Here's what you need to do:
Well, you can't there's no zeros so this will be the answer
18532.6×10^0=
That would be your answer.

Explanation:
Since we have given that
The prices of three t-shirts styles i.e $24, $30, $36 with their probability is given by

As we know that,


Now,

and

So,

So, the expected value of a t-shirt = $31.
Answer: C, because it is the only one that makes sense.
Hi!
Add the student together to get the number of students
10+4 = 14
Divide the girls by half
4/2 = 2
2/14 is the answer for this question.
<em>Hope this helps! Have an amazing day <3</em>
<em />
I think this problem gave the data for the radius of earth so that we will know the conversion ratio between statute miles and nautical miles. However, you can simply search that. There are 1.15 statute miles in 1 nautical mile. Hence,
Distance in nautical miles = <span>884884 statute miles * 1 nautical mile/1.15 statute miles
<em>Distance =769,464.35 nautical miles</em></span>