* im in a hurry * A recent internet survey showed per day lodging expenses at disney world in orlando range from less than $100 for a campsite to more than $2000 for a top luxury room . The cost - per day average of the top 10 most popular hotels at Disney World is $348 a - If the standard deviation is $80 , find the z score for a per - day expense of $400

b - if the data are approximately normally distributed with mean $348 and standard deviation $80 , what percent of the hotels have per day expenses greater than $268?

24

step-by-step explanation:

answer: solve 5\cdot 2^x=2405⋅2  

x

=2405, dot, 2, start superscript, x, end superscript, equals, 240.

solution

to solve for xxx, we must first isolate the exponential part. to do this, divide both sides by 555 as shown below. we do not multiply the 555 and the 222 as this goes against the order of operations!

\begin{aligned} 5\cdot 2^x& =240 2^x& =48 \\ \end{aligned}  

5⋅2  

x

2  

x

​    

=240

=48

​

step-by-step explanation:

10 - 75

step-by-step explanation:

there's nothing attached so it's most likely anything in that range.