), hence:
More can be learned about translation concepts at brainly.com/question/4521517
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