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Answer:
a) y=5ˣ-3
b) y=5ˣ⁻²
c) y=5^(x+1/2)
d) y=5^(x+2.5)+1
Step-by-step explanation:
b: vertical stretch or compression and reflection across x-axis
a: horizontal stretch or compression and reflection across y-axis
h: horizontal shift (left and right)
k: vertical shift (up and down)
a) y=5ˣ-3
b) y=5ˣ⁻²
c) y=5^(x+1/2)
d) y=5^(x+2.5)+1
Answer:
Step-by-step explanation:
Hello!
a)
The variable of interest is
X: monetary value of a grab bag. ($)
The possible values this variable can take are:
X: 15; 40; 60; 120
To calculate the relative frequency of each possible value of the variable you have to divide the number of times this value is observed by the total of observations (sample size)
There where a total of 300 bags created, from these, 40 cost $40, 20 cost $60, 1 cost $120, and the rest have a monetary value of $15.
There are then 300- (40+20+1)= 239 bags with a monetary value of $15
The frequencies for each value of X are:
X= 15 ⇒ p(X)= 239/300= 0.797
X= 40 ⇒ p(X)= 40/300= 0.133
X= 60 ⇒ p(X)= 20/300=0.067
X= 120⇒ p(X)= 1/300= 0.003
The probability distribution for X is:
Xi: 15; 40; 60; 120
p(Xi):0.797; 0.133; 0.067; 0.003
∑p(Xi)= 0.797+0.133+0.067+0.003=1
b)
To calculate the mean you have to use the following formula:
E(X)= ∑Xp(X)= 15*0.797+40*0.133+60*0.067+120*0.003= 21.655≅ $21.7
c)
The standard deviation is the square root of the variance:
Variance:
V(X)= ∑X²p(X)-(∑Xp(X))²= 676.525 - (21.655)²= 207.585
∑X²p(X)= 15²*0.797+40²*0.133+60²*0.067+120²*0.003= 676.525
Standard deviation:
√V(X)= √207.585= $14.407= $14.41
d)
Considered the probability distribution, it is more likely to buy a $15 worth bag than any of the others, it is therefore not worthwhile to spend $20 in a bag full of merchandise worth $15.
I hope this helps!
