2 solutions.
Hope this helps!!
Answer:
a) 
b) P(x>2) = 0.566
c) P(2<x<5) = 0.334
Step-by-step explanation:
Given 24% of U.S. adults say they are more likely to make purchases during a sales tax holiday
Probability 0f U.S. adults say they are more likely to make purchases during a sales tax holiday (p) = 0.24
n = 10
By using Poisson distribution
mean number of make purchases during a sales tax holiday
λ = np = 10 X 0.24 = 2.4
a)
The probability of getting exactly '2'
The probability 


b) The probability of getting more than '2'


= 0.090 + 0.2177+0.261 = 0.566
P(x>2) = 0.566
c) The probability of getting between two and five
P( 2<x<5) = P(x=3)+p(x=4) =
P(2<x<5) = 0.2090 + 0.125 = 0.334
I think its A not sure though
Log4 + logX + log13 = 0
logX = -log13 - log4
logX = -1.716
x= 10^[-1.716]
x= 0.019 A